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de28625ea8f7a58c6486249e978b88b9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484$ | There are 15 rows with 15 black tiles, and 16 rows with 16 black tiles, so the answer is $15^2+16^2 =225+256= \boxed{481}$ | B | 481 |
de28625ea8f7a58c6486249e978b88b9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484$ | We build the $31 \times 31$ checkerboard starting with a board of $30 \times 30$ that is exactly half black. There are $15 \cdot 30$ black tiles in this region.
Add to this $30 \times 30$ checkerboard a $1 \times 30$ strip on the bottom that has $15$ black tiles.
Add to this $31 \times 30$ checkerboard a $31 \times 1$ strip on the right that has $15 + 1$ black tiles.
In total, there are $15 \cdot 30 + 15 + 15 + 1 = 481$ tiles, giving an answer of $\boxed{481}$ | B | 481 |
de28625ea8f7a58c6486249e978b88b9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_14 | Chubby makes nonstandard checkerboards that have $31$ squares on each side. The checkerboards have a black square in every corner and alternate red and black squares along every row and column. How many black squares are there on such a checkerboard?
$\textbf{(A)}\ 480 \qquad\textbf{(B)}\ 481 \qquad\textbf{(C)}\ 482 \qquad\textbf{(D)}\ 483 \qquad\textbf{(E)}\ 484$ | Drawing smaller scale sketches, we notice that the odd columns of an $n \times n$ (where $n$ is odd) board have $\left \lceil \frac{n}{2} \right \rceil$ black tiles, while the even columns have $\left \lfloor \frac{n}{2} \right \rfloor$ black tiles. In our case, we have a $31 \times 31$ board. We have $16$ odd columns, and $15$ even columns, so the number of black tiles in total is $16^2 + 15^2 = \boxed{481}$ | B | 481 |
98b548ae2748f73c2ed64146525d14e1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | $AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$ . The hypotenuse of right triangle BED is $\sqrt{1^2+2^2}=\sqrt{5}$
\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \Rightarrow AC=2BC\]
\[\frac{AC}{AB}=\frac{BD}{BE} \Rightarrow \frac{AC}{1}=\frac{2}{\sqrt{5}} \Rightarrow AC=\frac{2}{\sqrt{5}}\]
Since $AC=2BC$ $BC=\frac{1}{\sqrt{5}}$ $\triangle ABC$ is a right triangle so the area is just $\frac12 \cdot AC \cdot BC = \frac12 \cdot \frac{2}{\sqrt{5}} \cdot \frac{1}{\sqrt{5}} = \boxed{15}$ | B | 15 |
98b548ae2748f73c2ed64146525d14e1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Let $\text{E}$ be the origin. Then, $\text{D}=(1, 0)$ $\text{A}=(0, 2)$ $\text{B}=(1, 2)$ $\text{F}=(2, 1)$
${EB}$ can be represented by the line $y=2x$ Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$
Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$ .
Thus, $x=\frac{4}{5}$ .
Plugging this into the first equation gives us $y=\frac{8}{5}$
Since $\text{C} (0.8, 1.6)$ $G$ is $(0.8, 2)$
${AB}=1$ and ${CG}=0.4$
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot 0.4=0.2=\frac{1}{5}$ . The answer is $\boxed{15}$ | B | 15 |
98b548ae2748f73c2ed64146525d14e1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Triangle $EAB$ is similar to triangle $EHI$ ; line $HI = 1/2$
Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$
Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\frac{1}{2} \cdot 1 \cdot \frac{2}{5}=\frac{1}{5}$ .
The answer is $\boxed{15}$ | B | 15 |
98b548ae2748f73c2ed64146525d14e1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_15 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | Let $L$ be the point where the diagonal and the end of the unit square meet, on the right side of the diagram. Let $K$ be the top right corner of the top right unit square, where segment $ABL$ is 2 units in length.
Because of the Pythagorean Theorem, since $AC = 2$ and $LK$ = 1, the diagonal of triangle $ALK$ is $\sqrt{5}$
Triangle $ALK$ is clearly a similar triangle to triangle $ABC$ . Segment $AB$ is the hypotenuse of triangle $ABC$ . So, we can write down:
\[AK/AB = LK/BC\] , which is equal to: \[\frac{\sqrt{5}}{1} = \frac{1}{BC}\] Solving this equation yields:
\[BC = \frac{1}{\sqrt{5}}\]
By Pythagorean theorem, we can now find segment $AC$ \[(\frac{1}{\sqrt{5}})^2 + AC^2 = 1\] Solving this yields:
\[AC^2 = \frac{4}{5}\] , so $AC = \frac{2}{\sqrt{5}}$
So then we can use \[A = \frac{1}{2} * b * a.\] So \[A = \frac{1}{2} * \frac{1}{\sqrt{5}} * \frac{2}{\sqrt{5}}\]
\[= \boxed{15}\] | B | 15 |
db5bc8395d540af9207c4d65b115abf8 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_16 | Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$ | First consider the first two runners. The faster runner will lap the slower runner exactly once, or run 500 meters farther. Let $x$ be the time these runners run in seconds.
\[4.8x-4.4x=500 \Rightarrow x=1250\]
Because $4.4(1250)=5500$ is a multiple of 500, it turns out they just meet back at the start line.
Now we must find a time that is a multiple of $1250$ and results in the 5.0 m/s runner to end up on the start line. Every $1250$ seconds, that fastest runner goes $5.0(1250)=6250$ meters. In $2(1250)=2500$ seconds, he goes $5.0(2500)=12500$ meters. Therefore the runners run $\boxed{2,500}$ seconds. | C | 2,500 |
db5bc8395d540af9207c4d65b115abf8 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_16 | Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$ | Working backwards from the answers starting with the smallest answer, if they had run $1000$ seconds, they would have run $4400, 4800, 5000$ meters, respectively. The first two runners have a difference of $400$ meters, which is not a multiple of $500$ (one lap), so they are not in the same place.
If they had run $1250$ seconds, the runners would have run $5500, 6000, 6250$ meters, respectively. The last two runners have a difference of $250$ meters, which is not a multiple of $500$
If they had run $2500$ seconds, the runners would have run $11000, 12000, 12500$ meters, respectively. The distance separating each pair of runners is a multiple of $500$ , so the answer is $\boxed{2,500}$ seconds. | C | 2,500 |
db5bc8395d540af9207c4d65b115abf8 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_16 | Three runners start running simultaneously from the same point on a 500-meter circular track. They each run clockwise around the course maintaining constant speeds of 4.4, 4.8, and 5.0 meters per second. The runners stop once they are all together again somewhere on the circular course. How many seconds do the runners run?
$\textbf{(A)}\ 1,000\qquad\textbf{(B)}\ 1,250\qquad\textbf{(C)}\ 2,500\qquad\textbf{(D)}\ 5,000\qquad\textbf{(E)}\ 10,000$ | Let $t$ be the time run in seconds, then the difference in meters run between the three runners is $0.2t, 0.4t, 0.6t$ . For them to be at the same location all of them need to be multiples of 500. It is now easy to see that $0.2t=500, 0.4t=1000, 0.6t=1500$ , so $t=\boxed{2,500}$ | C | 2,500 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Since $a$ and $b$ are relatively prime, $a^3-b^3$ and $(a-b)^3$ are both integers as well. Then, for the given fraction to simplify to $\frac{73}{3}$ , the denominator $(a-b)^3$ must be a multiple of $3.$ Thus, $a-b$ is a multiple of $3$ . Looking at the answer choices, the only multiple of $3$ is $\boxed{3}$ | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Using difference of cubes in the numerator and cancelling out one $(a-b)$ in the numerator and denominator gives $\frac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \frac{73}{3}$
Set $x = a^2 + b^2$ , and $y = ab$ . Then $\frac{x + y}{x - 2y} = \frac{73}{3}$ . Cross multiplying gives $3x + 3y = 73x - 146y$ , and simplifying gives $\frac{x}{y} = \frac{149}{70}$ . Since $149$ and $70$ are relatively prime, we let $x = 149$ and $y = 70$ , giving $a^2 + b^2 = 149$ and $ab = 70$ . Since $a>b>0$ , the only solution is $(a,b) = (10, 7)$ , which can be seen upon squaring and summing the various factor pairs of $70$
Thus, $a - b = \boxed{3}$ | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | The first step is the same as above which gives $\frac{a^2+ab+b^2}{a^2-2ab+b^2}=\frac{73}{3}$
Then we can subtract $3ab$ and then add $3ab$ to get $\frac{a^2-2ab+b^2+3ab}{a^2-2ab+b^2}=\frac{73}{3}$ , which gives $1+\frac{3ab}{(a-b)^2}=\frac{73}{3}$ $\frac{3ab}{(a-b)^2}=\frac{70}{3}$ .
Cross multiply $9ab=70(a-b)^2$ . Since $a>b$ , take the square root. $a-b=3\sqrt{\frac{ab}{70}}$ .
Since $a$ and $b$ are integers and relatively prime, $\sqrt{\frac{ab}{70}}$ is an integer. $ab$ is a multiple of $70$ , so $a-b$ is a multiple of $3$ .
Therefore $a=10$ and $b=7$ is a solution.
So $a-b=\boxed{3}$ | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Slightly expanding, we have that $\frac{(a-b)(a^2+ab+b^2)}{(a-b)(a-b)(a-b)}=\frac{73}{3}$
Canceling the $(a-b)$ , cross multiplying, and simplifying, we obtain that
$0=70a^2-149ab+70b^2$ .
Dividing everything by $b^2$ , we get that
$0=70(\frac{a}{b})^2-149(\frac{a}{b})+70$
Applying the quadratic formula....and following the restriction that $a>b>0$ ....
$\frac{a}{b}=\frac{10}{7}$
Hence, $7a=10b$
Since they are relatively prime, $a=10$ $b=7$
$10-7=\boxed{3}$ | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Note that the denominator, when simplified, gets $3.$ We now have to test the answer choices. If one has a good eye or by simply testing the answer choices the answer will be clearly $\boxed{3}$ ~mathboy282 | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Let us rewrite the expression as $\frac{(a-b)^2 + 3ab}{(a-b)^2}$ . Now letting $x = a - b$ , we simplify the expression to $\frac{70x^2 + 3ab}{x^2} = \frac{73}{3}$ . Cross multiplying and doing a bit of simplification, we obtain that $ab = \frac{70x^2}{9}$ . Since $a$ and $b$ are both integers, we know that $\frac{70x^2}{9}$ has to be an integer. Experimenting with values of $x$ , we get that $x = 3$ which means $ab = 70$ . We could prime factor from here to figure out possible values of $a$ and $b$ , but it is quite obvious that $a = 10$ and $b=7$ , so our desired answer is $\boxed{3}$ ~triggod | C | 3 |
b2900ec17f53a8fedda54ffa9545ff59 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_17 | Let $a$ and $b$ be relatively prime positive integers with $a>b>0$ and $\dfrac{a^3-b^3}{(a-b)^3} = \dfrac{73}{3}$ . What is $a-b$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad \textbf{(E)}\ 5$ | Since the two numbers are integers, and both $(a-b)^3$ and $a^3-b^3$ would yield integers, for the denominator to have a factor of 3, $(a-b)$ must have a factor of 3. Only choice $\boxed{3}$ has a factor of 3. ~hh99754539 | C | 3 |
3a2dae084a67dc79424402e051e84849 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_19 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$ | Let Paula work at a rate of $p$ , the two helpers work at a combined rate of $h$ , and the time it takes to eat lunch be $L$ , where $p$ and $h$ are in house/hours and L is in hours. Then the labor on Monday, Tuesday, and Wednesday can be represented by the three following equations:
\[(8-L)(p+h)=50\]
\[(6.2-L)h=24\]
\[(11.2-L)p=26\]
With three equations and three variables, we need to find the value of $L$ .
Adding the second and third equations together gives us $6.2h+11.2p-L(p+h)=50$ . Subtracting the first equation from this new one gives us $-1.8h+3.2p=0$ , so we get $h=\frac{16}{9}p$ .
Plugging into the second equation:
\[(6.2-L)\frac{16}{9}p=24\] \[(6.2-L)p=\frac{27}{2}\]
We can then subtract this from the third equation:
\[5p=26-\frac{27}{2}\] \[p=\frac{5}{2}\] Plugging $p$ into our third equation gives: \[L=\frac{4}{5}\]
Converting $L$ from hours to minutes gives us $L=48$ minutes, which is $\boxed{48}$ | D | 48 |
3a2dae084a67dc79424402e051e84849 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_19 | Paula the painter and her two helpers each paint at constant, but different, rates. They always start at 8:00 AM, and all three always take the same amount of time to eat lunch. On Monday the three of them painted 50% of a house, quitting at 4:00 PM. On Tuesday, when Paula wasn't there, the two helpers painted only 24% of the house and quit at 2:12 PM. On Wednesday Paula worked by herself and finished the house by working until 7:12 P.M. How long, in minutes, was each day's lunch break?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 60$ | Because Paula worked from \[8:00 \text{A.M.}\] to \[7:12 \text{P.M.}\] , she worked for 11 hours and 12 minutes = 672 minutes. Since there is $100-50-24=26$ % of the house left, we get the equation $26a=672$ . Because $672$ is $22$ mod $26$ , looking at our answer choices, the only answer that is $22$ $\text{mod}$ $26$ is $48$ . So the answer is $\boxed{48}$ | D | 48 |
d152039bc933545be4d6af324679dcb4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$
Thus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$
Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$ . Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.
Let $x = \sqrt{4m^2 - 847}$ ; note that this means $n = \frac{-1 + x}{2}$ . It can be rewritten as $x^2 = 4m^2 - 847$ , so $4m^2 - x^2 = 847$ . Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\cdot11^2$
Our goal is to find possible values for $x$ , then use the equation above to find $n$ . The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 121\cdot 7,$ and $77\cdot 11$ . The differences between these factors are $846$ $114$ , and $66$ - those are all possible values for $2x$ . Thus the possibilities for $x$ are $423$ $57$ , and $33$
Now plug in these values into the equation $n = \frac{-1 + x}{2}$ , so $n$ can equal $211$ $28$ , or $16$ , hence the answer is $\boxed{255}$ | A | 255 |
d152039bc933545be4d6af324679dcb4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$ . Clearly $m > n$ , so let $m = n + a$ , where $a$ is some positive integer. We have:
$(n+a)^2 = n(n+1) + 212$ .
Expanding, grouping like terms and factoring, we get: $n = \frac{(212 - a^2)}{(2a - 1)}$
We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ $14^2 = 196 < 212 < 15^2 = 225$ ). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$ . These gives $n$ values of $211, 28,$ and $16$ , respectively. $211 + 28 + 16 = 255$ . Hence, the answer is $\boxed{255}$ | A | 255 |
d152039bc933545be4d6af324679dcb4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_22 | The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$ | Using the closed forms for the sums, we get $m^2=n(n+1)+212$ , or $m^2=n^2+n+212$ . We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$ . Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$ . Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7\cdot11\cdot11$ . The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$ $7$ and $121$ $11$ and $77$ . In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$ . The sum of all possible values of $n$ is $211+28+16=\boxed{255}$ | A | 255 |
7cd8d0dcde4f9d829a69f9692d3f8c93 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_23 | Adam, Benin, Chiang, Deshawn, Esther, and Fiona have internet accounts. Some, but not all, of them are internet friends with each other, and none of them has an internet friend outside this group. Each of them has the same number of internet friends. In how many different ways can this happen?
$\text{(A)}\ 60\qquad\text{(B)}\ 170\qquad\text{(C)}\ 290\qquad\text{(D)}\ 320\qquad\text{(E)}\ 660$ | Note that if $n$ is the number of friends each person has, then $n$ can be any integer from $1$ to $4$ , inclusive.
One person can have at most 4 friends since they cannot be all friends (stated in the problem).
Also note that the cases of $n=1$ and $n=4$ are the same, since a map showing a solution for $n=1$ can correspond one-to-one with a map of a solution for $n=4$ by simply making every pair of friends non-friends and vice versa. The same can be said of configurations with $n=2$ when compared to configurations of $n=3$ . Thus, we have two cases to examine, $n=1$ and $n=2$ , and we count each of these combinations twice.
(Note: If you aren’t familiar with one-to-one correspondences, think of it like this: the number of ways to choose 4 friends is equal to number of ways to exclude one friend from your friend group. Hence, since the number of ways to choose 1 friend is the same thing as choosing 1 to not be friends with, $n=1$ and $n=4$ have the same number of ways. Similarly, $n=2$ and $n=3$ have the same number of ways as well. ~peelybonehead)
For $n=1$ , if everyone has exactly one friend, that means there must be $3$ pairs of friends, with no other interconnections. The first person has $5$ choices for a friend. There are $4$ people left. The next person has $3$ choices for a friend. There are two people left, and these remaining two must be friends. Thus, there are $15$ configurations with $n=1$
For $n=2$ , there are two possibilities. The group of $6$ can be split into two groups of $3$ , with each group creating a friendship triangle. The first person has $\binom{5}{2} = 10$ ways to pick two friends from the other five, while the other three are forced together. Thus, there are $10$ triangular configurations.
However, the group can also form a friendship hexagon, with each person sitting on a vertex, and each side representing the two friends that person has. The first person may be seated anywhere on the hexagon without loss of generality . This person has $\binom{5}{2} = 10$ choices for the two friends on the adjoining vertices. Each of the three remaining people can be seated "across" from one of the original three people, forming a different configuration. Thus, there are $10 \cdot 3! = 60$ hexagonal configurations, and in total $70$ configurations for $n=2$
As stated before, $n=3$ has $70$ configurations, and $n=4$ has $15$ configurations. This gives a total of $(70 + 15)\cdot 2 = 170$ configurations, which is option $\boxed{170}$ | B | 170 |
f4aac60acb95fc0044277fe50e45d791 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_24 | Let $a$ $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $a^2-b^2-c^2+ab=2011$ and $a^2+3b^2+3c^2-3ab-2ac-2bc=-1997$
What is $a$
$\textbf{(A)}\ 249\qquad\textbf{(B)}\ 250\qquad\textbf{(C)}\ 251\qquad\textbf{(D)}\ 252\qquad\textbf{(E)}\ 253$ | Add the two equations.
$2a^2 + 2b^2 + 2c^2 - 2ab - 2ac - 2bc = 14$
Now, this can be rearranged and factored.
$(a^2 - 2ab + b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2) = 14$
$(a - b)^2 + (a - c)^2 + (b - c)^2 = 14$
$a$ $b$ , and $c$ are all integers, so the three terms on the left side of the equation must all be perfect squares.
We see that the only is possibility is $14 = 9 + 4 + 1$
$(a-c)^2 = 9 \Rightarrow a-c = 3$ , since $a-c$ is the biggest difference. It is impossible to determine by inspection whether $a-b = 1$ or $2$ , or whether $b-c = 1$ or $2$
We want to solve for $a$ , so take the two cases and solve them each for an expression in terms of $a$ . Our two cases are $(a, b, c) = (a, a-1, a-3)$ or $(a, a-2, a-3)$ . Plug these values into one of the original equations to see if we can get an integer for $a$
$a^2 - (a-1)^2 - (a-3)^2 + a(a-1) = 2011$ , after some algebra, simplifies to $7a = 2021$ $2021$ is not divisible by $7$ , so $a$ is not an integer.
The other case gives $a^2 - (a-2)^2 - (a-3)^2 + a(a-2) = 2011$ , which simplifies to $8a = 2024$ . Thus, $a = 253$ and the answer is $\boxed{253}$ | E | 253 |
f573f6f983f5af0c41fb6e341b922e13 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Since $x,y,z$ are all reals located in $[0, n]$ , the number of choices for each one is continuous so we use geometric probability.
WLOG( Without loss of generality ), assume that $n\geq x \geq y \geq z \geq 0$ . Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes uniformly in this region. If this is not easy to understand, read Solution II.
The altitude of the tetrahedron is $n$ and the base is an isosceles right triangle with a leg length $n$ . The volume is $V_1=\dfrac{n^3}{6}$ , as shown in the first figure in red.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2,-1,2/3); // three - currentprojection, orthographic draw((1,1,0)--(0,1,0)--(0,0,0),dashed); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1),dashed); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); draw((0,0,0)--(1,0,0)--(1,1,0)--(1,1,1), red); draw((1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((1,1,1)--(1,0,0), red); [/asy]
Now we will find the region with points satisfying $|x-y|\geq1$ $|y-z|\geq1$ $|z-x|\geq1$
Since $n\geq x \geq y \geq z \geq 0$ , we have $x-y\geq1$ $y-z\geq1$
The region of points $(x,y,z)$ satisfying the condition is shown in the second figure in black. It is a tetrahedron, too.
[asy] import three; unitsize(10cm); size(150); currentprojection=orthographic(1/2, -1, 2/3); // three - currentprojection, orthographic draw((1, 1, 0)--(0, 1, 0)--(0, 0, 0), dashed+green); draw((0, 0, 0)--(0, 0, 1), green); draw((0, 1, 0)--(0, 1, 1), dashed+green); draw((1, 0, 0)--(1, 0, 1), green); draw((0, 0, 1)--(1, 0, 1)--(1, 1, 1)--(0, 1, 1)--cycle, green); draw((1,0,0)--(1,1,0)--(0,0,0)--(1,1,1), dashed+red); draw((0,0,0)--(1,0,0)--(1,1,1), red); draw((1,1,1)--(1,1,0)--(1,0.9,0), red); draw((1, 0.1, 0)--(1, 0.9, 0)--(1, 0.9, 0.8)--cycle); draw((0.2, 0.1, 0)--(1, 0.9, 0.8),dashed); draw((1, 0.1, 0)--(0.2, 0.1, 0)--(1, 0.9, 0),dashed); [/asy]
The volume of this region is $V_2=\dfrac{(n-2)^3}{6}$
So the probability is $p=\dfrac{V_2}{V_1}=\dfrac{(n-2)^3}{n^3}$
Substituting $n$ with the values in the choices, we find that when $n=10$ $p=\frac{512}{1000}>\frac{1}{2}$ , when $n=9$ $p=\frac{343}{729}<\frac{1}{2}$ . So $n\geq 10$
So the answer is $\boxed{10}$ | D | 10 |
f573f6f983f5af0c41fb6e341b922e13 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Because $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ , which means that $x$ $y$ , and $z$ distributes uniformly and independently in the interval $[0,n]$ . So the point $(x, y, z)$ distributes uniformly in the cubic $0\leqslant x, y, z \leqslant n$ , as shown in the figure below. The volume of this cubic is $V_0=n^3$
Cubic.png
As we want to find the probablity of the incident $A=\big\{ |x-y| \geq 1, |y-z| \geq1, |z-x| \geq 1 \big\}$ ,
we should find the volume of the region of points such that $|x-y|\geq 1$ $|y-z|\geq 1$ $|z-x|\geq 1$ and $0\leq x, y, z \leq n$
Now we will find the region $\big\{ (x,y,z)\ | \ 0\leq x, y, z \leq n, |x-y|\geq 1, |y-z|\geq 1, |z-x|\geq 1 \big\}$
The region can be generated by cutting off 3 slices corresponding to $|x-y|< 1$ $|y-z|< 1$ , and $|z-x|< 1$ , respectively, from the cubic.
After cutting off a slice corresponding to $|x-y|< 1$ , we get two triangular prisms, as shown in the figure.
2.png
In order to observe the object clearly, we rotate the object by the $z$ axis, as shown.
3.png
We can draw the slice corresponding to $|y-z|< 1$ on the object.
4B.png
After cutting off the slice corresponding to $|y-z|< 1$ , we have 4 pieces left.
5.png
After cutting off the slice corresponding to $|z-x|< 1$ , we have 6 congruent triangular prisms.
6B.png
Here we draw all the pictures in colors in order to explain the solution clearly. That does not mean that the students should do it in the examination. They can draw a figure with lines only, as shown below.
7.png
Every triangular pyramid has an altitude $n-2$ and a base of isoceless right triangle with leg length $n-2$ , so the volume is $(n-2)^3/6$ .
Then the volume of the region $\big\{ (x,y,z)\ | \ 0\leqslant x, y, z \leqslant n, |x-y|\geqslant 1, |y-z|\geqslant 1, |z-x|\geqslant 1 \big\}$ is $V_A=6\times(n-2)^3/6$ $(n-2)^3$
So the probability of the incident $A$ is $P(A)=\dfrac{V_A}{V_0}$ $\dfrac{(n-2)^3}{n^3}$
Then we can get the answer the same way as Solution I.
The answer is $\boxed{10}$ | D | 10 |
f573f6f983f5af0c41fb6e341b922e13 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | If $x$ $y$ , and $z$ are separated by at least one, then by subtracting the minimum space between the three variables (which is $2$ ), $x$ $y$ , and $z$ can be chosen randomly in the interval $[0,n-2]$ . The probability is hence $\dfrac{(n-2)^3}{n^3} > \dfrac{1}{2}.$ $\boxed{10}$ is the minimum value in the answer choices that satisfies this inequality. | D | 10 |
f573f6f983f5af0c41fb6e341b922e13 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10A_Problems/Problem_25 | Real numbers $x$ $y$ , and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$ . The probability that no two of $x$ $y$ , and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$ . What is the smallest possible value of $n$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Imagine Points $x$ $y$ $z$ as the "starting points" of three "blocks" of real numbers that have length $1$ . We are just trying to find the probability that those three "blocks" do not overlap. To do this we can set each unit of $1$ into $\mu$ equal little increments, and take the limit of the probability as $\mu$ approaches $\infty$ . (This is because there are indefinitely many real numbers in a given interval.) We see that the total number of arrangements for the three blocks of $1$ , without considering the rule that no two blocks shall overlap is $(n\mu + 1)^3$ . We see that the number of ways to arrange the three blocks such that no three of them are overlapping is simply $\dbinom{(n-2)\mu + 3}{3}$ . Taking the limit as $\mu$ approaches infinity, we obtain our closed form to be $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3 }.$ Dividing leading coefficients, we obtain $\lim_{\mu \to \infty} \frac{\dbinom{(n-2)\mu + 3}{3}}{ (n\mu + 1)^3 } = \frac{\mu(n-2)^3}{\mu n^3} = \frac{(n-2)^3}{n^3}$ . Solving the inequality $\frac{(n-2)^3}{n^3} > \frac{1}{2}$ , we get the least value of $n$ as $n=\boxed{10}$ | null | 10 |
bb85dfa013de473aa1c6831c82748d6c | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_1 | Each third-grade classroom at Pearl Creek Elementary has $18$ students and $2$ pet rabbits. How many more students than rabbits are there in all $4$ of the third-grade classrooms?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 56\qquad\textbf{(C)}\ 64\qquad\textbf{(D)}\ 72\qquad\textbf{(E)}\ 80$ | In each class, there are $18-2=16$ more students than rabbits. So for all classrooms, the difference between students and rabbits is $16 \times 4 = \boxed{64}$ | C | 64 |
cebadbf07b8f332e46be8b9bddcde4f6 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_2 | A circle of radius 5 is inscribed in a rectangle as shown. The ratio of the length of the rectangle to its width is 2:1. What is the area of the rectangle?
[asy] draw((0,0)--(0,10)--(20,10)--(20,0)--cycle); draw(circle((10,5),5));[/asy]
$\textbf{(A)}\ 50\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 125\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 200$ | Note that the diameter of the circle is equal to the shorter side of the rectangle. Since the radius is $5$ , the diameter is $2\cdot 5 = 10$ .
Since the sides of the rectangle are in a $2:1$ ratio, the longer side has length $2\cdot 10 = 20$ .
Therefore the area is $20\cdot 10 = 200$ or $\boxed{200}$ | E | 200 |
6e815c0d8d952e2067368f796a63c5b8 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_3 | The point in the $xy$ -plane with coordinates $(1000, 2012)$ is reflected across the line $y=2000$ . What are the coordinates of the reflected point?
$\textbf{(A)}\ (998,2012)\qquad\textbf{(B)}\ (1000,1988)\qquad\textbf{(C)}\ (1000,2024)\qquad\textbf{(D)}\ (1000,4012)\qquad\textbf{(E)}\ (1012,2012)$ | The line $y = 2000$ is a horizontal line located $12$ units beneath the point $(1000, 2012)$ . When a point is reflected about a horizontal line, only the $y$ - coordinate will change. The $x$ - coordinate remains the same. Since the $y$ -coordinate of the point is $12$ units above the line of reflection, the new $y$ - coordinate will be $2000 - 12 = 1988$ . Thus, the coordinates of the reflected point are $(1000, 1988)$ $\boxed{1000,1988}$ | B | 1000,1988 |
8189b59d6e1787de4717d70d7eddf698 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_4 | When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | In total, there were $3+4=7$ marbles left from both Ringo and Paul.We know that $7 \equiv 1 \pmod{6}$ . This means that there would be $1$ marble leftover, or $\boxed{1}$ | A | 1 |
8189b59d6e1787de4717d70d7eddf698 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_4 | When Ringo places his marbles into bags with 6 marbles per bag, he has 4 marbles left over. When Paul does the same with his marbles, he has 3 marbles left over. Ringo and Paul pool their marbles and place them into as many bags as possible, with 6 marbles per bag. How many marbles will be leftover?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Let $r$ be the number of marbles Ringo has and let $p$ be the number of marbles Paul has. we have the following equations: \[r \equiv 4 \mod{6}\] \[p \equiv 3 \mod{6}\] Adding these equations we get: \[p + r \equiv 7 \mod{6}\] We know that $7 \equiv 1 \mod{6}$ so therefore: \[p + r \equiv 7 \equiv 1 \mod{6} \implies p + r \equiv 1 \mod{6}\] Thus when Ringo and Paul pool their marbles, they will have $\boxed{1}$ marble left over. | A | 1 |
24d825ace4ec1b0806931ce7bda9a942 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_5 | Anna enjoys dinner at a restaurant in Washington, D.C., where the sales tax on meals is 10%. She leaves a 15% tip on the price of her meal before the sales tax is added, and the tax is calculated on the pre-tip amount. She spends a total of 27.50 dollars for dinner. What is the cost of her dinner without tax or tip in dollars?
$\textbf{(A)}\ 18\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$ | Let $x$ be the cost of her dinner.
$27.50=x+\frac{1}{10}*x+\frac{3}{20}*x$
$27+\frac{1}{2}=\frac{5}{4}*x$
$\frac{55}{2}=\frac{5}{4}x$
$\frac{55}{2}*\frac{4}{5}=x$
$x=22$
$\boxed{22}$ | D | 22 |
ae8df8b123c60bffe91ada4884128292 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_7 | For a science project, Sammy observed a chipmunk and a squirrel stashing acorns in holes. The chipmunk hid 3 acorns in each of the holes it dug. The squirrel hid 4 acorns in each of the holes it dug. They each hid the same number of acorns, although the squirrel needed 4 fewer holes. How many acorns did the chipmunk hide?
$\textbf{(A)}\ 30\qquad\textbf{(B)}\ 36\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 54$ | Let $x$ be the number of acorns that both animals had.
So by the info in the problem:
$\frac{x}{3}=\left( \frac{x}{4} \right)+4$
Subtracting $\frac{x}{4}$ from both sides leaves
$\frac{x}{12}=4$
$\boxed{48}$ | null | 48 |
5e414307582c83fcd64ccd3f84a8e683 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_8 | What is the sum of all integer solutions to $1<(x-2)^2<25$
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 25$ | $(x-2)^2$ = perfect square.
1 < perfect square < 25
Perfect square can equal: 4, 9, or 16
Solve for $x$
$(x-2)^2=4$
$x=4,0$
and
$(x-2)^2=9$
$x=5,-1$
and
$(x-2)^2=16$
$x=6,-2$
The sum of all integer solutions is
$4+5+6+0+(-1)+(-2)=\boxed{12}$ | B | 12 |
d07a0e22589150f44d7f6c2e48db0836 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_9 | Two integers have a sum of 26. When two more integers are added to the first two integers the sum is 41. Finally when two more integers are added to the sum of the previous four integers the sum is 57. What is the minimum number of odd integers among the 6 integers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | Out of the first two integers, it's possible for both to be even: for example, $10 + 16 = 26.$ But the next two integers, when added, increase the sum by $15,$ which is odd, so one of them must be odd and the other must be even: for example, $3 + 12 = 15.$ Finally, the next two integers increase the sum by $16,$ which is even, so we can have both be even: for example, $2 + 14 = 16.$ Therefore, $\boxed{1}$ is the minimum number of integers that must be odd. | A | 1 |
9d9286c35c4be9dadb59db8763899df1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_10 | How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Cross-multiplying gives $MN=36.$ We write $36$ as a product of two positive integers: \begin{align*} 36 &= 1\cdot36 \\ &= 2\cdot18 \\ &= 3\cdot12 \\ &= 4\cdot9 \\ &= 6\cdot6. \end{align*} The products $1\cdot36, 2\cdot18, 3\cdot12,$ and $4\cdot9$ each produce $2$ ordered pairs $(M,N),$ as we can switch the order of the factors. The product $6\cdot6$ produces $1$ ordered pair $(M,N).$ Together, we have $4\cdot2+1=\boxed{9}$ ordered pairs $(M,N).$ | D | 9 |
9d9286c35c4be9dadb59db8763899df1 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_10 | How many ordered pairs of positive integers $(M,N)$ satisfy the equation $\frac{M}{6}=\frac{6}{N}?$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Cross-multiplying gives $MN=36.$ From the prime factorization \[36=2^2\cdot3^2,\] we conclude that $36$ has $(2+1)(2+1)=9$ positive divisors. There are $9$ values of $M,$ and each value generates $1$ ordered pair $(M,N).$ So, there are $\boxed{9}$ ordered pairs $(M,N)$ in total. | D | 9 |
82fc698ce020aea17300f4fcf3efe22e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_11 | A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$ | Desserts must be chosen for $7$ days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday.
There are $3$ choices for dessert on Saturday: pie, ice cream, or pudding, as there must be cake on Friday and the same dessert may not be served two days in a row. Likewise, there are $3$ choices for dessert on Thursday. Once dessert for Thursday is selected, there are $3$ choices for dessert on Wednesday, once Wednesday's dessert is selected there are $3$ choices for dessert on Tuesday, etc. Thus, there are $3$ choices for dessert for each of $6$ days, so the total number of possible dessert menus is $3^6$ , or $\boxed{729}$ | A | 729 |
82fc698ce020aea17300f4fcf3efe22e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_11 | A dessert chef prepares the dessert for every day of a week starting with Sunday. The dessert each day is either cake, pie, ice cream, or pudding. The same dessert may not be served two days in a row. There must be cake on Friday because of a birthday. How many different dessert menus for the week are possible?
$\textbf{(A)}\ 729\qquad\textbf{(B)}\ 972\qquad\textbf{(C)}\ 1024\qquad\textbf{(D)}\ 2187\qquad\textbf{(E)}\ 2304$ | There are $4 \cdot 3^6$ ways for the desserts to be chosen. By symmetry, any of the desserts that are chosen on Friday share $\frac{1}{4}$ of the total arrangements. Therefore our answer is $\frac{4\cdot3^6}{4} = 3^6 = \boxed{729}.$ | null | 729 |
2df4902a1fc7a44e0fadb199c49d9539 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$ | Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Using $d = v t$ , the first statement can be translated to the equation $d = 60c$ . The second statement can be translated to $d = 24(c+s)$ . Since the same distance is being covered in each scenario, we can set the two equations equal and solve for $s$ . We find that $s = \dfrac{3c}{2}$ . The problem asks for the time it takes her to ride down the escalator when she just stands on it. Since $t = \dfrac{d}{s}$ and $d = 60c$ , we have $t = \dfrac{60c}{\dfrac{3c}{2}} = 40$ seconds. Answer choice $\boxed{40}$ is correct. | B | 40 |
2df4902a1fc7a44e0fadb199c49d9539 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$ | Let $s$ be the speed of the escalator and $c$ be the speed of Clea. Then without loss of generality , assume that the length of the escalator be 1. Then $c=\dfrac{1}{60}$ and $c+s=\dfrac{1}{24}$ , so $s=\dfrac{1}{24}-\dfrac{1}{60}=\dfrac{1}{40}$ . Thus the time it takes for Clea to ride down the operating escalator when she just stands on it is $\dfrac{1}{\dfrac{1}{40}}=\boxed{40}$ | B | 40 |
2df4902a1fc7a44e0fadb199c49d9539 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_13 | It takes Clea 60 seconds to walk down an escalator when it is not operating, and only 24 seconds to walk down the escalator when it is operating. How many seconds does it take Clea to ride down the operating escalator when she just stands on it?
$\textbf{(A)}\ 36\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 48\qquad\textbf{(E)}\ 52$ | WLOG, let the length of the escalator be $120$ ft. So, Clea's rate is $120/60=2$ feet per second (fps $*$ ), and the escalator and Clea's rate is $120/24=5$ fps. So, the escalator's rate is $5-2=3$ fps. Therefore, $120/3=\boxed{40}$ | B | 40 |
a6c6d3ea1b7995b93fbc580eb8c7ebdd | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_15 | In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$ | The total number of games (and wins) in the tournament is $\frac{6 \times 5}{2}= 15$ . A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:
The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of its matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins.
Thus, the answer is $\boxed{5}$ | D | 5 |
e4c809e466eb815b5b88b5824cb940e5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$ . Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | The last number that Bernardo says has to be between 950 and 999. Note that $1\rightarrow 2\rightarrow 52\rightarrow 104\rightarrow 154\rightarrow 308\rightarrow 358\rightarrow 716\rightarrow 766$ contains 4 doubling actions. Thus, we have $x \rightarrow 2x \rightarrow 2x+50 \rightarrow 4x+100 \rightarrow 4x+150 \rightarrow 8x+300 \rightarrow 8x+350 \rightarrow 16x+700$
Thus, $950<16x+700<1000$ . Then, $16x>250 \implies x \geq 16$
Because we are looking for the smallest integer $x$ $x=16$ . Our answer is $1+6=\boxed{7}$ , which is A. | null | 7 |
e4c809e466eb815b5b88b5824cb940e5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$ . Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | Work backwards. The last number Bernardo produces must be in the range $[950,999]$ . That means that before this, Silvia must produce a number in the range $[475,499]$ . Before this, Bernardo must produce a number in the range $[425,449]$ . Before this, Silvia must produce a number in the range $[213,224]$ . Before this, Bernardo must produce a number in the range $[163,174]$ . Before this, Silvia must produce a number in the range $[82,87]$ . Before this, Bernardo must produce a number in the range $[32,37]$ . Before this, Silvia must produce a number in the range $[16,18]$ . Silvia could not have added 50 to any number before this to obtain a number in the range $[16,18]$ , hence the minimum $N$ is 16 with the sum of digits being $\boxed{7}$ | A | 7 |
e4c809e466eb815b5b88b5824cb940e5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_20 | Bernardo and Silvia play the following game. An integer between $0$ and $999$ inclusive is selected and given to Bernardo. Whenever Bernardo receives a number, he doubles it and passes the result to Silvia. Whenever Silvia receives a number, she adds $50$ to it and passes the result to Bernardo. The winner is the last person who produces a number less than $1000$ . Let $N$ be the smallest initial number that results in a win for Bernardo. What is the sum of the digits of $N$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 11$ | If our first number is $N,$ then the sequence of numbers will be \[2N,~2N+50,~4N+100,~4N+150,~8N+300,~8N+350,~16N+700,~16N+750\] Note that we cannot go any further because doubling gives an extra $1500$ at the end, which is already greater than $1000.$ The smallest $N$ will be given if $16N+750>1000>16N+700 \implies 15<N<19.$ Since the problem asks for the smallest possible value of $N,$ we get $16,$ and the sum of its digits is $1+6=\boxed{7}$ | A | 7 |
588ae01b718b7c333af3ed715692bc4e | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_21 | Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$ $a$ $a$ $a$ $2a$ , and $b$ . What is the ratio of $b$ to $a$
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$ | For any $4$ non-collinear points with the given requirement, notice that there must be a triangle with side lengths $a$ $a$ $2a$ , which is not possible as $a+a=2a$ . Thus at least $3$ of the $4$ points must be collinear.
If all $4$ points are collinear, then there would only be $3$ lines of length $a$ , which wouldn't work.
If exactly $3$ points are collinear, the only possibility that works is when a $30^{\circ}-90^{\circ}-60^{\circ}$ triangle is formed.
Thus $b=\sqrt{3}a$ , or $\frac{b}{a}=\boxed{3}$ | A | 3 |
eb6d596d52e16acd21af1873447a45e9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$ | If we have 1 as the first number, then the only possible list is $(1,2,3,4,5,6,7,8,9,10)$
If we have 2 as the first number, then we have 9 ways to choose where the $1$ goes, and the numbers ascend from the first number, $2$ , with the exception of the $1$ .
For example, $(2,3,1,4,5,6,7,8,9,10)$ , or $(2,3,4,1,5,6,7,8,9,10)$ . There are $\dbinom{9}{1}$ ways to do so.
If we use 3 as the first number, we need to choose 2 spaces to be 2 and 1, respectively. There are $\dbinom{9}{2}$ ways to do this.
In the same way, the total number of lists is: $\dbinom{9}{0} +\dbinom{9}{1} + \dbinom{9}{2} + \dbinom{9}{3} + \dbinom{9}{4}.....\dbinom{9}{9}$
By the binomial theorem , this is $2^{9}$ $512$ , or $\boxed{512}$ | B | 512 |
eb6d596d52e16acd21af1873447a45e9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$ | Arrange the spaces and put arrows pointing either up or down between them. Then for each arrangement of arrows there is one and only one list that corresponds to up. For example, all arrows pointing up is $(1,2,3,4,5...10)$ . There are 9 arrows, so the answer is $2^{9}$ $512$ $\boxed{512}$ | B | 512 |
eb6d596d52e16acd21af1873447a45e9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$ | Notice that the answer to the problem is solely based on the length of the lists, i.e. 10. We can replace 10 with smaller values, such as 2 and 3, and try to find a pattern. If we replace it with 2, we can easily see that there are two possible lists, $(1, 2)$ and $(2, 1)$ . If we replace it with 3, there are four lists, $(1, 2, 3), (2, 1, 3), (2, 3, 1),$ and $(3, 2, 1)$ . Since 2 and 4 are both powers of 2, it is likely that the number of lists is $2^{n-1}$ , where $n$ is the length of the lists. $2^{10-1}=512=\boxed{512}$ | B | 512 |
eb6d596d52e16acd21af1873447a45e9 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_22 | Let ( $a_1$ $a_2$ , ... $a_{10}$ ) be a list of the first 10 positive integers such that for each $2\le$ $i$ $\le10$ either $a_i + 1$ or $a_i-1$ or both appear somewhere before $a_i$ in the list. How many such lists are there?
$\textbf{(A)}\ \ 120\qquad\textbf{(B)}\ 512\qquad\textbf{(C)}\ \ 1024\qquad\textbf{(D)}\ 181,440\qquad\textbf{(E)}\ \ 362,880$ | If $a_1=10$ , the sequence must be $10, 9, 8,7,6,5,4,3,2,1$ . If $a_2=10$ , then $a_1=9$ , and the sequence is $9, 10, 8, 7, 6, 5,4,3,2,1$ . If $a_3=10$ , then the possible sequences are \[9,8,10,7,6,5,4,3,2,1 \text{ and}\] \[8,9,10,7,6,5,4,3,2,1.\] In general, for an $n$ -length sequence, if $a_i=n$ , then $a_1$ through $a_{i-1}$ can be filled in $f(i-1)$ ways with $n-i+1$ through $n-1$ , and $a_{i+1}$ through $a_{n}$ must be sorted in decreasing order with the remaining numbers ( $1$ through $n-i$ ), in one way. Thus $f(n) = \sum_{i=0}^{n-1} f(i)$ , where $f(0)=1$
We can see (or prove by induction) that $f(n)=2^{n-1} ~\forall~ n \ge 1$ . Hence, $f(10)=2^9=\boxed{512}$ | B | 512 |
3185d1ecbd11fdca0592859ba9d9a0e4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105$ | Let the ordered triple $(a,b,c)$ denote that $a$ songs are liked by Amy and Beth, $b$ songs by Beth and Jo, and $c$ songs by Jo and Amy. The only possible triples are $(1,1,1), (2,1,1), (1,2,1)(1,1,2)$
To show this, observe these are all valid conditions. Second, note that none of $a,b,c$ can be bigger than 3. Suppose otherwise, that $a = 3$ . Without loss of generality, say that Amy and Beth like songs 1, 2, and 3. Then because there is at least one song liked by each pair of girls, we require either $b$ or $c$ to be at least 1. In fact, we require either $b$ or $c$ to equal 1, otherwise there will be a song liked by all three. Suppose $b = 1$ . Then we must have $c=0$ since no song is liked by all three girls, a contradiction.
Case 1 : How many ways are there for $(a,b,c)$ to equal $(1,1,1)$ ? There are 4 choices for which song is liked by Amy and Beth, 3 choices for which song is liked by Beth and Jo, and 2 choices for which song is liked by Jo and Amy. The fourth song can be liked by only one of the girls, or none of the girls, for a total of 4 choices. So $(a,b,c)=(1,1,1)$ in $4\cdot3\cdot2\cdot4 = 96$ ways.
Case 2 : To find the number of ways for $(a,b,c) = (2,1,1)$ , observe there are $\binom{4}{2} = 6$ choices of songs for the first pair of girls. There remain 2 choices of songs for the next pair (who only like one song). The last song is given to the last pair of girls. But observe that we let any three pairs of the girls like two songs, so we multiply by 3. In this case there are $6\cdot2\cdot3=36$ ways for the girls to like the songs.
That gives a total of $96 + 36 = 132$ ways for the girls to like the songs, so the answer is $\boxed{132}$ | B | 132 |
3185d1ecbd11fdca0592859ba9d9a0e4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105$ | Let $AB, BJ$ , and $AJ$ denote a song that is liked by Amy and Beth (but not Jo), Beth and Jo (but not Amy), and Amy and Jo (but not Beth), respectively. Similarly, let $A, B, J,$ and $N$ denote a song that is liked by only Amy, only Beth, only Jo, and none of them, respectively. Since we know that there is at least $1\: AB, BJ$ , and $AJ$ , they must be $3$ songs out of the $4$ that Amy, Beth, and Jo listened to. The fourth song can be of any type $N, A, B, J, AB, BJ$ , and $AJ$ (there is no $ABJ$ because no song is liked by all three, as stated in the problem.) Therefore, we must find the number of ways to rearrange $AB, BJ, AJ$ , and a song from the set $\{N, A, B, J, AB, BJ, AJ\}$
Case 1: Fourth song = $N, A, B, J$
Note that in Case 1, all four of the choices for the fourth song are different from the first three songs.
Number of ways to rearrange = $(4!)$ rearrangements for each choice $*\: 4$ choices = $96$
Case 2: Fourth song = $AB, BJ, AJ$
Note that in Case $2$ , all three of the choices for the fourth song repeat somewhere in the first three songs.
Number of ways to rearrange = $(4!/2!)$ rearrangements for each choice $*\: 3$ choices = $36$
$96 + 36 = \boxed{132}$ | B | 132 |
3185d1ecbd11fdca0592859ba9d9a0e4 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_24 | Amy, Beth, and Jo listen to four different songs and discuss which ones they like. No song is liked by all three. Furthermore, for each of the three pairs of the girls, there is at least one song liked by those two girls but disliked by the third. In how many different ways is this possible?
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 132\qquad\textbf{(C)}\ 671\qquad\textbf{(D)}\ 846\qquad\textbf{(E)}\ 1105$ | There are $\binom{4}{3}$ ways to choose the three songs that are liked by the three pairs of girls.
There are $3!$ ways to determine how the three songs are liked, or which song is liked by which pair of girls.
In total, there are $\binom{4}{3}\cdot3!$ possibilities for the first $3$ songs.
There are $3$ cases for the 4th song, call it song D.
Case $1$ : D is disliked by all $3$ girls $\implies$ there is only $1$ possibility.
Case $2$ : D is liked by exactly $1$ girl $\implies$ there are $3$ possibility.
Case $3$ : D is liked by exactly $2$ girls $\implies$ there are $3$ pairs of girls to choose from. However, there's overlap when the other song liked by the same pair of girl is counted as the 4th song at some point, in which case D would be counted as one of the first $3$ songs liked by the same girls.
Counting the overlaps, there are $3$ ways to choose the pair with overlaps and $4\cdot3=12$ ways to choose what the other $2$ pairs like independently. In total, there are $3\cdot12=36$ overlapped possibilities.
Finally, there are $\binom{4}{3}\cdot3!\cdot(3+1+3)-36=132$ ways for the songs to be likely by the girls. $\boxed{132}$ | B | 132 |
eb2d31ab9b106d5f969496693daeb3f5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
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$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | [asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); 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There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \cdot (4+4+8+8) = 120$ ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$ $\boxed{2400}$ | E | 2400 |
eb2d31ab9b106d5f969496693daeb3f5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | [asy] size(6cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)); draw((4.0, 0.0)--(6.0,0.0)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(3.0,-1.7320508075688772), red); draw((7.0,1.7320508075688772)--(6.0,-0.0)--(7.0,-1.7320508075688772), blue); dot((0,0)); label("$A$",(0,0),WNW); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); [/asy]
For every blue arrow, there are $2\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is $1$ path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by $5$ $\boxed{2400}$ is the only answer that is. | E | 2400 |
eb2d31ab9b106d5f969496693daeb3f5 | https://artofproblemsolving.com/wiki/index.php/2012_AMC_10B_Problems/Problem_25 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy] size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | We use casework.
The main thing to notice is that, if the bug does not go backwards, then every vertical set of arrows can be used, as the bug could travel straight downwards and then use any arrow to continue right.
Note: The motivation is quite weird so follow my numbers as they start from the left (point A) and go right (point B).
Case 1: Bug does not go backwards.
The number of cases for this is just each vertical set of arrows multiplied to each other (if you don't understand where I'm coming from, try to understand where these numbers come from!)
And so we have $2*2*4*4*4*2*2 = 2^{10}$ cases.
Case 2: The bug goes backwards once, either at the first arrow or third arrow.
Here, we have to count the fact that there is a horizontal midline that the bug could not cross, or otherwise it would be stepping on the same edge twice.
Back on first arrow: $2*1*2*4*4*2*2 = 2^{8}$ cases.
Back on third arrow: $2*2*4*4*4*1 = 2^{8}$ cases.
Case 3: The bug goes backwards once, at the second arrow.
Same thing as above, except since there are 4 arrows in the vertical set (plus one for the backwards arrow), then the calculations are a bit different.
We have $2*2*4*1*2*4*2*2 = 2^{9}$ cases.
Notice that the first and third back arrows decrease the number of cases by a factor of $2^2$ and the second back arrow decreases the number of cases by $2^1$ . Hence,
1st + 2nd = $2^7$
2nd + 3rd = $2^7$
1st + 3rd = $2^6$
1st + 2nd + 3rd = $2^5$
And so the number of cases in total is $2^{10} + 2^9 + 2^8 + 2^8 + 2^7 + 2^7 + 2^6 + 2^5 \Rightarrow 1024 + 512 + 256 + 256 + 128 + 128 + 64 + 32 = \boxed{2400}$ | null | 2400 |
b2df926a79222e35d583dac63a2c4a53 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_1 | A cell phone plan costs $20$ dollars each month, plus $5$ cents per text message sent, plus $10$ cents for each minute used over $30$ hours. In January Michelle sent $100$ text messages and talked for $30.5$ hours. How much did she have to pay?
$\textbf{(A)}\ 24.00 \qquad \textbf{(B)}\ 24.50 \qquad \textbf{(C)}\ 25.50 \qquad \textbf{(D)}\ 28.00 \qquad \textbf{(E)}\ 30.00$ | The base price of Michelle's cell phone plan is $20$ dollars.
If she sent $100$ text messages and it costs $5$ cents per text, then she must have spent $500$ cents for texting, or $5$ dollars. She talked for $30.5$ hours, but $30.5-30$ will give us the amount of time exceeded past 30 hours. $30.5-30=.5$ hours $=30$ minutes.
Since the price for phone calls is $10$ cents per minute, the additional amount Michelle has to pay for phone calls is $30*10=300$ cents, or $3$ dollars.
Adding $20+5+3$ dollars $\boxed{28}$ | D | 28 |
9edd201f0f1d85090a58be117f8ca4da | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_2 | A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | To find how many small bottles we need, we can simply divide $500$ by $35$ . This simplifies to $\frac{100}{7}=14 \frac{2}{7}.$ Since the answer must be an integer greater than $14$ , we have to round up to $15$ bottles, or $\boxed{15}$ | E | 15 |
9edd201f0f1d85090a58be117f8ca4da | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_2 | A small bottle of shampoo can hold $35$ milliliters of shampoo, whereas a large bottle can hold $500$ milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?
$\textbf{(A)}\ 11 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 13 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 15$ | We double $35$ to get $70.$ We see that $70\cdot7=490,$ which is very close to $500.$ Thus, $2\cdot7+1=\boxed{15}$ bottles are enough.
~Technodoggo | E | 15 |
eeee8cfad895483b6a65edbb550ab7b2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | We see that both sequences have equal numbers of terms, so reformat the sequence to look like:
\begin{align*} Y = \ &12 + 14 + \cdots + 100 + 102\\ X = 10 \ + \ &12 + 14 + \cdots + 100\\ \end{align*} From here it is obvious that $Y - X = 102 - 10 = \boxed{92}$ | A | 92 |
eeee8cfad895483b6a65edbb550ab7b2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | We see that every number in Y's sequence is two more than every corresponding number in X's sequence. Since there are 46 numbers in each sequence, the difference must be $46\cdot 2=\boxed{92}$ | null | 92 |
eeee8cfad895483b6a65edbb550ab7b2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | \begin{align*} X&=10+12+14+\cdots +100 \\ Y&=X-10+102 = X+92 \\ Y-X &= (X+92)-X \\ &= \boxed{92} $\blacksquare$ | A | 92 |
eeee8cfad895483b6a65edbb550ab7b2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_4 | Let X and Y be the following sums of arithmetic sequences:
\begin{eqnarray*}X &=& 10+12+14+\cdots+100,\\ Y &=& 12+14+16+\cdots+102.\end{eqnarray*}
What is the value of $Y - X?$
$\textbf{(A)}\ 92\qquad\textbf{(B)}\ 98\qquad\textbf{(C)}\ 100\qquad\textbf{(D)}\ 102\qquad\textbf{(E)}\ 112$ | In an actual contest, this would probably take too much time but is nevertheless a solution.
The general formula for computing sums of any arithmetic sequence where $x$ is the number of terms, $f$ is the first term and $l$ is the last term is $\frac{(f+l)x}{2}$ . If one uses that formula for both sequences, they will get $2530$ as the sum for $X$ and $2622$ as the sum for $Y$ .
Subtracting $X$ from $Y$ , one will get the answer $\boxed{92}$ .
- danfan | A | 92 |
9b2d72a98cf7b5364fad4e630c076aaf | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_6 | Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$
$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$ | $A \cup B$ will be smallest if $B$ is completely contained in $A$ , in which case all the elements in $B$ would be counted for in $A$ . So the total would be the number of elements in $A$ , which is $\boxed{20}$ | C | 20 |
9b2d72a98cf7b5364fad4e630c076aaf | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_6 | Set $A$ has $20$ elements, and set $B$ has $15$ elements. What is the smallest possible number of elements in $A \cup B$
$\textbf{(A)}5 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 35\qquad\textbf{(E)}\ 300$ | Assume WLOG that $A={1, 2, 3, 4, \cdots , 20}$ , and $B={6, 7, 8, 9, 10, \cdots , 20}$ . Then, all the integers $6$ through $20$ would be redundant in $A \cup B$ , so $A \cup B = 1, 2, 3, 4, \cdots, 20 \implies \boxed{20}$ | C | 20 |
ae59f1348d5b35566bdf21a3729421e2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_8 | Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$ | 75% of the total birds were not swans. Out of that 75%, there was $30\% / 75\% = \boxed{40}$ of the birds that were not swans that were geese. | C | 40 |
ae59f1348d5b35566bdf21a3729421e2 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_8 | Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?
$\textbf{(A)}\ 20\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 50\qquad\textbf{(E)}\ 60$ | WLOG, suppose there were 100 birds in total living on Town Lake, then 30 were geese, 25 were swans, 10 were herons, and 35 were ducks. $100-25 = 75$ of the birds are not swans and 30 of these are geese, so the answer is $\frac{30}{75} \times 100 = \boxed{40}$ | C | 40 |
33a6a4fef9a51e8ed5b526bc71121b74 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$ . What was the cost of a pencil in cents?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$ | The total cost of the pencils can be found by $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$
Since $1771$ is the product of three sets of values, we can begin with prime factorization, since it gives some insight into the values: $7, 11, 23$ . Since neither $(C)$ nor $(E)$ are any of these factors, they can be eliminated immediately, leaving $(A)$ $(B)$ , and $(D)$
Beginning with $(A) 7$ , we see that the number of pencils purchased by each student must be either $11$ or $23$ . However, the problem states that the price of each pencil must exceed the number of pencils purchased, so we can eliminate this.
Continuing with $(B) 11$ , we can conclude that the only case that fulfils the restrictions are that there are $23$ students who each purchased $7$ such pencils, so the answer is $\boxed{11}$ . We can apply the same logic to $(E)$ as we applied to $(A)$ if one wants to make doubly sure. | B | 11 |
33a6a4fef9a51e8ed5b526bc71121b74 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$ . What was the cost of a pencil in cents?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$ | We know the total cost of the pencils can be found by $(\text{students}\cdot\text{pencils purchased by each}\cdot\text{price of each pencil})$
Using prime factorization like in the solution above, we see that there is only one combination of three whole numbers whose product is equal to $1771$ $7, 11, 23$ (without using $1$ ). So we know that $7, 11$ , and $23$ must be the number of students, the number of pencils purchased by each student and the price of each pencil in cents.
We know that $23$ must be the number of students, as it is the only number that makes up the majority of 30.
We pick the greater of the remaining numbers for the price of each pencil in cents, which is $11$
Therefore, our answer is $\boxed{11}$ | B | 11 |
33a6a4fef9a51e8ed5b526bc71121b74 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_10 | A majority of the $30$ students in Ms. Demeanor's class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than $1$ . The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $$17.71$ . What was the cost of a pencil in cents?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 11 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 23 \qquad \textbf{(E)}\ 77$ | We let $s$ be the number of students that bought pencils at the bookstore, $c$ be the cost of each pencil in cents, and $n$ the number of pencils each student bought. Thus, we are looking for $c.$ Since a majority of the students in the class bought pencils at the bookstore, $s>\dfrac{30}2=15.$ (s>15) We also know that $n>1$ and $c>n>1.$ Finally, $s\cdot c\cdot n=1771.$ We can factor $1771$ as $7\cdot11\cdot23.$ Since $n>15$ and $c>n>1,$ none of them can be $1,$ and therefore $c,n,$ and $s$ are $7,11,$ and $23$ in some order. We know that $n>15,$ so $n$ must be $23.$ $c>n,$ so $c=11$ and $n=7.$ Thus, $c=\boxed{11}$ cents.
~Technodoggo | B | 11 |
059381fdfca715f4d0b8c878bb77442f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $61$ points. How many free throws did they make?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$ | For the points made from two-point shots and from three-point shots to be equal, the numbers of made shots are in a $3:2$ ratio. Therefore, assume they made $3x$ and $2x$ two- and three- point shots, respectively, and thus $3x+1$ free throws. The total number of points is \[2 \times (3x) + 3 \times (2x) + 1 \times (3x+1) = 15x+1\]
Set that equal to $61$ , we get $x = 4$ , and therefore the number of free throws they made $3 \times 4 + 1 = 13 \Rightarrow \boxed{13}$ | A | 13 |
059381fdfca715f4d0b8c878bb77442f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $61$ points. How many free throws did they make?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$ | Let $x$ be the number of free throws. Then the number of points scored by two-pointers is $2(x-1)$ and the same goes for three-pointers because they scored the same number of points with twos and threes. Thus, our equation is $x+4(x-1) = 61 \Rightarrow x=13$ , giving us $\boxed{13}$ for an answer. | A | 13 |
059381fdfca715f4d0b8c878bb77442f | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_12 | The players on a basketball team made some three-point shots, some two-point shots, and some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team's total score was $61$ points. How many free throws did they make?
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 17$ | We let $a$ be the number of $2$ -point shots, $b$ be the number of $3$ -point shots, and $x$ be the number of free throws. We are looking for $x.$ We know that $2a=3b$ , and that $x=a+1$ . Also, $2a+3b+1x=61$ . We can see
\begin{align*} a&=x-1 \\ 2a &= 2x-2 \\ 3a &= 2x-2. \\ \end{align*}
Plugging this into $2a+3b+1x=61$ , we see
\begin{align*} 2x-2+2x-2+x &= 61 \\ 5x-4 &= 61 \\ 5x &= 65 \\ x &= \boxed{13} | A | 13 |
3b9d6521b4be03489cde271246f2ac55 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We split up into cases of the hundreds digits being $2$ or $5$ . If the hundred digits is $2$ , then the units digits must be $8$ in order for the number to be even and then there are $4$ remaining choices ( $1,5,7,9$ ) for the tens digit, giving $1 \times 4 \times 1=4$ possibilities. Similarly, there are $1 \times 2 \times 4=8$ possibilities for the $5$ case, giving a total of $\boxed{12}$ possibilities. | A | 12 |
3b9d6521b4be03489cde271246f2ac55 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We see that the last digit of the $3$ -digit number must be even to have an even number. Therefore, the last digit must either be $2$ or $8$
Case $1$ -the last digit is $2$ . We must have the hundreds digit to be $5$ and the tens digit to be any $1$ of ${1,7,8,9}$ , thus obtaining $4$ numbers total.
Case $2$ -the last digit is $8$ . We now can have $2$ or $5$ to be the hundreds digit, and any choice still gives us $4$ choices for the tens digit. Therefore, we have $2 \cdot 4=8$ numbers.
Adding up our cases, we have $4+8=\boxed{12}$ numbers. | A | 12 |
3b9d6521b4be03489cde271246f2ac55 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_13 | How many even integers are there between $200$ and $700$ whose digits are all different and come from the set $\left\{1,2,5,7,8,9\right\}$
$\text{(A)}\,12 \qquad\text{(B)}\,20 \qquad\text{(C)}\,72 \qquad\text{(D)}\,120 \qquad\text{(E)}\,200$ | We see that there are $\dbinom63=20$ total possibilities for a 3-digit number whose digits do not repeat and are comprised of digits only from the set ${1,2,5,7,8,9}.$ Obviously, some of these (such as $987,$ for example) will not work, and thus the answer will be less than $20.$ The only possible option is $\boxed{12}.$ ~ Technodoggo | A | 12 |
84d01344a16e245ca3637b90a7a1a548 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\mathrm{(A)}\ 140 \qquad \mathrm{(B)}\ 240 \qquad \mathrm{(C)}\ 440 \qquad \mathrm{(D)}\ 640 \qquad \mathrm{(E)}\ 840$ | We know that $\frac{\text{total miles}}{\text{total gas}}=55$ . Let $x$ be the distance in miles the car traveled during the time it ran on gasoline, then the amount of gas used is $0.02x$ . The total distance traveled is $40+x$ , so we get $\frac{40+x}{0.02x}=55$ . Solving this equation, we get $x=400$ , so the total distance is $400 + 40 = \boxed{440}$ | C | 440 |
84d01344a16e245ca3637b90a7a1a548 | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_15 | Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first $40$ miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of $0.02$ gallons per mile. On the whole trip he averaged $55$ miles per gallon. How long was the trip in miles?
$\mathrm{(A)}\ 140 \qquad \mathrm{(B)}\ 240 \qquad \mathrm{(C)}\ 440 \qquad \mathrm{(D)}\ 640 \qquad \mathrm{(E)}\ 840$ | Let $d$ be the length of the trip in miles. Roy used no gasoline for the 40 first miles, then used 0.02 gallons of gasoline per mile on the remaining $d - 40$ miles, for a total of $0.02 (d - 40)$ gallons. Hence, his average mileage was \[\frac{d}{0.02 (d - 40)} = 55.\] Multiplying both sides by $0.02 (d - 40)$ , we get \[d = 55 \cdot 0.02 \cdot (d - 40) = 1.1d - 44.\] Then $0.1d = 44$ , so $d = \boxed{440}$ . The answer is $(C)$ | null | 440 |
0309f8e2dd512b605b428e5741d9c00b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Let $A=x$ . Then from $A+B+C=30$ , we find that $B=25-x$ . From $B+C+D=30$ , we then get that $D=x$ . Continuing this pattern, we find $E=25-x$ $F=5$ $G=x$ , and finally $H=25-x$ . So $A+H=x+25-x=25 \rightarrow \boxed{25}$ | C | 25 |
0309f8e2dd512b605b428e5741d9c00b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Given that the sum of 3 consecutive terms is 30, we have $(A+B+C)+(C+D+E)+(F+G+H)=90$ and $(B+C+D)+(E+F+G)=60$
It follows that $A+B+C+D+E+F+G+H=85$ because $C=5$
Subtracting, we have that $A+H=25\rightarrow \boxed{25}$ | C | 25 |
0309f8e2dd512b605b428e5741d9c00b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | From the given information, we can deduce the following equations:
$A+B=25, B+D=25, D+E=25, D+E+F=30, E+F+G =30$ , and $F+G+H=30$
We can then cleverly manipulate the equations above by adding and subtracting them to be left with the answer.
$(A+B)-(B+D)=25-25 \implies (A-D)=0$
$(A-D)+(D+E)=0+25 \implies (A+E)=25$
$(A+E)-(E+F+G)=25-30 \implies (A-F-G)=-5$ (Notice how we don't use $D+E+F=30$
$(A-F-G)+(F+G+H)=-5+30 \implies (A+H)=25$
Therefore, we have $A+H=25 \rightarrow \boxed{25}$ | C | 25 |
0309f8e2dd512b605b428e5741d9c00b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Since all of the answer choices are constants, it shouldn't matter what we pick $A$ and $B$ to be, so let $A = 20$ and $B = 5$ . Then $D = 30 - B -C = 20$ $E = 30 - D - C = 5$ $F = 30 - D - E =5$ , and so on until we get $H = 5$ . Thus $A + H = \boxed{25}$ | C | 25 |
0309f8e2dd512b605b428e5741d9c00b | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_17 | In the eight term sequence $A$ $B$ $C$ $D$ $E$ $F$ $G$ $H$ , the value of $C$ is $5$ and the sum of any three consecutive terms is $30$ . What is $A+H$
$\textbf{(A)}\ 17 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 25 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 43$ | Assume the sequence is \[15,10,5,15,10,5,15,10\]
Thus, $15+10=\boxed{25}$ | C | 25 |
f0380faa88cb03fa708d2290aac8b20e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$ | Let the population of the town in $1991$ be $p^2$ . Let the population in $2001$ be $q^2+9$ . It follows that $p^2+150=q^2+9$ . Rearrange this equation to get $141=q^2-p^2=(q-p)(q+p)$ . Since $q$ and $p$ are both positive integers with $q>p$ $(q-p)$ and $(q+p)$ also must be, and thus, they are both factors of $141$ . We have two choices for pairs of factors of $141$ $1$ and $141$ , and $3$ and $47$ . Assuming the former pair, since $(q-p)$ must be less than $(q+p)$ $q-p=1$ and $q+p=141$ . Solve to get $p=70, q=71$ . Since $p^2+300$ is not a perfect square, this is not the correct pair. Solve for the other pair to get $p=22, q=25$ . This time, $p^2+300=22^2+300=784=28^2$ . This is the correct pair. Now, we find the percent increase from $22^2=484$ to $28^2=784$ . Since the increase is $300$ , the percent increase is $\frac{300}{484}\times100\%\approx\boxed{62}$ | E | 62 |
f0380faa88cb03fa708d2290aac8b20e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$ | Since all the answer choices are around $50\%$ , we know the town's starting population must be around $600$ . We list perfect squares from $400$ to $1000$ \[441, 484, 529,576,625,676,729,784,841,900,961\] We see that $484$ and $784$ differ by $300$ , and we can confirm that $484$ is the correct starting number by noting that $484+150=634=25^2+9$ . Thus, the answer is $784/484-1\approx \boxed{62}$ | E | 62 |
f0380faa88cb03fa708d2290aac8b20e | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_19 | In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2011, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town's population during this twenty-year period?
$\textbf{(A)}\ 42 \qquad\textbf{(B)}\ 47 \qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 57\qquad\textbf{(E)}\ 62$ | Let the population of the town in 1991 be $a^2$ and the population in 2011 be $b^2$ . We know that $a^2+150+150=b^2 \implies a^2-b^2=-300 \implies b^2-a^2=300 \implies (b-a)(b+a)=300$ . Note that $b-a$ must be even. Testing, we see that $a=22$ and $b=28$ works, as $484+150-9=625=25^2$ , so $\frac{784-484}{484} \approx \boxed{62}$ | E | 62 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$ $C$ be one of the diagonal vertices to $B$ , and so on. We consider cases for this problem.
In the case that $C$ has the same color as $A$ $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$ . In this case, $A$ has $6$ choices, $B$ has $5$ choices (any color but the color of $A$ ), $C$ has $1$ choice, $D$ has $5$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600$ combinations.
In the case that $C$ has a different color from $A$ and $D$ has a different color from $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices (since $A$ and $B$ necessarily have different colors), $D$ has $4$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920$ combinations.
In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$ $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600$ combinations.
Adding all those combinations up, we get $600+1920+600=\boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | First, notice that there can be $3$ cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case $1$ : There are $6!$ ways of assigning each vertex a different color. $6! = 720$
Case $2$ : There are $\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\frac {6!}{2} * 5 = 1800$
Case $3$ : There are $\frac {\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$ spaces. But because the two pairs are indistinguishable, we divide by $2!$ $\frac {\frac {6!}{3!} * 10}{2!} = 600$
Adding all the possibilities up, we get $720+1800+600=\boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | There are $6$ ways to assign a color to $A$ . WLOG, give vertex $A$ a color; we can multiply by $6$ at the end. Since vertices $A$ and $C$ cannot have the same color, there are $5$ ways to assign colors to vertex $C$ . Using this same logic, there are $5$ ways to assign a color to vertices $E$ $B$ , and $D$ , giving a total of $5^4=625$ ways. However, vertex $D$ cannot be the same color as vertex $A$ . To use complementary counting, we need to find the amount of ways for $D$ and $A$ to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are $5^3=125$ ways, except we must subtract the number of ways for a triangle. Each time, there is $1$ less vertex, so $5$ times less ways to color. This process stops when there are only $2$ vertices left; in this case there are simply $5$ ways to color this figure.
So in conclusion, there are $6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120}$ ways. | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | This problem is a direct application of the chromatic polynomial of a graph, represented by $P_k(G)$ , which returns a polynomial representing the number of ways to color a graph $G$ with $k$ colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of $k$ where we can plug in $k=6$ to get our answer.
Well, if we want to find $P_k$ for a graph $G$ , we should probably first draw the graph, right? We can first draw five vertices (without any edges) in the shape of a pentagon. After connecting the diagonals, we get a star. We attempt to shift the vertices around to simplify the graph, which we quickly realize is isomorphic (can be turned into) the five-cycle, or the pentagon.
Finding $P_k$ on a graph with a cycle of length at most 3 is straightforward -- we first pick a vertex v. It has 0 colored (visited) neighbors, so we can color it in k ways. We then move on to the vertices adjacent to v, etc, and at the end we multiply all these together. For example, the chromatic polynomial of a triangle is $k(k-1)(k-2)$ . However, cycles with length > 3 introduce ambiguity, and thus the above technique fails. Thus, we need to use the recursive formula $P_k(G) = P_k(G-e)-P_k(G*e)$ , where $G-e$ represents removing e and $G*e$ represents contracting e, or collapsing the two endpoints of e into one. When we hit a graph where the longest cycle has length 3, we can use the first technique to quickly find $P_k$
After about four iterations of the algorithm, we get that the chromatic polynomial is $k(k-1)^4 - k(k-1)^3 + k(k-1)(k-2)$ . Plugging in 6 gives us $\boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | In pentagon $ABCDE$ , fix any vertex $A$ . Now draw diagonal $AC$ . There are six choices for vertex $A$ and $5$ choices for vertex $C$
Now draw diagonal $CE$ . Since $E$ cannot be the same color as vertex $C$ , we have $5$ choices for $E$ . Again, we have five choices for vertex $D$ (draw diagonal $AD$ ).
Thus there are $6 \cdot 5$ choices for vertices $A$ and $C$ and $5 \cdot 5$ combinations for $D$ and $E$
To determine the final count, we consider two cases for the final $25$ combinations of $D$ and $E$ , which uniquely determines $B$ . Then, we multiply by $30$ since the choices of $A$ and $C$ are independent from these two cases.
Case $1$ $D$ and $E$ are the same color. There are $4$ possible pairs (This is because $D$ and $E$ are not chosen from the same 5 colors. $D$ cannot be $A$ as they are on a diagonal, but $E$ can be. ~ primegn), and thus we have $5$ choices for $B$ . There are $20$ cases here.
Case $2$ $D$ and $E$ are different. There are $25-4 = 21$ possible combinations and we have $4$ choices for $B$ (not the color of $D$ nor $E$ ). In this case we have $21 \cdot 4 = 84$ cases.
Our final count is $30(84+20) = \boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | WLOG, draw $ABCDE$ such that point $A$ is at the top, and write the letters in counterclockwise order. WLOG, fill in $A$ first. There are $6$ ways to do so. From here we proceed with casework on the color of $B$
Summing up the cases, $600+600+1920=3120 \Longrightarrow \boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Notice that a minimum of, $3$ and a maximum of $5$ colours can be used.
This is straight forward and there are $6P5 = 720$ ways.
If $x$ and $y$ have same colours, they have to be adjacent. Let $x$ be directly right of $y$ $4$ consecutive points from $x$ clockwise have different colours. They can be coloured in $6P4=360$ ways. And $x$ can be chosen in $5$ ways, bringing the total to $360 \times 5=1800$
Note that $3$ points can't have the same colour as at least $1$ of the lines joining them will be a diagonal. Then the arrangement is $2-2-1$ . The point with distinct colour can be chosen in $5$ ways and can be coloured in $6$ ways. The next two points clockwise can be coloured in $5$ ways and the next two in $4$ ways. The total is $5 \times 6 \times 5 \times 4=600$
Finally, adding all cases, the answer is $600+1800+720=3120 \Longrightarrow \boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | Name the pentagon $ADBEC$ . Then $A,B;$ $B,C;$ $C,D;$ $D,E$ and $E,A$ have to be of different colours.
$A$ can be coloured in $6$ ways, $B$ in $5$ ways (Without the one used for $A$ ), $C$ in $5$ ways (Without the one used for $B$ ), $D$ in $5$ ways (Without the one used for $C$ ).
Now we find the expected value of how many ways can $E$ be coloured.
If $A$ and $C$ are of the same colour, $A$ and $D$ are always different. This happens $\frac{1}{5}$ times and leaves $4$ options for $E$
If $A$ and $C$ are of different colours $\left(\frac{4}{5} \ \text{of the time} \right)$ then there is a $\frac{1}{5}$ chance that $A$ and $D$ are of same colour and $\frac{4}{5}$ chance that they are different. The first one leaves $5$ ways to colour $E$ and the second one leaves $4$
Calculating all these, we find the expected value for the number of ways $E$ can be coloured is $\frac{1}{5} \times 4 + \frac{4}{5} \left( \frac{1}{5} \times 5 + \frac{4}{5} \times 4 \right) = 4.16$
Therefore total number or ways $= 6 \times 5 \times 5 \times 5 \times 4.16 = \boxed{3120}$ | C | 3120 |
11a0b8251608f423ddf1af38e7cbcfad | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_22 | Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
$\textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750$ | First, we know that there are $6$ ways to assign a color to vertex $A$ . Then, there are $6$ ways to assign a color to vertex $B$ , because it doesn't depend on vertex $A$ . There are also $5$ ways to assign a color to vertex $C$ , because it can't be the same color as vertex $A$ . Notice that both vertices $E$ and $D$ have 5 or 4 choices, depending on whether $A$ $B$ , and $C$ are the same color. This means that the we know that the amount of choices must be bigger than $6 \times 6 \times 5 \times 4 \times 4 = 2880$ , which eliminates options $A$ and $B$
Then, notice that the amount of choices must be smaller than $6 \times 6 \times 5 \times 4.5 \times 4.5 = 3645$ , because the expected amount of choices $E$ and $D$ have are less than 4.5. (There is a greater probability that $A$ $B$ , and $B$ $C$ are both different). This eliminates option $E$ , which leaves us with options $C$ and $D$
We know that the amount of choices must be a multiple of $6$ , so $D$ is eliminated, leaving us with option $\boxed{3120}$ | C | 3120 |
a4ba286964e674a8d5c3afd3bb18197c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$ | First look at the numbers Alice says. $1, 3, 4, 6, 7, 9 \cdots$ skipping every number that is congruent to $2 \pmod 3$ . Thus, Barbara says those numbers EXCEPT every second - being $2 + 3^1 \equiv 5 \pmod{3^2=9}$ . So Barbara skips every number congruent to $5 \pmod 9$ . We continue and see:
Alice skips $2 \pmod 3$ , Barbara skips $5 \pmod 9$ , Candice skips $14 \pmod {27}$ , Debbie skips $41 \pmod {81}$ , Eliza skips $122 \pmod {243}$ , and Fatima skips $365 \pmod {729}$
Since the only number congruent to $365 \pmod {729}$ and less than $1,000$ is $365$ , the correct answer is $\boxed{365}$ | C | 365 |
a4ba286964e674a8d5c3afd3bb18197c | https://artofproblemsolving.com/wiki/index.php/2011_AMC_10A_Problems/Problem_23 | Seven students count from 1 to 1000 as follows:
Alice says all the numbers, except she skips the middle number in each consecutive group of three numbers. That is, Alice says 1, 3, 4, 6, 7, 9, . . ., 997, 999, 1000.
Barbara says all of the numbers that Alice doesn't say, except she also skips the middle number in each consecutive group of three numbers.
Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers.
Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.
Finally, George says the only number that no one else says.
What number does George say?
$\textbf{(A)}\ 37\qquad\textbf{(B)}\ 242\qquad\textbf{(C)}\ 365\qquad\textbf{(D)}\ 728\qquad\textbf{(E)}\ 998$ | After Alice says all her numbers, the numbers not mentioned yet are \[\text{Alice: } 2,5,8,11,14,17,\cdots,998.\] After Barbara says all her numbers, the numbers that haven't been said yet are \[\text{Barbara: } 5,14,23,32,41,50,\cdots,995.\] After Candice, the list is \[\text{Candice: } 14,41,68,\cdots,986.\] Notice how each list is an arithmetic sequence where the common differene is thrice the common ratio of the previous list and the first term is the second term of the previous list. Now that the pattern is clear, we construct the rest of the lists: \[\text{Debbie: } 41,122,203,\cdots,959\] \[\text{Eliza: } 122,365,608,878\] \[\text{Fatima: } 365\]
Thus, George says $\boxed{365}$ | C | 365 |
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