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a6357ed0d92ba2c50d3ade3b9501df5b | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_7 | Two nonhorizontal, non vertical lines in the $xy$ -coordinate plane intersect to form a $45^{\circ}$ angle. One line has slope equal to $6$ times the slope of the other line. What is the greatest possible value of the product of the slopes of the two lines?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac23 \qquad\textbf{(C)}\ \frac32 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 6$ | If you have this formula memorized then it will be very easy to do: If $\theta$ is the angle between two lines with slopes $m_1$ and $m_2$ , then $\tan(\theta)=\frac{m_1-m_2}{1+m_1m_2}$
Now let the smaller slope be $m$ , thus the other slope is $6m$ . Using our formula above: \[\tan(45^\circ)=\frac{6m-m}{1+6m^2} \implies 6m^2-5m+1=0 \implies (2m-1)(3m-1)=0.\] Therefore the two possible values for $m$ are $\tfrac{1}{2}$ and $\tfrac{1}{3}$ . We choose the larger one and thus our answer is \[6m^2=6 \cdot \frac{1}{4} = \frac{3}{2} \implies \boxed{32}.\] | C | 32 |
248f0bfcf0347aae1e1053f27227e21f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Rearranging the terms and and completing the square for $y$ yields the result $x^{2020}+(y-1)^2=1$ . Then, notice that $x$ can only be $0$ $1$ and $-1$ because any value of $x^{2020}$ that is greater than 1 will cause the term $(y-1)^2$ to be less than $0$ , which is impossible as $y$ must be real. Therefore, plugging in the above values for $x$ gives the ordered pairs $(0,0)$ $(1,1)$ $(-1,1)$ , and $(0,2)$ gives a total of $\boxed{4}$ ordered pairs. | D | 4 |
248f0bfcf0347aae1e1053f27227e21f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Bringing all of the terms to the LHS, we see a quadratic equation \[y^2 - 2y + x^{2020} = 0\] in terms of $y$ . Applying the quadratic formula, we get \[y = \frac{2\pm\sqrt{4-4\cdot1\cdot x^{2020}}}{2}=\frac{2\pm\sqrt{4(1-x^{2020})}}{2}.\] In order for $y$ to be real, which it must be given the stipulation that we are seeking integral answers, we know that the discriminant, $4(1-x^{2020})$ must be nonnegative. Therefore, \[4(1-x^{2020}) \geq 0 \implies x^{2020} \leq 1.\] Here, we see that we must split the inequality into a compound, resulting in $-1 \leq x \leq 1$
The only integers that satisfy this are $x \in \{-1,0,1\}$ . Plugging these values back into the quadratic equation, we see that $x = \{-1,1\}$ both produce a discriminant of $0$ , meaning that there is only 1 solution for $y$ . If $x = \{0\}$ , then the discriminant is nonzero, therefore resulting in two solutions for $y$
Thus, the answer is $2 \cdot 1 + 1 \cdot 2 = \boxed{4}$ | D | 4 |
248f0bfcf0347aae1e1053f27227e21f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Set it up as a quadratic in terms of y: \[y^2-2y+x^{2020}=0\] Then the discriminant is \[\Delta = 4-4x^{2020}\] This will clearly only yield real solutions when $|x^{2020}| \leq 1$ , because the discriminant must be positive.
Then $x=-1,0,1$ . Checking each one: $-1$ and $1$ are the same when raised to the 2020th power: \[y^2-2y+1=(y-1)^2=0\] This has only has solutions $1$ , so $(\pm 1,1)$ are solutions.
Next, if $x=0$ \[y^2-2y=0 \Rightarrow y(y-2)=0\] Which has 2 solutions, so $(0,2)$ and $(0,0)$
These are the only 4 solutions, so our answer is $\boxed{4}$ | D | 4 |
248f0bfcf0347aae1e1053f27227e21f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Move the $y^2$ term to the other side to get $x^{2020}=2y-y^2 = y(2-y)$
Because $x^{2020} \geq 0$ for all $x$ , then $y(2-y) \geq 0 \Rightarrow y = 0,1,2$
If $y=0$ or $y=2$ , the right side is $0$ and therefore $x=0$
When $y=1$ , the right side become $1$ , therefore $x=1,-1$
Our solutions are $(0,2)$ $(0,0)$ $(1,1)$ $(-1,1)$ . These are the only $4$ solutions, so the answer is $\boxed{4}$ | D | 4 |
248f0bfcf0347aae1e1053f27227e21f | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_8 | How many ordered pairs of integers $(x, y)$ satisfy the equation \[x^{2020}+y^2=2y?\]
$\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 3 \qquad\textbf{(D) } 4 \qquad\textbf{(E) } \text{infinitely many}$ | Since $x^{2020}$ and $y^2$ are perfect squares, they are both nonnegative. That means $y^2$ plus a nonnegative number equals $2y$ , which means $y^2 \leq 2y.$ The only possible integer values for $y$ are $0, 1, 2$
For $y=0$ $x$ can only be $0$
For $y=1$ $x^2=1$ so $x=1, -1$
For $y=2$ $x$ can only be $0$ as well.
This gives us the solutions $(0, 0)$ $(1, 1)$ $(-1, 1)$ , and $(0, 2)$ . These are the only solutions, so there is a total of $\boxed{4}$ ordered pairs. | D | 4 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | Let $O$ be the center of the circle, and $X$ be the midpoint of $\overline{CD}$ . Let $CX=a$ and $EX=b$ . This implies that $DE = a - b$ . Since $CE = CX + EX = a + b$ , we now want to find $(a+b)^2+(a-b)^2=2(a^2+b^2)$ . Since $\angle CXO$ is a right angle, by Pythagorean theorem $a^2 + b^2 = CX^2 + OX^2 = (5\sqrt{2})^2=50$ . Thus, our answer is $2\times50=\boxed{100}$ | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | Let $O$ be the center of the circle, and $X$ be the midpoint of $CD$ . Draw triangle $OCD$ , and median $OX$ . Because $OC = OD$ $OCD$ is isosceles, so $OX$ is also an altitude of $OCD$ $OE = 5\sqrt2 - 2\sqrt5$ , and because angle $OEC$ is $45$ degrees and triangle $OXE$ is right, $OX = EX = \frac{5\sqrt2 - 2\sqrt5}{\sqrt2} = 5 - \sqrt{10}$ . Because triangle $OXC$ is right, $CX = \sqrt{(5\sqrt2)^2 - (5 - \sqrt{10})^2} = \sqrt{15 + 10\sqrt{10}}$ . Thus, $CD = 2\sqrt{15 + 10\sqrt{10}}$
We are looking for $CE^2$ $DE^2$ which is also $(CE + DE)^2 - 2 \cdot CE \cdot DE$
Because $CE + DE = CD = 2\sqrt{15 + 10\sqrt{10}}$ $(CE + DE)^2 = CD^2=4(15 + 10\sqrt{10}) = 60 + 40\sqrt{10}$
By Power of a Point, $CE \cdot DE = AE \cdot BE = 2\sqrt5\cdot(10\sqrt2 - 2\sqrt5) = 20\sqrt{10} - 20$ , so $2 \cdot CE \cdot DE = 40\sqrt{10} - 40$
Finally, $CE^2 + DE^2 = (CE+ED)^2-2\cdot CE \cdot DE=(60 + 40\sqrt{10}) - (40\sqrt{10} - 40) = \boxed{100}$ | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | Let $O$ be the center of the circle. Notice how $OC = OD = r$ , where $r$ is the radius of the circle. By applying the law of cosines on triangle $OCE$ \[r^2=CE^2+OE^2-2(CE)(OE)\cos{45}=CE^2+OE^2-(CE)(OE)\sqrt{2}.\]
Similarly, by applying the law of cosines on triangle $ODE$ \[r^2=DE^2+OE^2-2(DE)(OE)\cos{135}=DE^2+OE^2+(DE)(OE)\sqrt{2}.\]
By subtracting these two equations, we get \[CE^2-DE^2-(CE)(OE)\sqrt{2}-(DE)(OE)\sqrt{2}=0.\] We can rearrange it to get \[CE^2-DE^2=(CE)(OE)\sqrt{2}+(DE)(OE)\sqrt{2}=(CE+DE)(OE\sqrt{2}).\]
Because both $CE$ and $DE$ are both positive, we can safely divide both sides by $(CE+DE)$ to obtain $CE-DE=OE\sqrt{2}$ . Because $OE = OB - BE = 5\sqrt{2} - 2\sqrt{5}$ \[(CE-DE)^2 = CE^2+DE^2 - 2(CE)(DE) = (OE\sqrt{2})^2 =2(5\sqrt{2} - 2\sqrt{5})^2 = 140 - 40\sqrt{10}.\]
Through power of a point, we can find out that $(CE)(DE)=20\sqrt{10} - 20$ , so \[CE^2+DE^2 = (CE-DE)^2+ 2(CE)(DE)= (140 - 40\sqrt{10}) + 2(20\sqrt{10} - 20) = \boxed{100}.\] | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250); pair O, A, B, C, D, E, D1; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); D1 = (D.x,-D.y); draw(p); dot("$O$",O,1.5*S,linewidth(4)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$D'$",D1,1.5*dir(D1),linewidth(4)); draw(A--B^^C--D^^C--D1--O--cycle^^D1--E); [/asy] Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$ , we have that $\angle BED'=45$ . Since $\angle AEC=45$ $\angle CED'=90$ . Since $DE=ED'$ , by the Pythagorean Theorem, our desired solution is just $CD'^2$ .
Looking next to circle arcs, we know that $\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45$ , so $\overarc{AC}+\overarc{BD}=90$ . Since $\overarc{BD'}=\overarc{BD}$ , and $\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180$ $\overarc{CD'}=90$ . Thus, $\angle COD'=90$ .
Since $OC=OD'=5\sqrt{2}$ , by the Pythagorean Theorem, the desired $CD'^2= \boxed{100}$ | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | Basically, by PoP, you have that \[CE \times DE = (10\sqrt{2}-2\sqrt{5})(2\sqrt{5}) = 20\sqrt{10} - 20.\] Therefore, as $CE^2 + DE^2 = (CD)^2 - 2(CE \times DE),$ basically, once you find $CD^2,$ the problem is done. Now, this is an IMPORTANT concept: If you have a circle which you know the radius of and you want to find the length of a chord of that circle, drop an altitude from the center of the circle to the chord to find distance between the center of the circle and the chord.
In this case, let $M$ be the midpoint of chord $CD.$ Notice that now we can use our $45^\circ{}$ angle, since $OME$ is a $45^\circ{}-45^\circ{}-90^\circ{}$ triangle so that $ME = x$ and $OE = x\sqrt{2}.$ However, we have that $OE = 5\sqrt{2}-2\sqrt{5},$ so that $x = 5 - \sqrt{10}.$ Now, notice that $x^2 = 35 - 10\sqrt{10},$ so that \[CM^2 = 50 - x^2 = 50 - (35 - 10\sqrt{10}) = 15 + 10 \sqrt{10}\] and \[CD^2 = 60 + 4\sqrt{10}.\] Therefore, \[CE^2 + DE^2 = CD^2 - 2(CE \times DE) = (60+4\sqrt{10}) - 2(2\sqrt{10} - 20) = (60+4\sqrt{10}) + (40-4\sqrt{10}) = \boxed{100}.\] | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | [asy] /* Made by sofas103; edited by MRENTHUSIASM */ size(250); pair O, A, B, C, D, E, P, Q; O = origin; A = (-5*sqrt(2),0); B = (5*sqrt(2),0); E = (5*sqrt(2)-2*sqrt(5),0); P = (5*sqrt(2), -4.5); Q=(5*sqrt(2)-2*sqrt(5)-0.5, 0); path p; p = Circle(O,5*sqrt(2)); C = intersectionpoint(p,E--E+10*dir(135)); D = intersectionpoint(p,E--E+10*dir(-45)); draw(p); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); label("$45^{\circ}$",Q,NW); draw(A--B^^C--P--B); draw((5*sqrt(2), 0)--(5*sqrt(2)-0.4, 0)--(5*sqrt(2)-0.4, -0.4)--(5*sqrt(2), -0.4)); [/asy] For ease of notation, let $x = DE$ and $y=EC$ .
Extend $\overline{CD}$ to point $P$ until $\overline{BP}$ is perpendicular to $AB$ . It's given that $\angle AEC = 45^{\circ}$ , so, by vertical angles, we have $\angle BEP = \angle EPB = 45^{\circ}$
Since $PEB$ is a $45-45-90$ right triangle, we have $BE = BP = 2\sqrt{5}$ and $PE=2\sqrt{10}$ . Hence, $PD = 2\sqrt{10}-x.$
By Power of a Point, we have
\[PB^2 = PD\cdot PC\] \[20 = \left(2\sqrt{10}-x\right)\left(y + 2\sqrt{10}\right).\]
Isolating the variables after expanding gives $x-y = 2\sqrt{10}-10.$
Using Power of a Point again, we have
\[DE\cdot EC = BE\cdot EA\] \[xy = 2\sqrt{5}\left(10\sqrt{2}-2\sqrt{5} \right)\] \[xy = 20\sqrt{10}-20.\] To get $x^2 + y^2$ , we can perform the operation $(x-y)^2 + 2xy$ . Plugging these values in,
\[(x-y)^2 + 2xy = \left(2\sqrt{10}-10\right)^2 + 2\left(20\sqrt{10}-20\right)\] \[= 40 + 100 - 40\sqrt{10} + 40\sqrt{10} - 40 = \boxed{100}\] | E | 100 |
8c1cfe7c1a7cbb06b03a0e7081af2018 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | Perhaps not reliable in general, but very useful as a last resort.
The choice of the radius $5\sqrt2$ is strange, and is probably motivated by a nice answer in the end, so we only consider integer options. Notice that a 5 also appears in the condition $BE=2\sqrt5$ , therefore it will likely be present in the answer as well; the only integer containing a factor of 5 amongst the choices is 100, thus the answer is $\boxed{100}$ | E | 100 |
9c5832d142f40d56de1f7c277e59cdcf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_15 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | Consider the $10$ people to be standing in a circle, where two people opposite each other form a diameter of the circle.
Let us use casework on the number of pairs that form a diameter of the circle.
Case 1: $0$ diameters
There are $2$ ways: either $1$ pairs with $2$ $3$ pairs with $4$ , and so on or $10$ pairs with $1$ $2$ pairs with $3$ , etc.
Case 2: $1$ diameter
There are $5$ possible diameters to draw (everyone else pairs with the person next to them).
Note that there cannot be $2$ diameters since there would be one person on either side that will not have a pair adjacent to them. The only scenario forced is when the two people on either side would be paired up across a diameter. Thus, a contradiction will arise.
Case 3: $3$ diameters
There are $5$ possible sets of $3$ diameters to draw.
Notice we are technically choosing the number of ways to choose a pair of two diameters that are neighbors to each other. This means we can choose the first diameter in the pair, and have only two diameters to choose from for the second in the pair. This means we have $5*2=10$ possibilities for choosing 5 neighboring diameters. However, notice that there are duplicates, so we divide the $10$ possibilities by $2$ to get $5$
Note that there cannot be a case with $4$ diameters because then there would have to be $5$ diameters for the two remaining people as they have to be connected with a diameter. A contradiction arises.
Case 4: $5$ diameters
There is only $1$ way to do this.
Thus, in total there are $2+5+5+1=\boxed{13}$ possible ways.
- Minor edits by Pearl2008 | C | 13 |
9c5832d142f40d56de1f7c277e59cdcf | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_15 | There are $10$ people standing equally spaced around a circle. Each person knows exactly $3$ of the other $9$ people: the $2$ people standing next to her or him, as well as the person directly across the circle. How many ways are there for the $10$ people to split up into $5$ pairs so that the members of each pair know each other?
$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 13 \qquad\textbf{(D)}\ 14 \qquad\textbf{(E)}\ 15$ | If each person knows exactly $3$ people, that means we form " $4$ -person groups". That is, all the people in the group knows every other person. We can rotate until the circle is filled. We want pairs of people, so the first pair has $\dbinom{4}{2}=6$ . The $2$ nd pair is just $\dbinom{2}{2} =1$ . We need to multiply these together since these are $1$ group. The $3$ rd pair would be $\dbinom{4}{2}=6$ . The $4$ th pair is $\dbinom{2}{2}=1$ . We multiply these $2$ together and get $6$ . The final group would be $\dbinom{2}{2}=1$ . So we add these up and we have $6 + 6 + 1 = \boxed{13}$ possible ways. | C | 13 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Suppose that we have a deck, currently containing just one black card. We then insert $n$ red cards one-by-one into the deck at random positions. It is easy to see using induction, that the black card is randomly situated in the deck.
Now, suppose that we have this deck again, with only one black card. Each time we pick a red ball, we place a card above the black card, and each time we pick a blue ball, we place a card below the black card. It is easy to see that the probability that the card is inserted into the top part of the deck is simply equal to the number of red balls divided by the total number of balls, and the probability that the card is inserted into the bottom part of the deck is equal to the number of blue balls divided by the total number of balls. Therefore, this is equivalent to inserting the card randomly into the deck.
Finally, four more red cards will be inserted into the deck, and so the black card can be in five possible positions. Only one corresponds to having three balls of each type. Our probability is thus $\frac{1}{5}$ , and so the answer is $\boxed{15}$ | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Let $R$ denote the action where George selects a red ball and $B$ denote the action where he selects a blue one. Now, in order to get $3$ balls of each color, he needs $2$ more of both $R$ and $B$
There are 6 cases: $RRBB, RBRB, RBBR, BBRR, BRBR, BRRB$ (we can confirm that there are only $6$ since $\binom{4}{2}=6$ ). However we can clump $RRBB + BBRR$ $RBRB + BRBR$ , and $RBBR + BRRB$ together since they are equivalent by symmetry.
$\textbf{CASE 1: }$ $RRBB$ and $BBRR$
Let's find the probability that he picks the balls in the order of $RRBB$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks red again is now $\frac{2}{3}$
There are $3$ reds and $1$ blue now. The probability that he picks a blue is $\frac{1}{4}$
Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$
So the probability that the $RRBB$ case happens is $\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{1}{4} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BBRR$ case is the exact same by symmetry, case 1 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
$\textbf{CASE 2: }$ $RBRB$ and $BRBR$
Let's find the probability that he picks the balls in the order of $RBRB$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$
There are $2$ reds and $2$ blues now. The probability that he picks a red is $\frac{1}{2}$
Finally, there are $3$ reds and $2$ blues. The probability that he picks a blue is $\frac{2}{5}$
So the probability that the $RBRB$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BRBR$ case is the exact same by symmetry, case 2 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
$\textbf{CASE 3: }$ $RBBR$ and $BRRB$
Let's find the probability that he picks the balls in the order of $RBBR$
The probability that the first ball he picks is red is $\frac{1}{2}$
Now there are $2$ reds and $1$ blue in the urn. The probability that he picks blue is $\frac{1}{3}$
There are $2$ reds and $2$ blues now. The probability that he picks a blue is $\frac{1}{2}$
Finally, there are $2$ reds and $3$ blues. The probability that he picks a red is $\frac{2}{5}$
So the probability that the $RBBR$ case happens is $\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{2}{5} = \frac{1}{30}$ . However, since the $BRRB$ case is the exact same by symmetry, case 3 has a probability of $\frac{1}{30} \cdot 2 = \frac{1}{15}$ chance of happening.
Adding up the cases, we have $\frac{1}{15}+\frac{1}{15}+\frac{1}{15}=\boxed{15}$ ~quacker88 | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | We know that we need to find the probability of adding 2 red and 2 blue balls in some order.
There are 6 ways to do this, since there are $\binom{4}{2}=6$ ways to arrange $RRBB$ in some order.
We will show that the probability for each of these 6 ways is the same.
We first note that the denominators should be counted by the same number. This number is $2 \cdot 3 \cdot 4 \cdot 5=120$ . This is because 2, 3, 4, and 5 represent how many choices there are for the four steps. No matter what the $k-th$ step involves $k+1$ numbers to choose from.
The numerators are the number of successful operations. No matter the order, the first time a red is added will come from 1 choice and the second time will come from 2 choices, since that is how many reds are in the urn originally.
The same goes for the blue ones. The numerator must equal $(1 \cdot 2)^2$
Therefore, the probability for each of the orderings of $RRBB$ is $\frac{4}{120}=\frac{1}{30}$ . There are 6 of these, so the total probability is $\boxed{15}$ | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | First, notice that when George chooses a ball he just adds another ball of the same color. On George's first move, he either chooses the red or the blue with a $\frac{1}{2}$ chance each. We can assume he chooses Red(chance $\frac{1}{2}$ ), and then multiply the final answer by two for symmetry. Now, there are two red balls and one blue ball in the urn. Then, he can either choose another Red(chance $\frac{2}{3}$ ), in which case he must choose two blues to get three of each, with probability $\frac{1}{4}\cdot\frac{2}{5}=\frac{1}{10}$ or a blue for two blue and two red in the urn, with chance $\frac{1}{3}$ . If he chooses blue next, he can either choose a red then a blue, or a blue then a red. Each of these has a $\frac{1}{2}\cdot\frac{2}{5}$ for total of $2\cdot\frac{1}{5}=\frac{2}{5}$ . The total probability that he ends up with three red and three blue is $2\cdot\frac{1}{2}(\frac{2}{3}\cdot\frac{1}{10}+\frac{1}{3}\cdot\frac{2}{5})=\frac{1}{15}+\frac{2}{15}=\boxed{15}$ | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Let the probability that the urn ends up with more red balls be denoted $P(R)$ . Since this is equal to the probability there are more blue balls, the probability there are equal amounts is $1-2P(R)$ $P(R) =$ the probability no more blues are chosen plus the probability only 1 more blue is chosen. The first case, $P(\text{no more blues}) = \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}$
The second case, $P(\text{1 more blue}) = 4\cdot\frac{1\cdot1\cdot2\cdot3}{2\cdot3\cdot4\cdot5} = \frac{1}{5}$ . Thus, the answer is $1-2\left(\frac{1}{5}+\frac{1}{5}\right)=1-\frac{4}{5}=\boxed{15}$ | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | Here $X$ stands for R or B, and $Y$ for the remaining color.
After 3 rounds one can either have a $4+1$ configuration ( $XXXXY$ ), or $3+2$ configuration ( $XXXYY$ ). The probability of getting to $XXXYYY$ from $XXXYY$ is $\frac{2}{5}$ . Observe that the probability of arriving to $4+1$ configuration is \[\frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}\] $\frac{2}{3}$ to get from $XXY$ to $XXXY$ $\frac{3}{4}$ to get from $XXXY$ to $XXXXY$ ). Thus the probability of arriving to $3+2$ configuration is also $\frac{1}{2}$ , and the answer is \[\frac{1}{2} \cdot \frac{2}{5} = \boxed{15}.\] | B | 15 |
0eef8b82f97d26f43231d6c8051b60a7 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_16 | An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
$\textbf{(A) } \frac16 \qquad \textbf{(B) }\frac15 \qquad \textbf{(C) } \frac14 \qquad \textbf{(D) } \frac13 \qquad \textbf{(E) } \frac12$ | We can use dynamic programming to solve this problem.
We let $dp[i][j]$ be the probability that we end up with $i$ red balls and $j$ blue balls.
Notice that there are only two ways that we can end up with $i$ red balls and $j$ blue balls: one is by fetching a red ball from the urn when we have $i - 1$ red balls and $j$ blue balls and the other is by fetching a blue ball from the urn when we have $i$ red balls and $j - 1$ blue balls.
Then we have $dp[i][j] = \frac{i - 1}{i - 1 + j} dp[i - 1][j] + \frac{j - 1}{i - 1 + j} dp[i][j - 1]$
Then we start can with $dp[1][1] = 1$ and try to compute $dp[3][3]$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 1 & 2 & 3\\ \hline\hline 1 & 1 & 1/2 & 1/3\\ \hline 2 & 1/2 & 1/3 & 1/4\\ \hline 3 & 1/3 & 1/4 & 1/5\\ \hline \end{array}\] The answer is $\boxed{15}$ | B | 15 |
2bf214fa5caab8d392ad5010cbc10947 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$ | Let $P(x) = x^5+ax^4+bx^3+cx^2+dx+2020$ . We first notice that $\frac{-1+i\sqrt{3}}{2} = e^{2\pi i / 3}$ . That is because of Euler's Formula : $e^{ix} = \cos(x) + i \cdot \sin(x)$ $\frac{-1+i\sqrt{3}}{2}$ $-\frac{1}{2} + i \cdot \frac {\sqrt{3}}{2}$ $\cos(120^\circ) + i \cdot \sin(120^\circ) = e^{ 2\pi i / 3}$
In order $r$ to be a root of $P$ $re^{2\pi i / 3}$ must also be a root of P, meaning that 3 of the roots of $P$ must be $r$ $re^{i\frac{2\pi}{3}}$ $re^{i\frac{4\pi}{3}}$ . However, since $P$ is degree 5, there must be two additional roots. Let one of these roots be $w$ , if $w$ is a root, then $we^{2\pi i / 3}$ and $we^{4\pi i / 3}$ must also be roots. However, $P$ is a fifth degree polynomial, and can therefore only have $5$ roots. This implies that $w$ is either $r$ $re^{2\pi i / 3}$ , or $re^{4\pi i / 3}$ . Thus we know that the polynomial $P$ can be written in the form $(x-r)^m(x-re^{2\pi i / 3})^n(x-re^{4\pi i / 3})^p$ . Moreover, by Vieta's, we know that there is only one possible value for the magnitude of $r$ as $||r||^5 = 2020$ , meaning that the amount of possible polynomials $P$ is equivalent to the possible sets $(m,n,p)$ . In order for the coefficients of the polynomial to all be real, $n = p$ due to $re^{2\pi i / 3}$ and $re^{4 \pi i / 3}$ being conjugates and since $m+n+p = 5$ , (as the polynomial is 5th degree) we have two possible solutions for $(m, n, p)$ which are $(1,2,2)$ and $(3,1,1)$ yielding two possible polynomials. The answer is thus $\boxed{2}$ | C | 2 |
2bf214fa5caab8d392ad5010cbc10947 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_17 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$ , where $a$ $b$ $c$ , and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$ ? (Note that $i=\sqrt{-1}$
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$ | Let $x_1=r$ , then \[x_2=\frac{-1+i\sqrt{3}}{2} r,\] \[x_3=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 r =\left( \frac{-1-i\sqrt{3}}{2} \right) r,\] \[x_4=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 r=r,\] which means $x_4$ is the same as $x_1$
Now we have 3 different roots of the polynomial, $x_1$ $x_2$ , and $x_3$ . Next, we will prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there is one root $x_4=p$ which is different from the three roots we already know, then there must be two other roots, \[x_5=\left( \frac{-1+i\sqrt{3}}{2} \right) ^2 p =\left( \frac{-1-i\sqrt{3}}{2} \right) p,\] \[x_6=\left( \frac{-1+i\sqrt{3}}{2} \right) ^3 p=p,\] different from all known roots. We get 6 different roots for the polynomial, which contradicts the limit of 5 roots. Therefore the assumption of a different root is wrong, thus the roots must be chosen from $x_1$ $x_2$ , and $x_3$
The polynomial then can be written like $f(x)=(x-x_1)^m (x-x_2)^n (x-x_3)^q$ , where $m$ $n$ , and $q$ are non-negative integers and $m+n+q=5$ . Since $a$ $b$ $c$ and $d$ are real numbers, then $n$ must be equal to $q$ . Therefore $(m,n,q)$ can only be $(1,2,2)$ or $(3,1,1)$ , so the answer is $\boxed{2}$ | C | 2 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We can first consider the equation without a floor function:
\[\dfrac{n+1000}{70} = \sqrt{n}\]
Multiplying both sides by 70 and then squaring:
\[n^2 + 2000n + 1000000 = 4900n\]
Moving all terms to the left:
\[n^2 - 2900n + 1000000 = 0\]
Now we can determine the factors:
\[(n-400)(n-2500) = 0\]
This means that for $n = 400$ and $n = 2500$ , the equation will hold without the floor function.
Now we can simply check the multiples of 70 around 400 and 2500 in the original equation, which we abbreviate as $L=R$
For $n = 330$ $L=19$ but $18^2 < 330 < 19^2$ so $R=18$
For $n = 400$ $L=20$ and $R=20$
For $n = 470$ $L=21$ $R=21$
For $n = 540$ $L=22$ but $540 > 23^2$ so $R=23$
Now we move to $n = 2500$
For $n = 2430$ $L=49$ and $49^2 < 2430 < 50^2$ so $R=49$
For $n = 2360$ $L=48$ and $48^2 < 2360 < 49^2$ so $R=48$
For $n = 2290$ $L=47$ and $47^2 < 2360 < 48^2$ so $R=47$
For $n = 2220$ $L=46$ but $47^2 < 2220$ so $R=47$
For $n = 2500$ $L=50$ and $R=50$
For $n = 2570$ $L=51$ but $2570 < 51^2$ so $R=50$
Therefore we have 6 total solutions, $n = 400, 470, 2290, 2360, 2430, 2500 = \boxed{6}$ | C | 6 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | This is my first solution here, so please forgive me for any errors.
We are given that \[\frac{n+1000}{70}=\lfloor\sqrt{n}\rfloor\]
$\lfloor\sqrt{n}\rfloor$ must be an integer, which means that $n+1000$ is divisible by $70$ . As $1000\equiv 20\pmod{70}$ , this means that $n\equiv 50\pmod{70}$ , so we can write $n=70k+50$ for $k\in\mathbb{Z}$
Therefore, \[\frac{n+1000}{70}=\frac{70k+1050}{70}=k+15=\lfloor\sqrt{70k+50}\rfloor\]
Also, we can say that $\sqrt{70k+50}-1 < k+15$ and $k+15\leq\sqrt{70k+50}$
Squaring the second inequality, we get $k^{2}+30k+225\leq70k+50\implies k^{2}-40k+175\leq 0\implies (k-5)(k-35)\leq0\implies 5\leq k\leq 35$
Similarly solving the first inequality gives us $k < 19-\sqrt{155}$ or $k > 19+\sqrt{155}$
$\sqrt{155}$ is larger than $12$ and smaller than $13$ , so instead, we can say $k\leq 6$ or $k\geq 32$
Combining this with $5\leq k\leq 35$ , we get $k=5,6,32,33,34,35$ are all solutions for $k$ that give a valid solution for $n$ , meaning that our answer is $\boxed{6}$ .
-Solution By Qqqwerw | C | 6 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | We start with the given equation \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor\] From there, we can start with the general inequality that $\lfloor \sqrt{n} \rfloor \leq \sqrt{n} < \lfloor \sqrt{n} \rfloor + 1$ . This means that \[\dfrac{n+1000}{70} \leq \sqrt{n} < \dfrac{n+1070}{70}\] Solving each inequality separately gives us two inequalities: \[n - 70\sqrt{n} +1000 \leq 0 \rightarrow (\sqrt{n}-50)(\sqrt{n}-20)\leq 0 \rightarrow 20\leq \sqrt{n} \leq 50\] \[n-70\sqrt{n}+1070 > 0 \rightarrow \sqrt{n} < 35-\sqrt{155} , \sqrt{n} > 35+\sqrt{155}\] Simplifying and approximating decimals yields 2 solutions for one inequality and 4 for the other. Hence, the answer is $2+4 = \boxed{6}$ | C | 6 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Since the right-hand-side is an integer, so must be the left-hand-side. Therefore, we must have $n\equiv -20\pmod{70}$ ; let $n=70j-20$ . The given equation becomes \[j+14 = \lfloor \sqrt{70j-20} \rfloor\]
Since $\lfloor x \rfloor \leq x < \lfloor x \rfloor +1$ for all real $x$ , we can take $x=\sqrt{70j-20}$ with $\lfloor x \rfloor =j+14$ to get \[j+14 \leq \sqrt{70j-20} < j+15\] We can square the inequality to get \[196+28j+j^{2} \leq 70j-20 < 225 + 30j + j^{2}\] The left inequality simplifies to $(j-36)(j-6) \leq 0$ , which yields \[6 \le j \le 36.\] The right inequality simplifies to $(j-20)^2 - 155 > 0$ , which yields \[j < 20 - \sqrt{155} < 8 \quad \text{or} \quad j > 20 + \sqrt{155} > 32\]
Solving $j < 8$ , and $6 \le j \le 36$ , we get $6 \le j < 8$ , for $2$ values $j\in \{6, 7\}$
Solving $j >32$ , and $6 \le j \le 36$ , we get $32 < j \le 36$ , for $4$ values $k\in \{33, \ldots , 36\}$
Thus, our answer is $2 + 4 = \boxed{6}$ | C | 6 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | Set $x=\sqrt{n}$ in the given equation and solve for $x$ to get $x^2 = 70 \cdot \lfloor x \rfloor - 1000$ . Set $k = \lfloor x \rfloor \ge 0$ ; since $\lfloor x \rfloor^2 \le x^2 < (\lfloor x \rfloor + 1)^2$ , we get \[k^2 \le 70k - 1000 < k^2 + 2k + 1.\] The left inequality simplifies to $(k-20)(k-50) \le 0$ , which yields \[20 \le k \le 50.\] The right inequality simplifies to $(k-34)^2 > 155$ , which yields \[k < 34 - \sqrt{155} < 22 \quad \text{or} \quad k > 34 + \sqrt{155} > 46\] Solving $k < 22$ , and $20 \le k \le 50$ , we get $20 \le k < 22$ , for $2$ values $k\in \{20, 21\}$
Solving $k >46$ , and $20 \le k \le 50$ , we get $46 < k \le 50$ , for $4$ values $k\in \{47, \ldots , 50\}$
Thus, our answer is $2 + 4 = \boxed{6}$ | C | 6 |
f3f93a5bbe033d47ee5e9d5689fbe8fb | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_21 | How many positive integers $n$ satisfy \[\dfrac{n+1000}{70} = \lfloor \sqrt{n} \rfloor?\] (Recall that $\lfloor x\rfloor$ is the greatest integer not exceeding $x$ .)
$\textbf{(A) } 2 \qquad\textbf{(B) } 4 \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 30 \qquad\textbf{(E) } 32$ | If $n$ is a perfect square, we can write $n = k^2$ for a positive integer $k$ , so $\lfloor \sqrt{n} \rfloor = \sqrt{n} = k.$ The given equation turns into
\begin{align*}
\frac{k^2 + 1000}{70} &= k \\
k^2 - 70k + 1000 &= 0 \\
(k-20)(k-50) &= 0,
\end{align*}
so $k = 20$ or $k= 50$ , so $n = 400, 2500.$
If $n$ is not square, then we can say that, for a positive integer $k$ , we have
\begin{align*}
k^2 < &n < (k+1)^2 \\
k^2 + 1000 < &n + 1000 = 70\lfloor \sqrt{n} \rfloor = 70k< (k+1)^2 + 1000 \\
k^2 + 1000 < &70k < (k+1)^2 + 1000.
\end{align*}
To solve this inequality, we take the intersection of the two solution sets to each of the two inequalities $k^2 + 1000 < 70k$ and $70k < (k+1)^2 + 1000$ . To solve the first one, we have
\begin{align*}
k^2 - 70k + 1000 &< 0 \\
(k-20)(k-50) &< 0\\
\end{align*} $k\in (20, 50),$ because the portion of the parabola between its two roots will be negative.
The second inequality yields
\begin{align*}
70k &< k^2 + 2k + 1 + 1000 \\
0 &< k^2 -68k + 1001.
\end{align*}
This time, the inequality will hold for all portions of the parabola that are not on or between the its two roots, which are $34 + \sqrt{155}>46$ and $34-\sqrt{155}<22$ (they are roughly equal, but this is to ensure that we do not miss any solutions).
Notation wise, we need all integers $k$ such that
\[k \in \left(20, 50\right) \cap \left(-\infty,34 - \sqrt{155} \right)\] or \[k \in \left(20, 50\right) \cap \left(34 + \sqrt{155}, \infty \right).\]
For the first one, since our uppoer bound is a little less than $22$ , the $k$ that works is $21$ . For the second, our lower bound is a little more than $46$ , so the $k$ that work are $47, 48,$ and $49$
$\boxed{6}$ total solutions for $n$ , since each value of $k$ corresponds to exactly one value of $n$ | C | 6 |
296a675cb2eccb3f3daddea51d191d68 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_23 | How many integers $n \geq 2$ are there such that whenever $z_1, z_2, ..., z_n$ are complex numbers such that
\[|z_1| = |z_2| = ... = |z_n| = 1 \text{ and } z_1 + z_2 + ... + z_n = 0,\] then the numbers $z_1, z_2, ..., z_n$ are equally spaced on the unit circle in the complex plane?
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 5$ | For $n=2$ , we see that if $z_{1}+z_{2}=0$ , then $z_{1}=-z_{2}$ , so they are evenly spaced along the unit circle.
For $n=3$ , WLOG, we can set $z_{1}=1$ . Notice that now Re $(z_{2}+z_{3})=-1$ and Im $\{z_{2}\}$ $-$ Im $\{z_{3}\}$ . This forces $z_{2}$ and $z_{3}$ to be equal to $e^{i\frac{2\pi}{3}}$ and $e^{-i\frac{2\pi}{3}}$ , meaning that all three are equally spaced along the unit circle.
We can now show that we can construct complex numbers when $n\geq 4$ that do not satisfy the conditions in the problem.
Suppose that the condition in the problem holds for some $n=k$ . We can now add two points $z_{k+1}$ and $z_{k+2}$ anywhere on the unit circle such that $z_{k+1}=-z_{k+2}$ , which will break the condition. Now that we have shown that $n=2$ and $n=3$ works, by this construction, any $n\geq 4$ does not work, making the answer $\boxed{2}$ | B | 2 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 2^5 \cdot 3$ . Since there are at most six not necessarily distinct factors $>1$ multiplying to $96$ , we have six cases: $k=1, 2, ..., 6.$ Now we look at each of the six cases.
$k=1$ : We see that there is $1$ way, merely $96$
$k=2$ : This way, we have the $3$ in one slot and $2$ in another, and symmetry. The four other $2$ 's leave us with $5$ ways and symmetry doubles us so we have $10$
$k=3$ : We have $3, 2, 2$ as our baseline. We need to multiply by $2$ in $3$ places, and see that we can split the remaining three powers of $2$ in a manner that is $3-0-0$ $2-1-0$ or $1-1-1$ . A $3-0-0$ split has $6 + 3 = 9$ ways of happening ( $24-2-2$ and symmetry; $2-3-16$ and symmetry), a $2-1-0$ split has $6 \cdot 3 = 18$ ways of happening (due to all being distinct) and a $1-1-1$ split has $3$ ways of happening ( $6-4-4$ and symmetry) so in this case we have $9+18+3=30$ ways.
$k=4$ : We have $3, 2, 2, 2$ as our baseline, and for the two other $2$ 's, we have a $2-0-0-0$ or $1-1-0-0$ split. The former grants us $4+12=16$ ways ( $12-2-2-2$ and symmetry and $3-8-2-2$ and symmetry) and the latter grants us also $12+12=24$ ways ( $6-4-2-2$ and symmetry and $3-4-4-2$ and symmetry) for a total of $16+24=40$ ways.
$k=5$ : We have $3, 2, 2, 2, 2$ as our baseline and one place to put the last two: on another two or on the three. On the three gives us $5$ ways due to symmetry and on another two gives us $5 \cdot 4 = 20$ ways due to symmetry. Thus, we have $5+20=25$ ways.
$k=6$ : We have $3, 2, 2, 2, 2, 2$ and symmetry and no more twos to multiply, so by symmetry, we have $6$ ways.
Thus, adding, we have $1+10+30+40+25+6=\boxed{112}$ | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | As before, note that $96=2^5\cdot3$ , and we need to consider 6 different cases, one for each possible value of $k$ , the number of factors in our factorization. However, instead of looking at each individually, find a general form for the number of possible factorizations with $k$ factors. First, the factorization needs to contain one factor that is itself a multiple of $3$ . There are $k$ to choose from. That leaves $k-1$ slots left to fill, each of which must contain at least one factor of $2$ . Once we have filled in a $2$ to each of the remaining slots, we're left with $5-(k-1)=6-k$ twos.
Consider the remaining $6-k$ factors of $2$ left to assign to the $k$ factors. Using stars and bars, the number of ways to do this is: \[{{(6-k)+k-1}\choose{6-k}}={5\choose{6-k}}\] This makes $k{5\choose{6-k}}$ possibilities for each k.
To obtain the total number of factorizations, add all possible values for k: \[\sum_{k=1}^6 k{5\choose{6-k}}=1+10+30+40+25+6=\boxed{112}.\] | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Begin by examining $f_1$ $f_1$ can take on any value that is a factor of $96$ except $1$ . For each choice of $f_1$ , the resulting $f_2...f_k$ must have a product of $96/f_1$ . This means the number of ways the rest $f_a$ $1<a<=k$ can be written by the scheme stated in the problem for each $f_1$ is equal to $D(96/f_1)$ , since the product of $f_2 \cdot f_3... \cdot f_k=x$ is counted as one valid product if and only if $f_1 \cdot x=96$ , the product $x$ has the properties that factors are greater than $1$ , and differently ordered products are counted separately.
For example, say the first factor is $2$ . Then, the remaining numbers must multiply to $48$ , so the number of ways the product can be written beginning with $2$ is $D(48)$ . To add up all the number of solutions for every possible starting factor, $D(96/f_1)$ must be calculated and summed for all possible $f_1$ , except $96$ and $1$ , since a single $1$ is not counted according to the problem statement. The $96$ however, is counted, but only results in $1$ possibility, the first and only factor being $96$ . This means
$D(96)=D(48)+D(32)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$
Instead of calculating D for the larger factors first, reduce $D(48)$ $D(32)$ , and $D(24)$ into sums of $D(m)$ where $m<=16$ to ease calculation. Following the recursive definition $D(n)=($ sums of $D(c))+1$ where c takes on every divisor of n except for 1 and itself, the sum simplifies to
$D(96)=(D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1)$ $(D(16)+D(8)+D(4)+D(2)+1)+D(24)+D(16)+D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1.$
$D(24)=D(12)+D(8)+D(6)+D(4)+D(3)+D(2)+1$ , so the sum further simplifies to
$D(96)=3D(16)+4D(12)+5D(8)+4D(6)+5D(4)+4D(3)+5D(2)+5$ , after combining terms. From quick casework,
$D(16)=8, D(12)=8, D(8)=4, D(6)=3, D(4)=2, D(3)=1$ and $D(2)=1$ . Substituting these values into the expression above,
$D(96)=3 \cdot 8+4 \cdot 8+5 \cdot 4+4 \cdot 3+5 \cdot 2+4 \cdot 1+5 \cdot 1+5=\boxed{112}$ | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 3 \cdot 2^5$ , and that $D$ of a perfect power of a prime is relatively easy to calculate. Also note that you can find $D(96)$ from $D(32)$ by simply totaling the number of ways there are to insert a $3$ into a set of numbers that multiply to $32$
First, calculate $D(32)$ . Since $32 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2$ , all you have to do was find the number of ways to divide the $2$ 's into groups, such that each group has at least one $2$ . By stars and bars, this results in $1$ way with five terms, $4$ ways with four terms, $6$ ways with three terms, $4$ ways with two terms, and $1$ way with one term. (The total, $16$ , is not needed for the remaining calculations.)
Then, to get $D(96)$ , in each possible $D(32)$ sequence, insert a $3$ somewhere in it, either by placing it somewhere next to the original numbers (in one of $n+1$ ways, where $n$ is the number of terms in the $D(32)$ sequence), or by multiplying one of the numbers by $3$ (in one of $n$ ways). There are $2+1=3$ ways to do this with one term, $3+2=5$ with two, $7$ with three, $9$ with four, and $11$ with five.
The resulting number of possible sequences is $3 \cdot 1 + 5 \cdot 4 + 7 \cdot 6 + 9 \cdot 4 + 11 \cdot 1 = \boxed{112}$ . ~ emerald_block | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Consider the arrangement of the prime factors of 96 in a line $(2,2, 2, 2, 2, 3)$ . An arrangement of factors can be created by placing "dividers" to group primes. For example, $(2, 2, |, 2, 2, 2, |, 3)$ is equivalent to the arrangement $4 \cdot 8 \cdot 3$ . Because there are $6$ ways to order the prime factors, and $2^5$ ways to place dividers, this gives us an initial $6 \cdot 2^5$ ways to arrange divisors.
However, through this method, we overcount cases where $3$ is combined with another factor. For example, the arrangement $4 \cdot 6 \cdot 4$ can be written as $(2, 2, |, 2, 3, |, 2, 2)$ or $(2, 2, |, 3, 2, |, 2, 2)$ . Precisely, we double count any case with $6$ as a factor, triple count any case with $12$ , quadruple count any case with $24$ , etc.
Now, consider all cases where $3$ must be grouped with at least one $2$ . This can be expressed in the same "line" format as $(2, 2, 2, 2, 6)$ , where dividers can again be placed to group divisors. In this case, there are $5$ ways to order divisors, and $2^4$ ways to place dividers, so we have an $5 \cdot 2^4$ possible sequences for this case. Notice that in this format, we double count cases where $12$ is a factor, we triple count cases where $24$ is a factor, etc. Precisely, for any case counted $n$ times in the first step, it is counted $n - 1$ times in this step. Thus, if we subtract, we count each case exactly once.
So, we get:
$6 \cdot 2^5 - 5 \cdot 2^4 = \boxed{112}$ . ~ hdai1122 | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | First we factor $32$ into $m$ numbers $g_1, \cdots, g_m$ where $g_i>1,i=1,\ldots,m$ . By applying stars and bars there are $\binom{5-1}{m-1}$ ways. Then we can either insert $3$ into each of the $m+1$ spaces between (or beyond) $g_i$ 's, or multiply it to one of the $g_i$ 's, a total of $2m+1$ ways. Hence the answer to the problem is
\[\sum_{m=1}^5 (2m+1)\binom{5-1}{m-1} = \sum_{n=0}^4 (2n+3) \binom{4}{n} = 8\sum_{n=0}^4 \frac{n}{4} \binom{4}{n} + 3 \sum_{n=0}^4 \binom{4}{n} = 8 \sum_{n=0}^4 \binom{3}{n-1} + 3\sum_{n=0}^4 \binom{4}{n} = 8 \cdot 2^3 + 3\cdot 2^4 = \boxed{112}.\] | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Note that $96 = 2^5 \cdot 3$ $D(n)$ depends on dividing $2^5$ into different terms, which is the integer partition of $5$
Divide $96$ into $1$ term:
There is only one way. $\underline{\textbf{1}}$
Divide $96$ into $2$ terms:
$2 + 8 = \underline{\textbf{10}}$
Divide $96$ into $3$ terms:
$12 + 18 = \underline{\textbf{30}}$
Divide $96$ into $4$ terms:
$24 + 16 = \underline{\textbf{40}}$
Divide $96$ into $5$ terms:
When dividing $5$ into $5$ parts there are $2$ cases.
$20 + 5 = \underline{\textbf{25}}$
Divide $96$ into $6$ terms:
$5 = 1 + 1 + 1 + 1 + 1$ . The number of arrangements of $(2, 2, 2, 2, 2, 3)$ is $\frac{6!}{5!} = \underline{\textbf{6}}$
$1 + 10 + 30 + 40 + 25 + 6 = \boxed{112}$ | A | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Ignore the $3$ first and first count $2^{x_1+x_2+...+x^n}=32$ which $x_1+x_2+...+x_n=5$ . This implies that $n$ is less than or equal to $5$ . Now, we can see that $3$ can lie between the two $x_i, x_{i+1}$ , or contribute to one of them. This gives $2k+1$ if $x_1+...+x_k=5$ . Now, just sum up gives $\binom{5-1}{5-1}\cdot 11+\binom{5-1}{4-1}\cdot 9+\binom{5-1}{3-1}\cdot 7+\binom{5-1}{2-1}\cdot 5+\binom{5-1}{1-1}\cdot 3=\boxed{112}$ | null | 112 |
30b94434a671d8e4856bab045fb90f88 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_24 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product \[n = f_1\cdot f_2\cdots f_k,\] where $k\ge1$ , the $f_i$ are integers strictly greater than $1$ , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number $6$ can be written as $6$ $2\cdot 3$ , and $3\cdot2$ , so $D(6) = 3$ . What is $D(96)$
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | Consider how we partition the factors of $96 = 2^5\cdot 3$ . For each $k$ , there are two cases. Either we can put the $2$ s into $k$ nonzero parts, so that the $3$ shares a partition with some $2$ s, which can be done in $k\binom{5-(k-1)+(k-1)-1}{k-2} = k\binom{4}{k-2}$ ways, or we can put the $2$ s into $k-1$ nonzero parts and put the $3$ in its own partition, which can be done in $k\binom{5-k+k-1}{k-1} = k\binom{4}{k-1}$ ways. Summing over all $k$ , we have \[\sum_{k=1}^6 \binom{4}{k-2} + \sum_{k=1}^6 \binom{4}{k-1}.\]
But $\sum_{k=1}^6 \binom{4}{k-2} = 2\binom{4}{0}+3\binom{4}{1} + \dots + 6\binom{4}{4} = 4 \cdot \sum_{i=0}^4 \binom{4}{i} = 4 \cdot 2^4 = 64$ . Similarly, $\sum_{k=1}^6 \binom{4}{k-1} = 3\cdot 2^4 = 48$ . So our answer is $64+48 = \boxed{112}$ | A | 112 |
05b2e467fe1c6d65b590c5d456f6751f | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_1 | The area of a pizza with radius $4$ is $N$ percent larger than the area of a pizza with radius $3$ inches. What is the integer closest to $N$
$\textbf{(A) } 25 \qquad\textbf{(B) } 33 \qquad\textbf{(C) } 44\qquad\textbf{(D) } 66 \qquad\textbf{(E) } 78$ | The area of the larger pizza is $16\pi$ , while the area of the smaller pizza is $9\pi$ . Therefore, the larger pizza is $\frac{7\pi}{9\pi} \cdot 100\%$ bigger than the smaller pizza. $\frac{7\pi}{9\pi} \cdot 100\% = 77.777....$ , which is closest to $\boxed{78}$ | E | 78 |
713306a9ed1cb757abd58ddd9328a6d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Since $a=1.5b$ , that means $b=\frac{a}{1.5}$ . We multiply by $3$ to get a $3b$ term, yielding $3b=2a$ , and $2a$ is $\boxed{200}$ of $a$ | D | 200 |
713306a9ed1cb757abd58ddd9328a6d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Without loss of generality, let $b=100$ . Then, we have $a=150$ and $3b=300$ . Thus, $\frac{3b}{a}=\frac{300}{150}=2$ , so $3b$ is $200\%$ of $a$ . Hence the answer is $\boxed{200}$ | D | 200 |
713306a9ed1cb757abd58ddd9328a6d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | As before, $a = 1.5b$ . Multiply by 2 to obtain $2a = 3b$ . Since $2 = 200\%$ , the answer is $\boxed{200}$ | D | 200 |
713306a9ed1cb757abd58ddd9328a6d3 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_2 | Suppose $a$ is $150\%$ of $b$ . What percent of $a$ is $3b$
$\textbf{(A) } 50 \qquad \textbf{(B) } 66+\frac{2}{3} \qquad \textbf{(C) } 150 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 450$ | Without loss of generality, let $b=2$ . Then, we have $a=3$ and $3b=6$ . This gives $\frac{3b}{a}=\frac{6}{3}=2$ , so $3b$ is $200\%$ of $a$ , so the answer is $\boxed{200}$ | D | 200 |
4fa30c063a9aaf73df4fab03a53edb63 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_3 | A box contains $28$ red balls, $20$ green balls, $19$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls. What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least $15$ balls of a single color will be drawn?
$\textbf{(A) } 75 \qquad\textbf{(B) } 76 \qquad\textbf{(C) } 79 \qquad\textbf{(D) } 84 \qquad\textbf{(E) } 91$ | We try to find the worst case scenario where we can find the maximum number of balls that can be drawn while getting $<15$ of each color by applying the pigeonhole principle and through this we get a perfect guarantee.
Namely, we can draw up to $14$ red balls, $14$ green balls, $14$ yellow balls, $13$ blue balls, $11$ white balls, and $9$ black balls, for a total of $75$ balls, without drawing $15$ balls of any one color. Drawing one more ball guarantees that we will get $15$ balls of one color — either red, green, or yellow. Thus, the answer is $75 + 1 = \boxed{76}$ | B | 76 |
cf276db8b0ef4883013cc145ca3af401 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_4 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | We might at first think that the answer would be $9$ , because $1+2+3 \dots +n = 45$ when $n = 9$ . But note that the problem says that they can be integers, not necessarily positive. Observe also that every term in the sequence $-44, -43, \cdots, 44, 45$ cancels out except $45$ . Thus, the answer is, intuitively, $\boxed{90}$ integers. | D | 90 |
cf276db8b0ef4883013cc145ca3af401 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_4 | What is the greatest number of consecutive integers whose sum is $45?$
$\textbf{(A) } 9 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 45 \qquad\textbf{(D) } 90 \qquad\textbf{(E) } 120$ | To maximize the number of integers, we need to make the average of them as low as possible while still being positive. The average can be $\frac12$ if the middle two numbers are $0$ and $1$ , so the answer is $\frac{45}{\frac12}=\boxed{90}$ | D | 90 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Let's first work out the slope-intercept form of all three lines: $(x,y)=(2,2)$ and $y=\frac{x}{2} + b$ implies $2=\frac{2}{2} +b=1+b$ so $b=1$ , while $y=2x + c$ implies $2= 2 \cdot 2+c=4+c$ so $c=-2$ . Also, $x+y=10$ implies $y=-x+10$ . Thus the lines are $y=\frac{x}{2} +1, y=2x-2,$ and $y=-x+10$ .
Now we find the intersection points between each of the lines with $y=-x+10$ , which are $(6,4)$ and $(4,6)$ . Using the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle with base $2\sqrt{2}$ and height $3\sqrt{2}$ , whose area is $\boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in Solution 1, we determine the coordinates of the three vertices of the triangle. The coordinates that we get are: $(2,2)$ $(6,4)$ $(4,6)$ . Now, using the Shoelace Theorem , we can directly find that the area is $\boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, solve the systems of equations to see that the triangle's two other vertices are at $(4, 6)$ and $(6, 4)$ . Then apply Heron's Formula: the semi-perimeter will be $s = \sqrt{2} + \sqrt{20}$ , so the area reduces nicely to a difference of squares, making it $\boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in the other solutions, we find, either using algebra or simply by drawing the lines on squared paper, that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . We can now draw the bounding square with vertices $(2, 2)$ $(2, 6)$ $(6, 6)$ and $(6, 2)$ , and deduce that the triangle's area is $16-4-2-4=\boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . Using graph paper, we can see that this triangle has $6$ boundary lattice points and $4$ interior lattice points. By Pick's Theorem, the area is $\frac62 + 4 - 1 = \boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ , so the area is $\frac12 bc \sin A = \frac12 \cdot 2\sqrt5 \cdot 2\sqrt5 \cdot \frac35 = \boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . The area of the triangle is half the absolute value of the determinant of the matrix determined by these points. \[\frac12\begin{Vmatrix} 2&2&1\\ 4&6&1\\ 6&4&1\\ \end{Vmatrix} = \frac12|-12| = \frac12 \cdot 12 = \boxed{6}\] | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . Then vectors $\overrightarrow{AB} = \langle 2, 4 \rangle$ and $\overrightarrow{AC} = \langle 4, 2 \rangle$ . The area of the triangle is half the magnitude of the cross product of these two vectors. \[\frac12\begin{Vmatrix} i&j&k\\ 2&4&0\\ 4&2&0\\ \end{Vmatrix} = \frac12|-12k| = \frac12 \cdot 12 = \boxed{6}\] | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that the three points of intersection are $(2, 2)$ $(4, 6)$ , and $(6, 4)$ . By the Pythagorean theorem, this is an isosceles triangle with base $2\sqrt2$ and equal length $2\sqrt5$ . The area of an isosceles triangle with base $b$ and equal length $l$ is $\frac{b\sqrt{4l^2-b^2}}{4}$ . Plugging in $b = 2\sqrt2$ and $l = 2\sqrt5$ \[\frac{2\sqrt2 \cdot \sqrt{80-8}}{4} = \frac{\sqrt{576}}{4} = \frac{24}{4} = \boxed{6}\] | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find the three points of intersection. Label these $A (2, 2)$ $B (4, 6)$ , and $C (6, 4)$ . By the Pythagorean Theorem, $AB = AC = 2\sqrt5$ and $BC = 2\sqrt2$ . By the Law of Cosines, \[\cos A = \frac{(2\sqrt5)^2 + (2\sqrt5)^2 - (2\sqrt2)^2}{2 \cdot 2\sqrt5 \cdot 2\sqrt5} = \frac{20 + 20 - 8}{40} = \frac{32}{40} = \frac45\] Therefore, $\sin A = \sqrt{1 - \cos^2 A} = \frac35$ . By the extended Law of Sines, \[2R = \frac{a}{\sin A} = \frac{2\sqrt2}{\frac35} = \frac{10\sqrt2}{3}\] \[R = \frac{5\sqrt2}{3}\] Then the area is $\frac{abc}{4R} = \frac{2\sqrt2 \cdot 2\sqrt5^2}{4 \cdot \frac{5\sqrt2}{3}} = \boxed{6}$ | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | The area of a triangle formed by three lines, \[a_1x + a_2y + a_3 = 0\] \[b_1x + b_2y + b_3 = 0\] \[c_1x + c_2y + c_3 = 0\] is the absolute value of \[\frac12 \cdot \frac{1}{(b_1c_2-b_2c_1)(a_1c_2-a_2c_1)(a_1b_2-a_2b_1)} \cdot \begin{vmatrix} a_1&a_2&a_3\\ b_1&b_2&b_3\\ c_1&c_2&c_3\\ \end{vmatrix}^2\] Plugging in the three lines, \[-x + 2y - 2 = 0\] \[-2x + y + 2 = 0\] \[x + y - 10 = 0\] the area is the absolute value of \[\frac12 \cdot \frac{1}{(-2-1)(-1-2)(-1+4)} \cdot \begin{vmatrix} -1&2&-2\\ -2&1&2\\ 1&1&-10\\ \end{vmatrix}^2 = \frac12 \cdot \frac{1}{27} \cdot 18^2 = \boxed{6}\] Source: Orrick, Michael L. “THE AREA OF A TRIANGLE FORMED BY THREE LINES.” Pi Mu Epsilon Journal, vol. 7, no. 5, 1981, pp. 294–298. JSTOR, www.jstor.org/stable/24336991. | C | 6 |
f4341c0be1d7dcaee16a5c967e68affd | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_5 | Two lines with slopes $\dfrac{1}{2}$ and $2$ intersect at $(2,2)$ . What is the area of the triangle enclosed by these two lines and the line $x+y=10 ?$
$\textbf{(A) } 4 \qquad\textbf{(B) } 4\sqrt{2} \qquad\textbf{(C) } 6 \qquad\textbf{(D) } 8 \qquad\textbf{(E) } 6\sqrt{2}$ | Like in other solutions, we find that our triangle is isosceles with legs of $2\sqrt5$ and base $2\sqrt2$ . Then, the semi - perimeter of our triangle is, \[\frac{4\sqrt5+2\sqrt2}{2} = 2\sqrt5 + \sqrt2.\] Applying Heron's formula, we find that the area of this triangle is equivalent to \[\sqrt{{(2\sqrt5+\sqrt2)}{(2\sqrt5-\sqrt2)}{(2)}} = \sqrt{{(20-2)}{(2)}} = \boxed{6}.\] | C | 6 |
2f8efa6f69cbec136bb123420190c6eb | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_8 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | It is possible to obtain $0$ $1$ $3$ $4$ $5$ , and $6$ points of intersection, as demonstrated in the following figures: [asy] unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0)); draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6)); draw((-1/3,1-d)--(-1/3,-1-d),Arrows); draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d)); draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d)); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d)); dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); [/asy]
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$ , since each pair of lines can intersect at most once. We now prove that it is impossible to obtain two intersections.
We proceed by contradiction. Assume a configuration of four lines exists such that there exist only two intersection points. Let these intersection points be $A$ and $B$ . Consider two cases:
Case 1 : No line passes through both $A$ and $B$
Then, since an intersection point is obtained by an intersection between at least two lines, two lines pass through each of $A$ and $B$ . Then, since there can be no additional intersections, the 2 lines that pass through $A$ cant intersect the 2 lines that pass through $B$ , and so 2 lines passing through $A$ must be parallel to 2 lines passing through $B$ . Then the two lines passing through $B$ are parallel to each other by transitivity of parallelism, so they coincide, contradiction.
Case 2 : There is a line passing through $A$ and $B$
Then there must be a line $l_a$ passing through $A$ , and a line $l_b$ passing through $B$ . These lines must be parallel. The fourth line $l$ must pass through either $A$ or $B$ . Without loss of generality, suppose $l$ passes through $A$ . Then since $l$ and $l_a$ cannot coincide, they cannot be parallel. Then $l$ and $l_b$ cannot be parallel either, so they intersect, contradiction.
All possibilities have been exhausted, and thus we can conclude that two intersections is impossible. Our answer is given by the sum $0+1+3+4+5+6=\boxed{19}$ | D | 19 |
508a1373b5cddadf1c2298118d631863 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | We can expand the fraction $0.\overline{23}_k$ as follows: $0.\overline{23}_k = 2\cdot k^{-1} + 3 \cdot k^{-2} + 2 \cdot k^{-3} + 3 \cdot k^{-4} + \cdots$
Notice that this is equivalent to \[2( k^{-1} + k^{-3} + k^{-5} + ... ) + 3 (k^{-2} + k^{-4} + k^{-6} + \cdots )\]
By summing the geometric series and simplifying, we have $\frac{2k+3}{k^2-1} = \frac{7}{51}$ . Solving this quadratic equation (or simply testing the answer choices) yields the answer $k = \boxed{16}$ | D | 16 |
508a1373b5cddadf1c2298118d631863 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Let $a = 0.2323\dots_k$ . Therefore, $k^2a=23.2323\dots_k$
From this, we see that $k^2a-a=23_k$ , so $a = \frac{23_k}{k^2-1} = \frac{2k+3}{k^2-1} = \frac{7}{51}$
Now, similar to in Solution 1, we can either test if $2k+3$ is a multiple of 7 with the answer choices, or actually solve the quadratic, so that the answer is $\boxed{16}$ | D | 16 |
508a1373b5cddadf1c2298118d631863 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Just as in Solution 1, we arrive at the equation $\frac{2k+3}{k^2-1}=\frac{7}{51}$
Therefore now, we can rewrite this as $\frac{2k+3}{(k-1)(k+1)}=\frac{7}{51}=\frac{7}{3\cdot 17}$ . Notice that $2k+3=2(k+1)+1=2(k-1)+5$ . As $17$ is a prime number, we therefore must have that one of $k-1$ and $k+1$ is divisible by $17$ . Now, checking each of the answer choices, this will lead us to the answer $\boxed{16}$ | D | 16 |
508a1373b5cddadf1c2298118d631863 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Assuming you are familiar with the rules for basic repeating decimals, $0.232323... = \frac{23}{99}$ . Now we want our base, $k$ , to conform to $23\equiv7\pmod k$ and $99\equiv51\pmod k$ , the reason being that we wish to convert the number from base $10$ to base $k$ . Given the first equation, we know that $k$ must equal 9, 16, 23, or generally, $7n+2$ . The only number in this set that is one of the multiple choices is $16$ . When we test this on the second equation, $99\equiv51\pmod k$ , it comes to be true. Therefore, our answer is $\boxed{16}$ | D | 16 |
508a1373b5cddadf1c2298118d631863 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_11 | For some positive integer $k$ , the repeating base- $k$ representation of the (base-ten) fraction $\frac{7}{51}$ is $0.\overline{23}_k = 0.232323..._k$ . What is $k$
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$ | Note that the LHS equals \[\bigg(\frac{2}{k} + \frac{2}{k^3} + \cdots \bigg) + \bigg(\frac{3}{k^2} + \frac{3}{k^4} + \cdots \bigg) = \frac{\frac{2}{k}}{1 - \frac{1}{k^2}} + \frac{\frac{3}{k^2}}{1 - \frac{1}{k^2}} = \frac{2k+3}{k^2-1},\] from which we see our equation becomes \[\frac{2k+3}{k^2-1} = \frac{7}{51}, \ \ \implies \ \ 51(2k+3) = 7(k^2-1).\]
Note that $17$ therefore divides $k^2 - 1,$ but as $17$ is prime this therefore implies \[k \equiv \pm 1 \pmod{17}.\] (Warning: This would not be necessarily true if $17$ were composite.) Note that $\boxed{16}$ is the only answer choice congruent satisfying this modular congruence, thus completing the problem. $\square$ | D | 16 |
56171f165a3d66224821b9363e08e99c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | Let $\log_2{x} = \log_y{16}=k$ , so that $2^k=x$ and $y^k=16 \implies y=2^{\frac{4}{k}}$ . Then we have $(2^k)(2^{\frac{4}{k}})=2^{k+\frac{4}{k}}=2^6$
We therefore have $k+\frac{4}{k}=6$ , and deduce $k^2-6k+4=0$ . The solutions to this are $k = 3 \pm \sqrt{5}$
To solve the problem, we now find \begin{align*} (\log_2\tfrac{x}{y})^2&=(\log_2 x - \log_2 y)^2\\ &=(k-\tfrac{4}{k})^2=(3 \pm \sqrt{5} - \tfrac{4}{3 \pm \sqrt{5}})^2 \\ &= (3 \pm \sqrt{5} - [3 \mp \sqrt{5}])^2\\ &= (3 \pm \sqrt{5} - 3 \pm \sqrt{5})^2\\ &=(\pm 2\sqrt{5})^2 \\ &= \boxed{20} ~Edits by BakedPotato66 | B | 20 |
56171f165a3d66224821b9363e08e99c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | After obtaining $k + \frac{4}{k} = 6$ , notice that the required answer is $\left(k - \frac{4}{k}\right)^{2} = k^2 - 8 + \frac{16}{k^2} = \left(k^2 + 8 + \frac{16}{k^2}\right) - 16 = \left(k+\frac{4}{k}\right)^2 - 16 = 6^2 - 16 = \boxed{20}$ , as before. | B | 20 |
56171f165a3d66224821b9363e08e99c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_12 | Positive real numbers $x \neq 1$ and $y \neq 1$ satisfy $\log_2{x} = \log_y{16}$ and $xy = 64$ . What is $(\log_2{\tfrac{x}{y}})^2$
$\textbf{(A) } \frac{25}{2} \qquad\textbf{(B) } 20 \qquad\textbf{(C) } \frac{45}{2} \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 32$ | From the given data, $\log_2(x) = \frac{1}{\log_{16}(y)}$ , or $\log_2(x) = \frac{4}{{\log_{2}(y)}}$
We know that $xy=64$ , so $x= \frac{64}{y}$
Thus $\log_2\left(\frac{64}{y}\right) = \frac{4}{{\log_{2}(y)}}$ , so $6-\log_2(y) = \frac{4}{{\log_{2}(y)}}$ , so $6(\log_2(y))-(\log_2(y))^2=4$
Solving for $\log_2(y)$ , we obtain $\log_2(y)=3+\sqrt{5}$
Easy resubstitution further gives $\log_2(x)=\frac{4}{3+\sqrt{5}}$ . Simplifying, we obtain $\log_2(x)= 3-\sqrt{5}$
Looking back at the original problem, we have What is $(\log_2{\tfrac{x}{y}})^2$
Deconstructing this expression using log rules, we get $(\log_2{x}-\log_2{y})^2$
Plugging in our known values, we get $((3-\sqrt{5})-(3+\sqrt{5}))^2$ or $(-2\sqrt{5})^2$
Our answer is $\boxed{20}$ | B | 20 |
e600f683ed3a78da0872230982aa35a1 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | The $5$ and $7$ can be painted with no restrictions because the set of integers does not contain a multiple or proper factor of $5$ or $7$ . There are 3 ways to paint each, giving us $\underline{9}$ ways to paint both. The $2$ is the most restrictive number. There are $\underline{3}$ ways to paint $2$ , but without loss of generality, let it be painted red. $4$ cannot be the same color as $2$ or $8$ , so there are $\underline{2}$ ways to paint $4$ , which automatically determines the color for $8$ $6$ cannot be painted red, so there are $\underline{2}$ ways to paint $6$ , but WLOG, let it be painted blue. There are $\underline{2}$ choices for the color for $3$ , which is either red or green in this case. Lastly, there are $\underline{2}$ ways to choose the color for $9$
$9 \cdot 3 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = \boxed{432}$ | E | 432 |
e600f683ed3a78da0872230982aa35a1 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | We note that the primes can be colored any of the $3$ colors since they don't have any proper divisors other than $1$ , which is not in the list. Furthermore, $6$ is the only number in the list that has $2$ distinct prime factors (namely, $2$ and $3$ ), so we do casework on $6$
Case 1 $2$ and $3$ are the same color
In this case, we have $3$ primes to choose the color for ( $2$ $5$ , and $7$ ). Afterwards, $4$ $6$ , and $9$ have two possible colors, which will determine the color of $8$ . Thus, there are $3^3\cdot 2^3=216$ possibilities here.
Case 2 $2$ and $3$ are different colors
In this case, we have $4$ primes to color. Without loss of generality, we'll color the $2$ first, then the $3$ . Then there are $3$ color choices for $2,5,7$ , and $2$ color choices for $3$ . This will determine the color of $6$ as well. After that, we only need to choose the color for $4$ and $9$ , which each have $2$ choices. Thus, there are $3^3\cdot 2^3=216$ possibilities here as well.
Adding up gives $216+216=\boxed{432}$ | E | 432 |
e600f683ed3a78da0872230982aa35a1 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_13 | How many ways are there to paint each of the integers $2, 3, \cdots , 9$ either red, green, or blue so that each number has a different color from each of its proper divisors?
$\textbf{(A)}~144\qquad\textbf{(B)}~216\qquad\textbf{(C)}~256\qquad\textbf{(D)}~384\qquad\textbf{(E)}~432$ | $2,4,8$ require different colors each, so there are $6$ ways to color these.
$5$ and $7$ can be any color, so there are $3\times 3$ ways to color these.
$6$ can have $2$ colors once $2$ is colored, and thus $3$ also has $2$ colors following $6$ , which leaves another $2$ for $9$
All together: $6\times 3 \times 3 \times 2 \times 2 \times 2 = 432 \Rightarrow \boxed{432}$ | E | 432 |
e019e06fa70d63925d839d2dfec7961c | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_14 | For a certain complex number $c$ , the polynomial \[P(x) = (x^2 - 2x + 2)(x^2 - cx + 4)(x^2 - 4x + 8)\] has exactly 4 distinct roots. What is $|c|$
$\textbf{(A) } 2 \qquad \textbf{(B) } \sqrt{6} \qquad \textbf{(C) } 2\sqrt{2} \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \sqrt{10}$ | The polynomial can be factored further broken down into
$P(x) = (x - [1 - i])(x - [1 + i])(x - [2 - 2i])(x - [2 + 2i])(x^2 - cx + 4)$
by using the quadratic formula on each of the quadratic factors. Since the first four roots are all distinct, the term $(x^2 - cx + 4)$ must be a product of any combination of two (not necessarily distinct) factors from the set: $(x - [1 - i]), (x - [1 + i]), (x - [2 - 2i]),$ and $(x - [2 + 2i])$ . We need the two factors to yield a constant term of $4$ when multiplied together. The only combinations that work are $(x - [1 - i])$ and $(x - [2 + 2i])$ , or $(x - [1+i])$ and $(x - [2-2i])$ . When multiplied together, the polynomial is either $(x^2 + [-3 + i]x + 4)$ or $(x^2+[-3-i]x+4)$ . Therefore, $c = 3 \pm i$ and $|c| = \boxed{10}$ | E | 10 |
5b44a5786fef655c77453d0524bb1f83 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$ | Let $p, q$ , and $r$ be the roots of the polynomial. Then,
$p^3 - 5p^2 + 8p - 13 = 0$
$q^3 - 5q^2 + 8q - 13 = 0$
$r^3 - 5r^2 + 8r - 13 = 0$
Adding these three equations, we get
$(p^3 + q^3 + r^3) - 5(p^2 + q^2 + r^2) + 8(p + q + r) - 39 = 0$
$s_3 - 5s_2 + 8s_1 = 39$
$39$ can be written as $13s_0$ , giving
$s_3 = 5s_2 - 8s_1 + 13s_0$
We are given that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ is satisfied for $k = 2$ $3$ $....$ , meaning it must be satisfied when $k = 2$ , giving us $s_3 = a \, s_2 + b \, s_1 + c \, s_0$
Therefore, $a = 5, b = -8$ , and $c = 13$ by matching coefficients.
$5 - 8 + 13 = \boxed{10}$ | D | 10 |
5b44a5786fef655c77453d0524bb1f83 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$ | Let $p, q$ , and $r$ be the roots of the polynomial. By Vieta's Formulae, we have
$p+q+r = 5$
$pq+qr+rp = 8$
$pqr=13$
We know $s_k = p^k + q^k + r^k$ . Consider $(p+q+r)(s_k) =5s_k$
$5s_k = [p^{k+1} + q^{k+1} + r^{k+1}] + p^k q + p^k r + pq^k + q^k r + pr^k + qr^k$
Using $pqr = 13$ and $s_{k-2} = p^{k-2} + q^{k-2} + r^{k-2}$ , we see $13s_{k-2} = p^{k-1}qr + pq^{k-1}r + pqr^{k-1}$
We have \[\begin{split} 5s_k + 13s_{k-2} &= s_{k+1} + (p^k q + p^k r + p^{k-1}qr) + (pq^k + pq^{k-1}r + q^k r) + (pqr^{k-1} + pr^k + qr^k) \\ &= s_{k+1} + p^{k-1} (pq + pr + qr) + q^{k-1} (pq + pr + qr) + r^{k-1} (pq + pr + qr) \\ &= s_{k+1} + (p^{k-1} + q^{k-1} + r^{k-1})(pq + pr + qr) \\ &= 5s_k + 13s_{k-2} = s_{k+1} + 8s_{k-1}\end{split}\]
Rearrange to get $s_{k+1} = 5s_k - 8s_{k-1} + 13s_{k-2}$
So, $a+ b + c = 5 -8 + 13 = \boxed{10}$ | D | 10 |
5b44a5786fef655c77453d0524bb1f83 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_17 | Let $s_k$ denote the sum of the $\textit{k}$ th powers of the roots of the polynomial $x^3-5x^2+8x-13$ . In particular, $s_0=3$ $s_1=5$ , and $s_2=9$ . Let $a$ $b$ , and $c$ be real numbers such that $s_{k+1} = a \, s_k + b \, s_{k-1} + c \, s_{k-2}$ for $k = 2$ $3$ $....$ What is $a+b+c$
$\textbf{(A)} \; -6 \qquad \textbf{(B)} \; 0 \qquad \textbf{(C)} \; 6 \qquad \textbf{(D)} \; 10 \qquad \textbf{(E)} \; 26$ | Let $r,s,t$ be the roots of $x^3-5x^2+8x-13$ . Then:
$r^3=5r^2-8r+13$ \\ $s^3=5s^2-8s+13$ \\ $t^3=5t^2-8t+13$
If we multiply both sides of the equation by $r^k$ , where $k$ is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find $r^4+s^4+t^4$ , but that is just to check. So then with the above information about $r^3,s^3,t^3$ , we see that:
$r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}$ $s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}$ $t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}$
$s_k=r^k+s^k+t^k$
Then: $s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}$
This means that $s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}$ , as expected. So we have $a=5, b=-8, c=13$ . So our answer is $5-8+13=\boxed{10}$ | D | 10 |
01d243b06c4c4384199a7f898c440679 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | Notice that by the Law of Sines, $a:b:c = \sin{A}:\sin{B}:\sin{C}$ , so let's flip all the cosines using $\sin^{2}{x} + \cos^{2}{x} = 1$ $\sin{x}$ is positive for $0^{\circ} < x < 180^{\circ}$ , so we're good there).
$\sin A=\frac{3\sqrt{15}}{16}, \qquad \sin B= \frac{\sqrt{15}}{8}, \qquad \text{and} \qquad\sin C=\frac{\sqrt{15}}{4}$
These are in the ratio $3:2:4$ , so our minimal triangle has side lengths $2$ $3$ , and $4$ $\boxed{9}$ is our answer. | A | 9 |
01d243b06c4c4384199a7f898c440679 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | $\angle ACB$ is obtuse since its cosine is negative, so we let the foot of the altitude from $C$ to $AB$ be $H$ . Let $AH=11x$ $AC=16x$ $BH=7y$ , and $BC=8y$ . By the Pythagorean Theorem, $CH=\sqrt{256x^2-121x^2}=3x\sqrt{15}$ and $CH=\sqrt{64y^2-49y^2}=y\sqrt{15}$ . Thus, $y=3x$ . The sides of the triangle are then $16x$ $11x+7(3x)=32x$ , and $24x$ , so for some integers $a,b$ $16x=a$ and $24x=b$ , where $a$ and $b$ are minimal. Hence, $\frac{a}{16}=\frac{b}{24}$ , or $3a=2b$ . Thus the smallest possible positive integers $a$ and $b$ that satisfy this are $a=2$ and $b=3$ , so $x=\frac{1}{8}$ . The sides of the triangle are $2$ $3$ , and $4$ , so $\boxed{9}$ is our answer. | A | 9 |
01d243b06c4c4384199a7f898c440679 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_19 | In $\triangle ABC$ with integer side lengths, $\cos A = \frac{11}{16}$ $\cos B = \frac{7}{8}$ , and $\cos C = -\frac{1}{4}$ . What is the least possible perimeter for $\triangle ABC$
$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 23 \qquad \textbf{(D) } 27 \qquad \textbf{(E) } 44$ | Using the law of cosines, we get the following equations:
\[c^2=a^2+b^2+\frac{ab}{2}\] \[b^2=a^2+c^2-\frac{7ac}{4}\] \[a^2=b^2+c^2-\frac{11bc}{8}\]
Substituting $a^2+c^2-\frac{7ac}{4}$ for $b^2$ in $a^2=b^2+c^2-\frac{11bc}{8}$ and simplifying, we get the following: \[14a+11b=16c\]
Note that since $a, b, c$ are integers, we can solve this for integers. By some trial and error, we get that $(a,b,c) = (3,2,4)$ . Checking to see that this fits the triangle inequality, we find out that this indeed works. Hence, our answer is $3+2+4 = \boxed{9}$ | A | 9 |
2707db51e08d5c91ef46b063bdc25cef | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$ | Note that $z = \mathrm{cis }(45^{\circ})$
Also note that $z^{k} = z^{k + 8}$ for all positive integers $k$ because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo $8$
$1^2, 5^2,$ and $9^2$ are all $1 \pmod{8}$
$2^2, 6^2,$ and $10^2$ are all $4 \pmod{8}$
$3^2, 7^2,$ and $11^2$ are all $1 \pmod{8}$
$4^2, 8^2,$ and $12^2$ are all $0 \pmod{8}$
Therefore,
$z^{1^2} = z^{5^2} = z^{9^2} = \mathrm{cis }(45^{\circ})$
$z^{2^2} = z^{6^2} = z^{10^2} = \mathrm{cis }(180^{\circ}) = -1$
$z^{3^2} = z^{7^2} = z^{11^2} = \mathrm{cis }(45^{\circ})$
$z^{4^2} = z^{8^2} = z^{12^2} = \mathrm{cis }(0^{\circ}) = 1$
The term thus $\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)$ simplifies to $6\mathrm{cis }(45^{\circ})$ , while the term $\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)$ simplifies to $\frac{6}{\mathrm{cis }(45^{\circ})}$ . Upon multiplication, the $\mathrm{cis }(45^{\circ})$ cancels out and leaves us with $\boxed{36}$ | C | 36 |
2707db51e08d5c91ef46b063bdc25cef | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$ | It is well known that if $|z|=1$ then $\bar{z}=\frac{1}{z}$ . Therefore, we have that the desired expression is equal to \[\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)\] We know that $z=e^{\frac{i\pi}{4}}$ so $\bar{z}=e^{\frac{i7\pi}{4}}$ . Then, by De Moivre's Theorem, we have \[\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)\] which can easily be computed as $\boxed{36}$ | null | 36 |
2707db51e08d5c91ef46b063bdc25cef | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$ | We first calculate that $z^4 = -1$ . After a bit of calculation for the other even powers of $z$ , we realize that they cancel out add up to zero. Now we can simplify the expression to $\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)$ . Then, we calculate the first few odd powers of $z$ . We notice that $z^1 = z^9$ , so the values cycle after every 8th power. Since all of the odd squares are a multiple of $8$ away from each other, $z^1 = z^9 = z^{25} = ... = z^{121}$ , so $z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}$ , and $\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}$ . When multiplied together, we get $6 \cdot 6 = \boxed{36}$ as our answer. | C | 36 |
2707db51e08d5c91ef46b063bdc25cef | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_21 | Let \[z=\frac{1+i}{\sqrt{2}}.\] What is \[\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?\]
$\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2$ | $z=\mathrm{cis }(\pi/4)$
Perfect squares mod 8: $1,4,1,0,1,4,1,0,1,4,1,0$
$1/z=\overline{z}=\mathrm{cis }(7\pi/4)$
$6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}$ | null | 36 |
2a3916001883c2095a301590d24a4b63 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_22 | Circles $\omega$ and $\gamma$ , both centered at $O$ , have radii $20$ and $17$ , respectively. Equilateral triangle $ABC$ , whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$ , has vertex $A$ on $\omega$ , and the line containing side $\overline{BC}$ is tangent to $\gamma$ . Segments $\overline{AO}$ and $\overline{BC}$ intersect at $P$ , and $\dfrac{BP}{CP} = 3$ . Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$ $n$ $p$ $q$ with $\text{gcd}(m,n) = \text{gcd}(p,q) = 1$ . What is $m+n+p+q$ $\phantom{}$
$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$ | [asy] size(20cm); draw(circle((0,0), 20)); label("$\omega$", (0,0), 4.05*20*dir(149)*20/21); draw(circle((0,0), 17)); label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21); dot((0,0)); label("$O$", (0,0), E); pair aa = (-20, 0); dot(aa); label("$A$", aa, W); draw((-20,0)--(0,0)); real a = (-20 + (80/sqrt(13) - 34/sqrt(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2)); real ans = (80/sqrt(13) - 34/sqrt(3)); dot((a,0)); label("$P$", (a, 0), dir(290)*0.58); pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13); dot(s); label("$S$", s, dir(135)); pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13)); dot(c); label("$C$", c, dir(110)); pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13)); dot(m); label("$M$", m, dir(285)); pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13)); dot(b); label("$B$", b, S); draw(b--s); draw(s--(0,0)); draw(aa--b); draw(aa--c); draw(aa--m); markscalefactor=0.1; draw(rightanglemark(s,m,aa,3.4)); draw(rightanglemark((0,0),s,m,3.4)); [/asy]
Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$ , and $M$ be the midpoint of $\overline{BC}$ . Note that $AM \perp BS$ and $OS \perp BS$ . This implies that $\angle OAM \cong \angle AOS$ , and $\angle AMP \cong \angle OSP$ . Thus, $\triangle PMA \sim \triangle PSO$
If we let $s$ be the side length of $\triangle ABC$ , then it follows that $AM = \frac{\sqrt{3}}{2}s$ and $PM = \frac{s}{4}$ . This implies that $AP = \frac{\sqrt{13}}{4}s$ , so $\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}$ . Furthermore, $\frac{AM + SO}{AO} = \frac{AM}{AP}$ (because $\triangle PMA \sim \triangle PSO$ ) so this gives us the equation \[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\] to solve for the side length $s$ , or $AB$ . Thus, \[\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}\] \[\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}\] \[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB\] The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{130}$ | E | 130 |
88dbb94d1c2b747c7e6ce58ee1d61b06 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | By definition, the recursion becomes $a_n = \left(n^{\frac1{\log_7(n-1)}}\right)^{\log_7(a_{n-1})}=n^{\frac{\log_7(a_{n-1})}{\log_7(n-1)}}$ . By the change of base formula, this reduces to $a_n = n^{\log_{n-1}(a_{n-1})}$ . Thus, we have $\log_n(a_n) = \log_{n-1}(a_{n-1})$ . Thus, for each positive integer $m \geq 3$ , the value of $\log_m(a_m)$ must be some constant value $k$
We now compute $k$ from $a_3$ . It is given that $a_3 = 3\,\heartsuit\,2 = 3^{\frac1{\log_7(2)}}$ , so $k = \log_3(a_3) = \log_3\left(3^{\frac1{\log_7(2)}}\right) = \frac1{\log_7(2)} = \log_2(7)$
Now, we must have $\log_{2019}(a_{2019}) = k = \log_2(7)$ . At this point, we simply switch some bases around. For those who are unfamiliar with logarithms, we can turn the logarithms into fractions which are less intimidating to work with.
$\frac{\log{a_{2019}}}{\log{2019}} = \frac{\log{7}}{\log{2}}\implies \frac{\log{a_{2019}}}{\log{7}} = \frac{\log{2019}}{\log{2}}\implies \log_7(a_{2019}) =\log_2(2019)$
We conclude that $\log_7(a_{2019}) = \log_2(2019) \approx \boxed{11}$ | D | 11 |
88dbb94d1c2b747c7e6ce58ee1d61b06 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | Using the recursive definition, $a_4 = (4 \, \heartsuit \, 3) \, \diamondsuit\, (3 \, \heartsuit\, 2)$ or $a_4 = (4^{m})^{k}$ where $m = \frac{1}{\log_{7}(3)}$ and $k = \log_{7}(3^{\frac{1}{\log_{7}(2)}})$ . Using logarithm rules, we can remove the exponent of the 3 so that $k = \frac{\log_{7}(3)}{\log_{7}(2)}$ . Therefore, $a_4 = 4^{\frac{1}{\log_{7}(2)}}$ , which is $4 \, \heartsuit \, 2$
We claim that $a_n = n \, \heartsuit \, 2$ for all $n \geq 3$ . We can prove this through induction.
Clearly, the base case where $n = 3$ holds.
$a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, ((n-1) \, \heartsuit \, 2)$
This can be simplified as $a_n = (n^{\log_{n-1}(7)}) \, \diamondsuit \, ((n-1)^{\log_{2}(7)})$
Applying the diamond operation, we can simplify $a_n = n^h$ where $h = \log_{n-1}(7) \cdot \log_{7}(n-1)^{\log_{2}(7)}$ . By using logarithm rules to remove the exponent of $\log_{7}(n-1)$ and after cancelling, $h = \frac{1}{\log_{7}(2)}$
Therefore, $a_n = n^{\frac{1}{\log_{7}(2)}} = n \, \heartsuit \, 2$ for all $n \geq 3$ , completing the induction.
We have $a_{2019} = 2019^{\log_{2}(7)}$ . Taking $\log_{2019}$ of both sides gives us ${\log_{2019}(a_{2019})} = {\log_{2}(7)}$ . Then, by changing to base $7$ and after cancellation, we arrive at ${\log_{7}(a_{2019})} = {\log_{2}(2019)}$ . Because $2^{11} = 2048$ and $2^{10} = 1024$ , our answer is $\boxed{11}$ | D | 11 |
88dbb94d1c2b747c7e6ce58ee1d61b06 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_23 | Define binary operations $\diamondsuit$ and $\heartsuit$ by \[a \, \diamondsuit \, b = a^{\log_{7}(b)} \qquad \text{and} \qquad a \, \heartsuit \, b = a^{\frac{1}{\log_{7}(b)}}\] for all real numbers $a$ and $b$ for which these expressions are defined. The sequence $(a_n)$ is defined recursively by $a_3 = 3\, \heartsuit\, 2$ and \[a_n = (n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] for all integers $n \geq 4$ . To the nearest integer, what is $\log_{7}(a_{2019})$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | We are given that \[a_n=(n\, \heartsuit\, (n-1)) \,\diamondsuit\, a_{n-1}\] \[a_n=(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Since we are asked to find $\log_7(a_{2019})$ , we directly apply \[\log_7(a_n)=\log_7(n^{\frac{1}{\log_7(n-1)}})^{\log_7(a_{n-1})}\] Using the property that $\log_ab^c=c\log_ab$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7(n^{\frac{1}{\log_7(n-1)}}))\] Now using the property that $\frac{1}{\log_ab}=\log_ba$ \[\log_7(a_n)=(\log_7a_{n-1})(\log_7n^{\log_{n-1}7})\] Once again applying the first property yields \[\log_7(a_n)=(\log_7a_{n-1})(\log_{n-1}7)(\log_7n)\] Rearranging the expression, \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7a_{n-1})\]
Now expressing $\log_7a_{n-1}$ in a similar expression as $\log_7a_n$
\[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7a_{n-2})\] \[\log_7(a_n)=(\log_7n)(\log_{n-1}7)(\log_7n-1)(\log_{n-2}7)(\log_7n-2)(\log_{n-3}7)...(\log_74)(\log_37)(\log_7a_3)\]
Because of the fact that $(\log_ab)(\log_ba)=1$ , we can cancel out the terms to get
\[\log_7(a_n)=(\log_7n)(\log_37)(\log_7a_3)\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_7(3^{\frac{1}{\log_72}}))\] \[\log_7(a_n)=(\log_7n)(\log_37)(\log_27)(\log_73)\] \[\log_7(a_n)=(\log_27)(\log_7n)\]
Using the Chain Rule for Logarithm, $(\log_ab)(\log_bc)=\log_ac$ , yields
\[\log_7(a_n)=(\log_2n)\] Finally, substituting in $n=2019$ , we have \[\log_7(a_{2019})=(\log_22019)\] \[\log_7(a_{2019})\approx11\boxed{11}\] | D | 11 |
911a858e726fd0e0dc391dfa08bfcbf2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | The main insight is that
\[\frac{(n^2)!}{(n!)^{n+1}}\]
is always an integer. This is true because it is precisely the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n$ . Thus,
\[\frac{(n^2-1)!}{(n!)^n}=\frac{(n^2)!}{(n!)^{n+1}}\cdot\frac{n!}{n^2}\]
is an integer if $n^2 \mid n!$ , or in other words, if $\frac{(n-1)!}{n}$ , is an integer. This condition is false precisely when $n=4$ or $n$ is prime, by Wilson's Theorem . There are $15$ primes between $1$ and $50$ , inclusive, so there are $15 + 1 = 16$ terms for which
\[\frac{(n^2-1)!}{(n!)^{n}}\]
is potentially not an integer. It can be easily verified that the above expression is not an integer for $n=4$ as there are more factors of $2$ in the denominator than the numerator. Similarly, it can be verified that the above expression is not an integer for any prime $n=p$ , as there are more factors of p in the denominator than the numerator. Thus all $16$ values of n make the expression not an integer and the answer is $50-16=\boxed{34}$ | D | 34 |
911a858e726fd0e0dc391dfa08bfcbf2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | We can use the P-Adic Valuation (more info could be found here: Mathematicial notation ) of n to solve this problem (recall the P-Adic Valuation of 'n' is denoted by $v_p (n)$ and is defined as the greatest power of some prime 'p' that divides n. For example, $v_2 (6)=1$ or $v_7 (245)=2$ .) Using Legendre's formula, we know that :
\[v_p (n!)= \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor\]
Seeing factorials involved in the problem, this prompts us to use Legendre's Formula where n is a power of a prime.
We also know that , $v_p (m^n) = n \cdot v_p (m)$ .
Knowing that $a\mid b$ if $v_p (a) \le v_p (b)$ , we have that :
\[n \cdot v_p (n!) \le v_p ((n^2 -1 )!)\] and we must find all n for which this is true.
If we plug in $n=p$ , by Legendre's we get two equations:
\[v_p ((n^2 -1)!) = \sum_{i=1}^\infty \lfloor \frac {n^2 -1}{p^i} \rfloor = (p-1)+0+...+0 = p-1\]
And we also get :
\[v_p ((n!)^n) = n \cdot v_p (n!)= n \cdot \sum_{i=1}^\infty \lfloor \frac {n}{p^i} \rfloor = p \cdot ( 1+0+...0) = p\]
But we are asked to prove that $n \cdot v_p (n!) \le v_p ((n^2 -1 )!) \Longrightarrow p \le p-1$ which is false for all 'n' where n is prime.
Now we try the same for $n=p^2$ , where p is a prime. By Legendre we arrive at:
\[v_p ((p^4 -1)!) = p^3 + p^2 + p -3\]
(as $v_p(p^4!) = p^3 + p^2 + p + 1$ and $p^4$ contains 4 factors of $p$ ) and
\[p^2 \cdot v_p (p^2 !) = p^3 + p^2\]
Then we get:
\[p^2 \cdot v_p (p!) \le v_p ((n^4 -1)!) \Longrightarrow p^3 + p^2 \le p^3 + p^2 + p -3\] Which is true for all primes except for 2, so $2^2 = 4$ doesn't work. It can easily be verified that for all $n=p^i$ where $i$ is an integer greater than 2, satisfies the inequality : \[n \cdot v_p (n!) \le v_p ((n^2 -1 )!).\]
Therefore, there are 16 values that don't work and $50-16 = \boxed{34}$ values that work. | D | 34 |
911a858e726fd0e0dc391dfa08bfcbf2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_24 | For how many integers $n$ between $1$ and $50$ , inclusive, is \[\frac{(n^2-1)!}{(n!)^n}\] an integer? (Recall that $0! = 1$ .)
$\textbf{(A) } 31 \qquad \textbf{(B) } 32 \qquad \textbf{(C) } 33 \qquad \textbf{(D) } 34 \qquad \textbf{(E) } 35$ | Notice all $15$ primes don't work as there are $n$ factors of $n$ in the denominator and $n-1$ factors of $n$ in the numerator. Further experimentation finds that $n=4$ does not work as there are 11 factors of 2 in the numerator and 12 in the denominator. We also find that it seems that all other values of $n$ work. So we get $50-15-1=\boxed{34}$ and that happens to be right. | null | 34 |
dd1d2bbbe9a7c4283eb656dee5c2117f | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$ | For all nonnegative integers $n$ , let $\angle C_nA_nB_n=x_n$ $\angle A_nB_nC_n=y_n$ , and $\angle B_nC_nA_n=z_n$
Note that quadrilateral $A_0B_0A_1B_1$ is cyclic since $\angle A_0A_1B_0=\angle A_0B_1B_0=90^\circ$ ; thus, $\angle A_0A_1B_1=\angle A_0B_0B_1=90^\circ-x_0$ . By a similar argument, $\angle A_0A_1C_1=\angle A_0C_0C_1=90^\circ-x_0$ . Thus, $x_1=\angle A_0A_1B_1+\angle A_0A_1C_1=180^\circ-2x_0$ . By a similar argument, $y_1=180^\circ-2y_0$ and $z_1=180^\circ-2z_0$
Therefore, for any positive integer $n$ , we have $x_n=180^\circ-2x_{n-1}$ (identical recurrence relations can be derived for $y_n$ and $z_n$ ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to $n$ (and the coefficient of $x_0$ is $(-2)^n$ ). Hence, we let $x_n=pq^n+r+(-2)^nx_0$ . We will solve for $p$ $q$ , and $r$ by iterating the recurrence to obtain $x_1=180^\circ-2x_0$ $x_2=4x_0-180^\circ$ , and $x_3=540-8x_0$ . Letting $n=1,2,3$ respectively, we have \begin{align} pq+r&=180 \\ pq^2+r&=-180 \\ pq^3+r&=540 \end{align}
Subtracting $(1)$ from $(3)$ , we have $pq(q^2-1)=360$ , and subtracting $(1)$ from $(2)$ gives $pq(q-1)=-360$ . Dividing these two equations gives $q+1=-1$ , so $q=-2$ . Substituting back, we get $p=-60$ and $r=60$
We will now prove that for all positive integers $n$ $x_n=-60(-2)^n+60+(-2)^nx_0=(-2)^n(x_0-60)+60$ via induction. Clearly the base case of $n=1$ holds, so it is left to prove that $x_{n+1}=(-2)^{n+1}(x_0-60)+60$ assuming our inductive hypothesis holds for $n$ . Using the recurrence relation, we have \begin{align*} x_{n+1}&=180-2x_n \\ &=180-2((-2)^n(x_0-60)+60) \\ &=(-2)^{n+1}(x_0-60)+60 \end{align*}
Our induction is complete, so for all positive integers $n$ $x_n=(-2)^n(x_0-60)+60$ . Identical equalities hold for $y_n$ and $z_n$
The problem asks for the smallest $n$ such that either $x_n$ $y_n$ , or $z_n$ is greater than $90^\circ$ . WLOG, let $x_0=60^\circ$ $y_0=59.999^\circ$ , and $z_0=60.001^\circ$ . Thus, $x_n=60^\circ$ for all $n$ $y_n=-(-2)^n(0.001)+60$ , and $z_n=(-2)^n(0.001)+60$ . Solving for the smallest possible value of $n$ in each sequence, we find that $n=15$ gives $y_n>90^\circ$ . Therefore, the answer is $\boxed{15}$ | E | 15 |
dd1d2bbbe9a7c4283eb656dee5c2117f | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12A_Problems/Problem_25 | Let $\triangle A_0B_0C_0$ be a triangle whose angle measures are exactly $59.999^\circ$ $60^\circ$ , and $60.001^\circ$ . For each positive integer $n$ , define $A_n$ to be the foot of the altitude from $A_{n-1}$ to line $B_{n-1}C_{n-1}$ . Likewise, define $B_n$ to be the foot of the altitude from $B_{n-1}$ to line $A_{n-1}C_{n-1}$ , and $C_n$ to be the foot of the altitude from $C_{n-1}$ to line $A_{n-1}B_{n-1}$ . What is the least positive integer $n$ for which $\triangle A_nB_nC_n$ is obtuse?
$\textbf{(A) } 10 \qquad \textbf{(B) }11 \qquad \textbf{(C) } 13\qquad \textbf{(D) } 14 \qquad \textbf{(E) } 15$ | We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of $\triangle A_nB_nC_n$ and $60^{\circ}$ (we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometric sequence with an integer common ratio. From here, we get that $S_n=0.001S^n$ and we need $S_n>30$ . The first power of two greater than $30,000$ is $2^{15}$ thus our answer is $\boxed{15}$ | E | 15 |
b466309ecc2d95ae7e7053ca2baa14ff | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_2 | Consider the statement, "If $n$ is not prime, then $n-2$ is prime." Which of the following values of $n$ is a counterexample to this statement?
$\textbf{(A) } 11 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 19 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 27$ | Since a counterexample must be a value of $n$ which is not prime, $n$ must be composite, so we eliminate $\text{A}$ and $\text{C}$ . Now we subtract $2$ from the remaining answer choices, and we see that the only time $n-2$ is not prime is when $n = \boxed{27}$ | E | 27 |
42a78e10a596f2d8e6a99c199120eb4d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | \[\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}\]
Solving by the quadratic formula, $n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19$ (since clearly $n \geq 0$ ). The answer is therefore $1 + 9 = \boxed{10}$ | C | 10 |
42a78e10a596f2d8e6a99c199120eb4d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Dividing both sides by $n!$ gives \[(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.\] Since $n$ is non-negative, $n=19$ . The answer is $1 + 9 = \boxed{10}$ | C | 10 |
42a78e10a596f2d8e6a99c199120eb4d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Dividing both sides by $n!$ as before gives $(n+1)+(n+1)(n+2)=440$ . Now factor out $(n+1)$ , giving $(n+1)(n+3)=440$ . By considering the prime factorization of $440$ , a bit of experimentation gives us $n+1=20$ and $n+3=22$ , so $n=19$ , so the answer is $1 + 9 = \boxed{10}$ | C | 10 |
42a78e10a596f2d8e6a99c199120eb4d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Since $(n+1)! + (n+2)! = (n+1)n! + (n+2)(n+1)n! = 440 \cdot n!$ , the result can be factored into $(n+1)(n+3)n!=440 \cdot n!$ and divided by $n!$ on both sides to get $(n+1)(n+3)=440$ . From there, it is easier to complete the square with the quadratic $(n+1)(n+3) = n^2 + 4n + 3$ , so $n^2+4n+4=441 \Rightarrow (n+2)^2=441$ . Solving for $n$ results in $n=19,-23$ , and since $n>0$ $n=19$ and the answer is $1 + 9 = \boxed{10}$ | C | 10 |
42a78e10a596f2d8e6a99c199120eb4d | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_4 | There is a positive integer $n$ such that $(n+1)! + (n+2)! = n! \cdot 440$ . What is the sum of the digits of $n$
$\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12$ | Rewrite $(n+1)! + (n+2)! = 440 \cdot n!$ as $(n+1)n! + (n+2)(n+1)n! = 440 \cdot n!.$ Factoring out the $n!$ we get $n!(n + 1 + (n+1)(n+2)) = 440 \cdot n!.$ Expand this to get $n!(n^2 + 4n + 3) = 440 \cdot n!.$ Factor this and divide by $n!$ to get $(n + 1)(n + 3) = 440.$ If we take the prime factorization of $440$ we see that it is $2^3 * 5 * 11.$ Intuitively, we can find that $n + 1 = 20$ and $n + 3 = 22.$ Therefore, $n = 19.$ Since the problem asks for the sum of the didgits of $n$ , we finally calculate $1 + 9 = 10$ and get answer choice $\boxed{10}$ | C | 10 |
d623a36ca4900dd839c7f5e63fcae3a2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$ | If he has enough money to buy $12$ pieces of red candy, $14$ pieces of green candy, and $15$ pieces of blue candy, then the smallest amount of money he could have is $\text{lcm}{(12,14,15)} = 420$ cents. Since a piece of purple candy costs $20$ cents, the smallest possible value of $n$ is $\frac{420}{20} = \boxed{21}$ | B | 21 |
d623a36ca4900dd839c7f5e63fcae3a2 | https://artofproblemsolving.com/wiki/index.php/2019_AMC_12B_Problems/Problem_5 | Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either $12$ pieces of red candy, $14$ pieces of green candy, $15$ pieces of blue candy, or $n$ pieces of purple candy. A piece of purple candy costs $20$ cents. What is the smallest possible value of $n$
$\textbf{(A) } 18 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 24\qquad \textbf{(D) } 25 \qquad \textbf{(E) } 28$ | We simply need to find a value of $20n$ that is divisible by $12$ $14$ , and $15$ . Observe that $20 \cdot 18$ is divisible by $12$ and $15$ , but not $14$ $20 \cdot 21$ is divisible by $12$ $14$ , and $15$ , meaning that we have exact change (in this case, $420$ cents) to buy each type of candy, so the minimum value of $n$ is $\boxed{21}$ | B | 21 |
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