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138d8679cdb4abb8b19933844d32166e | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_1 | Sandwiches at Joe's Fast Food cost $$3$ each and sodas cost $$2$ each. How many dollars will it cost to purchase $5$ sandwiches and $8$ sodas?
$\textbf{(A)}\ 31\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 34\qquad\textbf{(E)}\ 35$ | The $5$ sandwiches cost $5\cdot 3=15$ dollars. The $8$ sodas cost $8\cdot 2=16$ dollars. In total, the purchase costs $15+16=\boxed{31}$ dollars. | A | 31 |
f415f3b02325bba93650bb37daf6e0c4 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_3 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | Let $m$ be Mary's age. Then $\frac{m}{30}=\frac{3}{5}$ . Solving for $m$ , we obtain $m=\boxed{18}.$ | B | 18 |
f415f3b02325bba93650bb37daf6e0c4 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_3 | The ratio of Mary's age to Alice's age is $3:5$ . Alice is $30$ years old. How old is Mary?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 50$ | We can see this is a combined ratio of $8$ $(5+3)$ . We can equalize by doing $30\div5=6$ , and $6\cdot3=\boxed{18}$ . With the common ratio of $8$ and difference ratio of $6$ , we see $6\cdot8=30+18$ . Therefore, we can see our answer is correct. | B | 18 |
1e437341ecbbe116dfee862e045be5b8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | From the greedy algorithm , we have $9$ in the hours section and $59$ in the minutes section. $9+5+9=\boxed{23}$ | E | 23 |
1e437341ecbbe116dfee862e045be5b8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | With a matrix, we can see $\begin{bmatrix} 1+2&9&6&3\\ 1+1&8&5&2\\ 1+0&7&4&1 \end{bmatrix}$ The largest single digit sum we can get is $9$ .
For the minutes digits, we can combine the largest $2$ digits, which are $9,5 \Rightarrow 9+5=14$ , and finally $14+9=\boxed{23}$ | E | 23 |
1e437341ecbbe116dfee862e045be5b8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_4 | A digital watch displays hours and minutes with AM and PM. What is the largest possible sum of the digits in the display?
$\textbf{(A)}\ 17\qquad\textbf{(B)}\ 19\qquad\textbf{(C)}\ 21\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 23$ | We first note that since the watch displays time in AM and PM, the value for the hours section varies from $00-12$ . Therefore, the maximum value of the digits for the hours is when the watch displays $09$ , which gives us $0+9=9$
Next, we look at the value of the minutes section, which varies from $00-59$ . Let this value be a number $ab$ . We quickly find that the maximum value for $a$ and $b$ is respectively $5$ and $9$
Adding these up, we get $9+5+9=\boxed{23}$ | E | 23 |
c182a24dc15c7b8f0891178f7449e8c3 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_5 | Doug and Dave shared a pizza with $8$ equally-sized slices. Doug wanted a plain pizza, but Dave wanted anchovies on half the pizza. The cost of a plain pizza was $8$ dollars, and there was an additional cost of $2$ dollars for putting anchovies on one half. Dave ate all the slices of anchovy pizza and one plain slice. Doug ate the remainder. Each paid for what he had eaten. How many more dollars did Dave pay than Doug?
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Dave and Doug paid $8+2=10$ dollars in total. Doug paid for three slices of plain pizza, which cost $\frac{3}{8}\cdot 8=3$ . Dave paid $10-3=7$ dollars. Dave paid $7-3=\boxed{4}$ more dollars than Doug. | D | 4 |
2d757b0ec9f99f513c86f8d780bc7461 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW); label("$12$",E--B,N); label("$12$",D--F,S); label("$4$",E--A,W); label("$4$",(12.4,-1.75),E); label("$8$",A--D,W); label("$8$",(12.4,4),E); label("$y$",A--H,S); label("$y$",G--C,N); [/asy]
As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{6}$ | A | 6 |
2d757b0ec9f99f513c86f8d780bc7461 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
[asy] size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$",A--H,S); label("$y$",G--C,N); [/asy]
As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$y$",(9,-2),NW); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{6}$ | A | 6 |
2d757b0ec9f99f513c86f8d780bc7461 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_6 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$
[asy] unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4));[/asy]
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$ | A | 6 |
940a07645680df250a0bbdb8859062b0 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_8 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Notice that if the consecutive positive integers have a sum of $15$ , then their average (which could be a fraction) must be a divisor of $15$ . If the number of integers in the list is odd, then the average must be either $1, 3,$ or $5$ , and $1$ is clearly not possible. The other two possibilities both work:
If the number of integers in the list is even, then the average will have a $\frac{1}{2}$ . The only possibility is $\frac{15}{2}$ , from which we get:
Thus, the correct answer is $\boxed{3}.$ | C | 3 |
940a07645680df250a0bbdb8859062b0 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_8 | How many sets of two or more consecutive positive integers have a sum of $15$
$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$ | Any set will form a arithmetic progression with the first term say $a$ . Since the numbers are consecutive the common difference $d = 1$
The sum of the AP has to be 15. So,
\[S_n = \frac{n}{2} \cdot (2a + (n-1)d)\] \[S_n = \frac{n}{2} \cdot (2a + (n-1)1)\] \[15 = \frac{n}{2} \cdot (2a + n - 1)\] \[2an + n^2 - n = 30\] \[n^2 + n(2a - 1) - 30 = 0\]
Now we need to find the number of possible sets of values of a, n which satisfy this equation. Now $a$ cannot be 15 as we need 2 terms. So a can only be less the 15.
Trying all the values of a from 1 to 14 we observe that $a = 4$ $a = 7$ and $a = 1$ provide the only real solutions to the above equation.The three possibilites of a and n are.
\[(a,n) = (4,3),(7, 2),(1, 6)\]
The above values are obtained by solving the following equations obtained by substituting the above mentioned values of a into $n^2 + n(2a - 1) - 30 = 0$
\[n^2 + 7n - 30 = 0\] \[n^2 + 13n - 30 = 0\] \[n^2 - n - 30 = 0\]
Since there are 3 possibilities the answer is $\boxed{3}.$ | C | 3 |
e02ee415de88f118a922259c01c63af1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_10 | For how many real values of $x$ is $\sqrt{120-\sqrt{x}}$ an integer?
$\textbf{(A) } 3\qquad \textbf{(B) } 6\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 11$ | For $\sqrt{120-\sqrt{x}}$ to be an integer, $120-\sqrt{x}$ must be a perfect square.
Since $\sqrt{x}$ can't be negative, $120-\sqrt{x} \leq 120$
The perfect squares that are less than or equal to $120$ are $\{0,1,4,9,16,25,36,49,64,81,100\}$ , so there are $11$ values for $120-\sqrt{x}$
Since every value of $120-\sqrt{x}$ gives one and only one possible value for $x$ , the number of values of $x$ is $\boxed{11}$ | E | 11 |
97cc7ca02b87488ee3766a53dfcded95 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_12 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | The inside diameters of the rings are the positive integers from $1$ to $18$ . The total distance needed is the sum of these values plus $2$ for the top of the first ring and the bottom of the last ring. Using the formula for the sum of an arithmetic series , the answer is $\frac{18 \cdot 19}{2} + 2 = \boxed{173}$ | B | 173 |
97cc7ca02b87488ee3766a53dfcded95 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_12 | A number of linked rings, each $1$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
$\textbf{(A) } 171\qquad\textbf{(B) } 173\qquad\textbf{(C) } 182\qquad\textbf{(D) } 188\qquad\textbf{(E) } 210\qquad$ | Alternatively, the sum of the consecutive integers from 3 to 20 is $\frac{1}{2}(18)(3+20) = 207$ . However, the 17 intersections between the rings must be subtracted, and we also get $207 - 2(17) = \boxed{173}$ | B | 173 |
ff8b260c6e48fc087b54f3971592eb5d | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_14 | problem_id
ff8b260c6e48fc087b54f3971592eb5d Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | The problem can be restated as an equation of the form $300p + 210g = x$ , where $p$ is the number of pigs, $g$ is the number of goats, and $x$ is the positive debt. The problem asks us to find the lowest possible. $p$ and $g$ must be integers, which makes the equation a Diophantine equation Bezout's Lemma tells us that the smallest $c$ for the Diophantine equation $am + bn = c$ to have solutions is when $c$ is the GCD ( greatest common divisor ) of $a$ and $b$ . Therefore, the answer is $gcd(300,210)=\boxed{30}.$ | C | 30 |
ff8b260c6e48fc087b54f3971592eb5d | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_14 | problem_id
ff8b260c6e48fc087b54f3971592eb5d Two farmers agree that pigs are worth $300$ do...
ff8b260c6e48fc087b54f3971592eb5d Let us simplify this problem. Dividing by $30...
Name: Text, dtype: object | Alternatively, note that $300p + 210g = 30(10p + 7g)$ is divisible by $30$ no matter what $p$ and $g$ are, so our answer must be divisible by $30$ . Since we want the smallest integer, we can suppose that the answer is $30$ and go on from there. Note that three goats minus two pigs gives us $630 - 600 = 30$ exactly. Since our supposition can be achieved, the answer is $\boxed{30}$ | C | 30 |
240d176802561ac514df753ff829a8bc | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_18 | The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and
$f(x)+f\left(\frac{1}{x}\right)=x$
What is the largest set of real numbers that can be in the domain of $f$
$\mathrm{(A) \ } \{x|x\ne 0\}\qquad \mathrm{(B) \ } \{x|x<0\}$
$\mathrm{(C) \ } \{x|x>0\}$ $\mathrm{(D) \ } \{x|x\ne -1\;\rm{and}\; x\ne 0\;\rm{and}\; x\ne 1\}$
$\mathrm{(E) \ } \{-1,1\}$ | We know that $f(x) + f \left(\frac{1}{x}\right) = x.$ Plugging in $x = \frac{1}{x}$ we get \[f \left(\frac{1}{x}\right) + f \left(\frac{1}{\frac{1}{x}}\right) = \frac{1}{x}\] \[f \left(\frac{1}{x}\right) + f(x) = \frac{1}{x}.\]
Also notice \[f \left(\frac{1}{x}\right) + f(x) = x\] by the commutative property(this is the same as the equation given in the problem. We are just rearranging). So we can set $\frac{1}{x} = x$ which gives us $x = \pm 1$ which is answer option $\boxed{1,1}.$ | E | 1,1 |
0127c836a9a9e73402799dba5e2121f1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_21 | Let $S_1=\{(x,y)|\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)\}$ and $S_2=\{(x,y)|\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)\}$
What is the ratio of the area of $S_2$ to the area of $S_1$
$\mathrm{(A) \ } 98\qquad \mathrm{(B) \ } 99\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 101\qquad \mathrm{(E) \ } 102$ | Looking at the constraints of $S_1$
$x+y > 0$
$\log_{10}(1+x^2+y^2)\le 1+\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10} 10 +\log_{10}(x+y)$
$\log_{10}(1+x^2+y^2)\le \log_{10}(10x+10y)$
$1+x^2+y^2 \le 10x+10y$
$x^2-10x+y^2-10y \le -1$
$x^2-10x+25+y^2-10y+25 \le 49$
$(x-5)^2 + (y-5)^2 \le (7)^2$
$S_1$ is a circle with a radius of $7$ . So, the area of $S_1$ is $49\pi$
Looking at the constraints of $S_2$
$x+y > 0$
$\log_{10}(2+x^2+y^2)\le 2+\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10} 100 +\log_{10}(x+y)$
$\log_{10}(2+x^2+y^2)\le \log_{10}(100x+100y)$
$2+x^2+y^2 \le 100x+100y$
$x^2-100x+y^2-100y \le -2$
$x^2-100x+2500+y^2-100y+2500 \le 4998$
$(x-50)^2 + (y-50)^2 \le (7\sqrt{102})^2$
$S_2$ is a circle with a radius of $7\sqrt{102}$ . So, the area of $S_2$ is $4998\pi$
So the desired ratio is $\frac{4998\pi}{49\pi} = 102 \Rightarrow \boxed{102}$ | E | 102 |
30fa320a11387826a8a9ed1f9f371895 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_24 | The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$ | By the Multinomial Theorem , the summands can be written as
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^ay^bz^c}\]
and
\[\sum_{a+b+c=2006}{\frac{2006!}{a!b!c!}x^a(-y)^b(-z)^c},\]
respectively. Since the coefficients of like terms are the same in each expression, each like term either cancel one another out or the coefficient doubles. In each expansion there are:
\[{2006+2\choose 2} = 2015028\]
terms without cancellation. For any term in the second expansion to be negative, the parity of the exponents of $y$ and $z$ must be opposite. Now we find a pattern:
if the exponent of $y$ is $1$ , the exponent of $z$ can be all even integers up to $2004$ , so there are $1003$ terms.
if the exponent of $y$ is $3$ , the exponent of $z$ can go up to $2002$ , so there are $1002$ terms.
$\vdots$
if the exponent of $y$ is $2005$ , then $z$ can only be 0, so there is $1$ term.
If we add them up, we get $\frac{1003\cdot1004}{2}$ terms. However, we can switch the exponents of $y$ and $z$ and these terms will still have a negative sign. So there are a total of $1003\cdot1004$ negative terms.
By subtracting this number from 2015028, we obtain $\boxed{1,008,016}$ or $1,008,016$ as our answer. | D | 1,008,016 |
30fa320a11387826a8a9ed1f9f371895 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_24 | The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$ | Define $P$ such that $P=y+z$ . Then the expression in the problem becomes: $(x+P)^{2006}+(x-P)^{2006}$
Expanding this using binomial theorem gives $(x^n+Px^{n-1}+...+P^{n-1}x+P^n)+(x^n-Px^{n-1}+...-P^{n-1}x+P^n)$ , where $n=2006$ (we may omit the coefficients, as we are seeking for the number of terms, not the terms themselves).
Simplifying gives: $2(x^n+x^{n-2}P^2+...+x^2P^{n-2}+P^n)$ . Note that two terms that come out of different powers of $P$ cannot combine and simplify, as their exponent of $x$ will differ. Therefore, we simply add the number of terms produced from each addend. By the Binomial Theorem, $P^k=(y+z)^k$ will have $k+1$ terms, so the answer is $1+3+5+...+2007=1004^2=1,008,016 \implies \boxed{1,008,016}$ | D | 1,008,016 |
30fa320a11387826a8a9ed1f9f371895 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_24 | The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$ | Using stars and bars we know that $(x+y+z)^{2006}$ has ${2006+2\choose 2}$ or $2015028$ different terms. Now we need to find out how many of these terms are canceled out by $(x-y-z)^{2006}$ . We know that for any term(let's say $x^{a}(-y)^{b}(-z)^{c}$ ) where $a+b+c=2006$ of the expansion of $(x-y-z)^{2006}$ is only going to cancel out with the corresponding term $x^{a}y^{b}z^{c}$ if only $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. Now let's do some casework to see how many terms fit this criteria:
Case 1: $a$ is even
Now we know that $a$ is even and $a+b+c=2006$ . Thus $b+c$ is also even or both $b$ and $c$ are odd or both $b$ and $c$ are even. This case clearly fails the above criteria and there are 0 possible solutions.
Case 2: $a$ is odd
Now we know that $a$ is odd and $a+b+c=2006$ . Thus $b+c$ is odd and $b$ is odd and $c$ is even or $b$ is even and $c$ is odd. All terms that have $a$ being odd work.
We now need to figure out how many of the terms have $a$ as a odd number. We know that $a$ can be equal to any number between 0 and 2006. There are 1003 odd numbers between this range and 2007 total numbers. Thus $\frac{1003}{2007}$ of the $2015028$ terms will cancel out and $\frac{1004}{2007}$ of the terms will work. Thus there are $(\frac{1004}{2007})(2015028)$ terms. This number comes out to be $1,008,016$ $\implies \boxed{1,008,016}$ (Author: David Camacho) | D | 1,008,016 |
30fa320a11387826a8a9ed1f9f371895 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_24 | The expression
\[(x+y+z)^{2006}+(x-y-z)^{2006}\]
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ } 2,015,028$ | Noticing how $y$ and $z$ are negative in the second part of the expression, let $x=a$ and $y+z = b$ . Then we get \[(a+b)^{2006} + (a-b)^{2006}\] We know that the terms that don't cancel out have even powers of $a$ and $b$ . Our expansion will be in the form of \[2a^{2006} + x_1a^{2004}b^{2} + x_2a^{2002}b^{4} + \cdots + 2b^{2006}\] Note that $b^n = (x+y)^n$ has $n+1$ terms. Furthermore, the current expression is irreducible as each term has a different $x$ power. Thus, when we write $a$ and $b$ back to their original terms, we will have $1+3+5+ \cdots + 2007 = 1004^2 = 1,008,016 \implies \boxed{1,008,016}$ | D | 1,008,016 |
e343bfd5714a46552e0923858c2dfa43 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405$ | This question can be solved fairly directly by casework and pattern-finding. We give a somewhat more general attack, based on the solution to the following problem:
How many ways are there to choose $k$ elements from an ordered $n$ element set without choosing two consecutive members?
You want to choose $k$ numbers out of $n$ with no consecutive numbers. For each configuration, we can subtract $i-1$ from the $i$ -th element in your subset. This converts your configuration into a configuration with $k$ elements where the largest possible element is $n-k+1$ , with no restriction on consecutive numbers. Since this process is easily reversible, we have a bijection .
Without consideration of the second condition, we have: ${15 \choose 1} + {14 \choose 2} + {13 \choose 3} + ... + {9 \choose 7} + {8 \choose 8}$
Now we examine the second condition. It simply states that no element in our original configuration (and hence also the modified configuration, since we don't move the smallest element) can be less than $k$ , which translates to subtracting $k - 1$ from the "top" of each binomial coefficient .
Now we have, after we cancel all the terms ${n \choose k}$ where $n < k$ ${15 \choose 1} + {13 \choose 2} + {11 \choose 3} + {9 \choose 4} + {7 \choose 5}= 15 + 78 + 165 + 126 + 21 = \boxed{405}$ | E | 405 |
e343bfd5714a46552e0923858c2dfa43 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405$ | Let $s$ be the numbers of elements in a subset. First we examine the second condition. No elements less than $s$ can be put in a subset of size $s$ , therefore the "lowest" element that can go into the subset is $s$ , whereas the "highest" element that can go into the subset is $15$ . This is a total of $15-s+1$ or $16-s$ possible elements.
Now we consider the first condition. No consecutive elements are allowed. This means that if an element $e$ is put into the subset, both $e-1$ and $e+1$ are no longer possibilities. Assume that all subsets are ordered from least to greatest (we are looking for the number of combinations, so we can order these combinations however we want). Then the first element will be $s$ (as a reminder, the lowest possible element in a subset is $s$ ), the second element will be at least $s+2$ , and so on. After $s$ elements are chosen, we will have skipped $s-1$ elements (these are the elements that were "eliminated" as they were consecutive). Therefore, we ignore exactly $s-1$ elements (if we ignore more or less elements, then $s$ changes) Since we must ignore $s-1$ elements, we can simply remove those beforehand. ( $16-s)-(s-1) = 17-2s$ possible elements
Now we look for the bounds of $s$ . We are looking for non-empty subsets, so $s\ge1$ . If $s$ is too large, there will not be enough non-consecutive terms between $s$ and $15$ . Specifically, the highest element in a subset using "optimal" selection will be $s+2(s-1)$ or $3s-2$ . If $3s-1>15$ , that means s is too large. Therefore $3s-2\le15$ ; solving for $s$ yields $s\le5$ . Now we know that $1 \le s \le 5$
We want to know the number of ways to arrange $17-2s$ "balls" into $s$ identical "boxes" with at most $1$ ball per box, for $s=1$ to $s=5$ . This is equivalent to $\sum_{s=1}^{5}{17-2s \choose s} = 405$ , or $\boxed{405}$ | E | 405 |
e343bfd5714a46552e0923858c2dfa43 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12A_Problems/Problem_25 | How many non- empty subsets $S$ of $\{1,2,3,\ldots ,15\}$ have the following two properties?
$(1)$ No two consecutive integers belong to $S$
$(2)$ If $S$ contains $k$ elements , then $S$ contains no number less than $k$
$\mathrm{(A) \ } 277\qquad \mathrm{(B) \ } 311\qquad \mathrm{(C) \ } 376\qquad \mathrm{(D) \ } 377\qquad \mathrm{(E) \ } 405$ | We will split the problem into cases, and maybe one could then generalize this to arbitrary $n$
$\bold{Case 1}:$ $n=15, k=1$ ). Then this is easy. We have $\binom{15}{1}=15$
$\bold{Case 2}:$ $n=14, k=2$ ). Now we have something tricky. To get a good grasp on this case, let us consider the smallest element; $2$ , in the first spot. We then have $15-1=14$ numbers left. However, we cannot have the digit $3$ . Hence we have to choose $2$ numbers from $14-1=13$ integers left. This can be done in $\binom{13}{2}=78$ ways.
$\bold{Case 3}:$ $n=13, k=3$ ). As in the last case, we can use the smallest element to get a good grasp on this case. Put the digit $3$ in the first spot. Then there are $11$ integers left for the second spot. But in total, we have to avoid $2$ elements for each subset of cardinality $3$ ${a,b,c}$ . So we have to choose $3$ elements from $13-2=11$ elements, which can be done in $\binom{11}{3}=165$ ways.
$\bold{Case 4}:$ $n=12, k=4$ ). As in the previous case, we have to avoid $3$ numbers, and choose $4$ from them. Which is choosing $4$ elements from $12-3=9$ elements, which can be done in $\binom{9}{4}=126$ ways.
$\bold{Case 5}:$ $n=11, k=4$ ). Now you are accustomed to the strategy. We remove $4$ integers leaving $\binom{7}{5}=21$ ways.
Adding up all of the cases yields $405$ , as in $\boxed{405}$ | E | 405 |
513b33f8d626953c695b00ff17f4cd86 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_3 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$ | If the Cougars won by a margin of 14 points, then the Panthers' score would be half of (34-14). That's 10 $\Rightarrow \boxed{10}$ | A | 10 |
513b33f8d626953c695b00ff17f4cd86 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_3 | A football game was played between two teams, the Cougars and the Panthers. The two teams scored a total of 34 points, and the Cougars won by a margin of 14 points. How many points did the Panthers score?
$\text {(A) } 10 \qquad \text {(B) } 14 \qquad \text {(C) } 17 \qquad \text {(D) } 20 \qquad \text {(E) } 24$ | Let the Panthers' score be $x$ . The Cougars then scored $x+14$ . Since the teams combined scored $34$ , we get $x+x+14=34 \\ \rightarrow 2x+14=34 \\ \rightarrow 2x=20 \\ \rightarrow x = 10$
and the answer is $\boxed{10}$ | A | 10 |
403967e45de5ddc1ba5a6669465e16c1 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_6 | Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade. There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar. Water contains no calories. How many calories are in 200 grams of her lemonade?
$\text {(A) } 129 \qquad \text {(B) } 137 \qquad \text {(C) } 174 \qquad \text {(D) } 223 \qquad \text {(E) } 411$ | Francesca makes a total of $100+100+400=600$ grams of lemonade, and in those $600$ grams, there are $25$ calories from the lemon juice and $386$ calories from the sugar, for a total of $25+386=411$ calories per $600$ grams. We want to know how many calories there are in $200=600/3$ grams, so we just divide $411$ by $3$ to get $137\implies\boxed{137}$ | B | 137 |
1042a90387c3538dda9d8debc7eaac02 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_7 | Mr. and Mrs. Lopez have two children. When they get into their family car, two people sit in the front, and the other two sit in the back. Either Mr. Lopez or Mrs. Lopez must sit in the driver's seat. How many seating arrangements are possible?
$\text {(A) } 4 \qquad \text {(B) } 12 \qquad \text {(C) } 16 \qquad \text {(D) } 24 \qquad \text {(E) } 48$ | There are only two possible occupants for the driver's seat. After the driver is chosen, any of the remaining three people can sit in the front, and there are two arrangements for the other two people in the back. Thus, there are $2\cdot 3\cdot 2 = \boxed{12}$ possible seating arrangements. ~ aopsav (Credit to AoPS Alcumus) | null | 12 |
8a9eb72a435e01ac765a02702a96a4a8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Let the integer have digits $a$ $b$ , and $c$ , read left to right. Because $1 \leq a<b<c$ , none of the digits can be zero and $c$ cannot be 2. If $c=4$ , then $a$ and $b$ must each be chosen from the digits 1, 2, and 3. Therefore there are $\binom{3}{2}=3$ choices for $a$ and $b$ , and for each choice there is one acceptable order. Similarly, for $c=6$ and $c=8$ there are, respectively, $\binom{5}{2}=10$ and $\binom{7}{2}=21$ choices for $a$ and $b$ . Thus there are altogether $3+10+21=\boxed{34}$ such integers. | null | 34 |
8a9eb72a435e01ac765a02702a96a4a8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Let's set the middle (tens) digit first. The middle digit can be anything from 2-7 (If it was 1 we would have the hundreds digit to be 0, if it was more than 7, the ones digit cannot be even).
If it was 2, there is 1 possibility for the hundreds digit, 3 for the ones digit.
If it was 3, there are 2 possibilities for the hundreds digit, 3 for the ones digit.
If it was 4, there are 3 possibilities for the hundreds digit, and 2 for the ones digit,
and so on.
So, the answer is $3(1+2)+2(3+4)+1(5+6)=\boxed{34} \Rightarrow B$ | null | 34 |
8a9eb72a435e01ac765a02702a96a4a8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | The last digit is 4, 6, or 8.
If the last digit is $x$ , the possibilities for the first two digits correspond to 2-element subsets of $\{1,2,\dots,x-1\}$
Thus the answer is ${3\choose 2} + {5\choose 2} + {7\choose 2} = 3 + 10 + 21 = \boxed{34}$ | null | 34 |
8a9eb72a435e01ac765a02702a96a4a8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | The answer must be half of a triangular number (evens and decreasing/increasing) so $\boxed{34}$ or the letter B.
- | null | 34 |
8a9eb72a435e01ac765a02702a96a4a8 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_9 | How many even three-digit integers have the property that their digits, all read from left to right, are in strictly increasing order?
$\text {(A) } 21 \qquad \text {(B) } 34 \qquad \text {(C) } 51 \qquad \text {(D) } 72 \qquad \text {(E) } 150$ | Casework:
For the sake of simplicity, we are going to call the number $\overline{abc}$
1. If $a=1$
a. $c=2$ . No such number exists.
b. $c=4$ $b=2, 3$
c. $c=6$ $b=2, 3, 4, 5$
d. $c=8$ $b=2, 3, 4, 5, 6, 7$
2. If $a=2$ : continue as above.
We can count up that there are 34 such integers, so select $\boxed{34}$ | B | 34 |
2f42bda307d9d3aadc0a044194c4153c | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_10 | In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$ | If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43 \Rightarrow \boxed{43}$ | A | 43 |
8ca29b34d80b164013eba2122789bb7d | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_12 | The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$ -intercept $(0,-p)$ , where $p\ne 0$ . What is $b$
$\text {(A) } -p \qquad \text {(B) } 0 \qquad \text {(C) } 2 \qquad \text {(D) } 4 \qquad \text {(E) } p$ | A parabola with the given equation and with vertex $(p,p)$ must have equation $y=a(x-p)^2+p$ . Because the $y$ -intercept is $(0,-p)$ and $p\ne 0$ , it follows that $a=-2/p$ . Thus \[y=-\frac{2}{p}(x^2-2px+p^2)+p=-\frac{2}{p}x^2+4x-p,\] so $\boxed{4}$ | null | 4 |
4555d333bf0962531842c80020832769 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathrm{(B)}\ 121 \qquad \mathrm{(C)}\ 221 \qquad \mathrm{(D)}\ 230 \qquad \mathrm{(E)}\ 231$ | Let the starting point be $(0,0)$ . After $10$ steps we can only be in locations $(x,y)$ where $|x|+|y|\leq 10$ . Additionally, each step changes the parity of exactly one coordinate. Hence after $10$ steps we can only be in locations $(x,y)$ where $x+y$ is even. It can easily be shown that each location that satisfies these two conditions is indeed reachable.
Once we pick $x\in\{-10,\dots,10\}$ , we have $11-|x|$ valid choices for $y$ , giving a total of $\boxed{121}$ possible positions. | null | 121 |
4555d333bf0962531842c80020832769 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_18 | An object in the plane moves from one lattice point to another. At each step, the object may move one unit to the right, one unit to the left, one unit up, or one unit down. If the object starts at the origin and takes a ten-step path, how many different points could be the final point?
$\mathrm{(A)}\ 120 \qquad \mathrm{(B)}\ 121 \qquad \mathrm{(C)}\ 221 \qquad \mathrm{(D)}\ 230 \qquad \mathrm{(E)}\ 231$ | $10$ moves results in a lot of possible endpoints, so we try small cases first.
If the object only makes $1$ move, it is obvious that there are only 4 possible points that the object can move to.
If the object makes $2$ moves, it can move to $(0, 2)$ $(1, 1)$ $(2, 0)$ $(1, -1)$ $(0, -2)$ $(-1, -1)$ $(-2, 0)$ as well as $(0, 0)$ , for a total of $9$ moves.
If the object makes $3$ moves, it can end up at $(0, 3)$ $(2, 1)$ $(1, 2)$ $(3, 0)$ $(2, -1)$ $(1, -2)$ $(0, -3)$ $(-1, -2)$ $(-2, -1)$ $(0, -3)$ etc. for a total of $16$ moves.
At this point we can guess that for n moves, there are $(n + 1)^2$ different endpoints. Thus, for 10 moves, there are $11^2 = 121$ endpoints, and the answer is $\boxed{121}$ | B | 121 |
3e47c13ad3b758f2aa79c9fc6b7ad770 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15 \qquad \mathrm{(E)}\ \frac 14$ | Let $k$ be an arbitrary integer. For which $x$ do we have $\lfloor\log_{10}4x\rfloor = \lfloor\log_{10}x\rfloor = k$
The equation $\lfloor\log_{10}x\rfloor = k$ can be rewritten as $10^k \leq x < 10^{k+1}$ . The second one gives us $10^k \leq 4x < 10^{k+1}$ . Combining these, we get that both hold at the same time if and only if $10^k \leq x < \frac{10^{k+1}}4$
Hence for each integer $k$ we get an interval of values for which $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ . These intervals are obviously pairwise disjoint.
For any $k\geq 0$ the corresponding interval is disjoint with $(0,1)$ , so it does not contribute to our answer. On the other hand, for any $k<0$ the entire interval is inside $(0,1)$ . Hence our answer is the sum of the lengths of the intervals for $k<0$
For a fixed $k$ the length of the interval $\left[ 10^k, \frac{10^{k+1}}4 \right)$ is $\frac 32\cdot 10^k$
This means that our result is $\frac 32 \left( 10^{-1} + 10^{-2} + \cdots \right) = \frac 32 \cdot \frac 19 = \boxed{16}$ | null | 16 |
3e47c13ad3b758f2aa79c9fc6b7ad770 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_20 | Let $x$ be chosen at random from the interval $(0,1)$ . What is the probability that $\lfloor\log_{10}4x\rfloor - \lfloor\log_{10}x\rfloor = 0$ ?
Here $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$
$\mathrm{(A)}\ \frac 18 \qquad \mathrm{(B)}\ \frac 3{20} \qquad \mathrm{(C)}\ \frac 16 \qquad \mathrm{(D)}\ \frac 15 \qquad \mathrm{(E)}\ \frac 14$ | The largest value for $x$ is $10^{0}$ . If $x > 10^{-1}$ , then $\lfloor\log_{10}4x\rfloor$ doesn't fulfill the condition unless $10^{-2} \leq x < 0.25 * 10^{-1}$ . The same holds when you get smaller, because $x = 0.25 * 10^{n}$ for $n \leq 0$ is the lowest value such that $4x$ becomes a higher power of 10.
Recognize that this is a geometric sequence. The probability of choosing $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-1$ is $(9/10)* (15/90) =15/100$ , because there is a 90 percent chance of choosing $x > 10^{-1}$ , and only values of $x$ between $10^{-1}$ and $0.25*10^{0}$ work in this case. Then, for $x$ such that $\lfloor\log_{10}4x\rfloor$ and $\lfloor\log_{10}x\rfloor$ both equal $-2$ , you have $(1/10) * ((9/10) *(15/90))$ . This is a geometric series with ratio $1/10$ . Using $a/(1-r)$ for the sum of an infinite geometric sequence, we get $(15/100)/(1-(1/10)) = \boxed{16}$ | null | 16 |
acb34f4e1473c6ee18665990da67e979 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$ | The power of $10$ for any factorial is given by the well-known algorithm \[\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \cdots\] It is rational to guess numbers right before powers of $5$ because we won't have any extra numbers from higher powers of $5$ . As we list out the powers of 5, it is clear that $5^{4}=625$ is less than 2006 and $5^{5}=3125$ is greater. Therefore, set $a$ and $b$ to be 624. Thus, c is $2006-(624\cdot 2)=758$ . Applying the algorithm, we see that our answer is $152+152+188= \boxed{492}$ | null | 492 |
acb34f4e1473c6ee18665990da67e979 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_22 | Suppose $a$ $b$ and $c$ are positive integers with $a+b+c=2006$ , and $a!b!c!=m\cdot 10^n$ , where $m$ and $n$ are integers and $m$ is not divisible by $10$ . What is the smallest possible value of $n$
$\mathrm{(A)}\ 489 \qquad \mathrm{(B)}\ 492 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 498 \qquad \mathrm{(E)}\ 501$ | Clearly, the power of $2$ that divides $n!$ is larger or equal than the power of $5$ which divides
it. Hence we are trying to minimize the power of $5$ that will divide $a!b!c!$
Consider $n! = 1\cdot 2 \cdot \dots \cdot n$ . Each fifth term is divisible by $5$ , each $25$ -th one
by $25$ , and so on. Hence the total power of $5$ that divides $n$ is $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \cdots$ . (For any $n$ only finitely many terms in the sum
are
non-zero.)
In our case we have $a<2006$ , so the largest power of $5$ that will be less than $a$ is at most $5^4 = 625$ . Therefore the power of $5$ that divides $a!$ is equal to $\left\lfloor \frac a{5}\right\rfloor + \left\lfloor \frac a{25}\right\rfloor + \left\lfloor \frac a{125}\right\rfloor + \left\lfloor \frac a{625}\right\rfloor$ . The same
is true for $b$ and $c$
Intuition may now try to lure us to split $2006$ into $a+b+c$ as evenly as possible, giving $a=b=669$ and $c=668$ . However, this solution is not optimal.
To see how we can do better, let's rearrange the terms as follows:
\begin{align*} \text{result} & = \Big\lfloor \frac a{5}\Big\rfloor + \Big\lfloor \frac b{5}\Big\rfloor + \Big\lfloor \frac c{5}\Big\rfloor \\ & + \Big\lfloor \frac a{25}\Big\rfloor + \Big\lfloor \frac b{25}\Big\rfloor + \Big\lfloor \frac c{25}\Big\rfloor \\ & + \Big\lfloor \frac a{125}\Big\rfloor + \Big\lfloor \frac b{125}\Big\rfloor + \Big\lfloor \frac c{125}\Big\rfloor \\ & + \Big\lfloor \frac a{625}\Big\rfloor + \Big\lfloor \frac b{625}\Big\rfloor + \Big\lfloor \frac c{625}\Big\rfloor \end{align*}
The idea is that the rows of the above equation are roughly equal to $\left\lfloor \frac n{5}\right\rfloor$ $\left\lfloor \frac n{25}\right\rfloor$ , etc.
More precisely, we can now notice that for any positive integers $a,b,c,k$ we can write $a,b,c$ in the form $a=a_0k + a_1$ $b=b_0k+b_1$ $c=c_0k + c_1$ , where all $a_i,b_i,c_i$ are integers and $0\leq a_1,b_1,c_1<k$
It follows that \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor = a_0+b_0+c_0\] and \[\Big\lfloor \frac {a+b+c}k\Big\rfloor = a_0 + b_0 + c_0 + \Big\lfloor \frac {a_1+b_1+c_1}k\Big\rfloor \leq a_0 + b_0 + c_0 + 2\]
Hence we get that for any positive integers $a,b,c,k$ we have \[\Big\lfloor \frac a{k}\Big\rfloor + \Big\lfloor \frac b{k}\Big\rfloor + \Big\lfloor \frac c{k}\Big\rfloor \quad \geq \quad \Big\lfloor \frac {a+b+c}k\Big\rfloor - 2\]
Therefore for any $a,b,c$ the result is at least $\left\lfloor \frac n{5}\right\rfloor + \left\lfloor \frac n{25}\right\rfloor + \left\lfloor \frac n{125}\right\rfloor + \left\lfloor \frac n{625}\right\rfloor - 8 = 401 + 80 + 16 + 3 - 8 = 500 - 8 = 492$
If we now show how to pick $a,b,c$ so that we'll get the result $492$ , we will be done.
Consider the row with $625$ in the denominator. We need to achieve sum $1$ in this row,
hence we need to make two of the numbers smaller than $625$ . Choosing $a=b=624$ does this, and it will give us the largest possible remainders for $a$ and $b$ in
the other three rows, so this is a pretty good candidate. We can compute $c=2006-a-b=758$ and verify that this triple gives the desired result $\boxed{492}$ | null | 492 |
3647e641244147876ca7e15fa1535d27 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$ | [asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("\alpha",C,5*dir(80),f); MP("90^\circ-\alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); [/asy] Using the Law of Cosines on $\triangle PBC$ , we have:
\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}
Using the Law of Cosines on $\triangle PAC$ , we have: \begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}
Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$ \begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*}
Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$ . Thus, $a+b=85+42=\boxed{127}$ | null | 127 |
3647e641244147876ca7e15fa1535d27 | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_23 | Isosceles $\triangle ABC$ has a right angle at $C$ . Point $P$ is inside $\triangle ABC$ , such that $PA=11$ $PB=7$ , and $PC=6$ . Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$ , where $a$ and $b$ are positive integers. What is $a+b$
[asy] pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); [/asy]
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$ | Let point $P$ have coordinates $(x,y)$ and $C$ have coordinates $(0,0).$ Then, $A$ has $(s,0)$ and $B$ has $(0,s)$
By distance formula, we have \[x^2+y^2=36 \tag{1}.\] \[x^2+(y-s)^2=49 \tag{2}.\] \[(x-s)^2+y^2=121 \tag{3}\]
Expanding $(2)$ and $(3)$ gives \[x^2+y^2-2ys+s^2=49,\] and \[x^2+y^2-2sx+s^2=121,\] respectively. Then, using equation $(1)$ , we have that \[-2ys+s^2=49-36=13,\] and \[-2sx+s^2=121-36=85.\]
Then, solving for $x$ and $y$ gives \[x=\frac{85-s^2}{-2s}=\frac{s^2-85}{2s},\] and \[y=\frac{13-s^2}{-2s}=\frac{s^2-13}{2s}.\] Plugging these values of $x$ and $y$ into equation $(1)$ yields \[\left(\frac{s^2-85}{2s}\right)^2+\left(\frac{s^2-13}{2s}\right)^2=36.\]
We multiply both sides by $4s^2$ and expand, yielding the equation \[s^4-170s^2+7225+s^4-26s^2+169=144s^2.\] Simplifying gives the equation \[2s^4-340s^2+7394=0.\] Solving this quadratic gives $s^2=85\pm 42\sqrt{2}.$ Now, if this were the actual test, we stop here, noting that the question tells us $a$ and $b$ are positive, so $s^2$ must be $85+42\sqrt{2}$ , and our solution is $127$
However, here is why $s^2$ cannot be $85-42\sqrt{2}$
If $s^2=85-42\sqrt{2}$ , using $1.4$ as an approximation for $\sqrt{2}$ $s^2\approx 85-42\cdot 1.4=85-58.8=26.2$ , and $s$ is slightly greater than $5$ . Also note that this implies that $AB\approx 7$
Note that at least one of $\angle BPA$ $\angle APC$ and $\angle BPC$ must be obtuse, since they sum to $360^{\circ}$ . Then, note the well known fact that if $\angle A$ is the largest angle in $\Delta ABC$ $BC$ must be the largest side. However, combined with the first fact, implies that either $AB$ is the largest side of $\Delta ABP$ $BC$ is the largest side of $\Delta BPC$ , or $AC$ is the largest side of $\Delta APC$ . By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so $s^2$ cannot equal $85-42\sqrt{2}$ , and $s^2=85+42\sqrt{2}$
The answer is $85+42=\boxed{127}$ | null | 127 |
fd75d1c59c5b9b3e560d7d82aa740ffc | https://artofproblemsolving.com/wiki/index.php/2006_AMC_12B_Problems/Problem_25 | A sequence $a_1,a_2,\dots$ of non-negative integers is defined by the rule $a_{n+2}=|a_{n+1}-a_n|$ for $n\geq 1$ . If $a_1=999$ $a_2<999$ and $a_{2006}=1$ , how many different values of $a_2$ are possible?
$\mathrm{(A)}\ 165 \qquad \mathrm{(B)}\ 324 \qquad \mathrm{(C)}\ 495 \qquad \mathrm{(D)}\ 499 \qquad \mathrm{(E)}\ 660$ | We say the sequence $(a_n)$ completes at $i$ if $i$ is the minimal positive integer such that $a_i = a_{i + 1} = 1$ . Otherwise, we say $(a_n)$ does not complete.
Note that if $d = \gcd(999, a_2) \neq 1$ , then $d|a_n$ for all $n \geq 1$ , and $d$ does not divide $1$ , so if $\gcd(999, a_2) \neq 1$ , then $(a_n)$ does not complete. (Also, $a_{2006}$ cannot be 1 in this case since $d$ does not divide $1$ , so we do not care about these $a_2$ at all.)
From now on, suppose $\gcd(999, a_2) = 1$
We will now show that $(a_n)$ completes at $i$ for some $i \leq 2006$ . We will do this with 3 lemmas.
Lemma: If $a_j \neq a_{j + 1}$ , and neither value is $0$ , then $\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})$
Proof: There are 2 cases to consider.
If $a_j > a_{j + 1}$ , then $a_{j + 2} = a_j - a_{j + 1}$ , and $a_{j + 3} = |a_j - 2a_{j + 1}|$ . So $a_j > a_{j + 2}$ and $a_j > a_{j + 3}$
If $a_j < a_{j + 1}$ , then $a_{j + 2} = a_{j + 1} - a_j$ , and $a_{j + 3} = a_j$ . So $a_{j + 1} > a_{j + 2}$ and $a_{j + 1} > a_{j + 3}$
In both cases, $\max(a_j, a_{j + 1}) > \max(a_{j + 2}, a_{j + 3})$ , as desired.
Lemma: If $a_i = a_{i + 1}$ , then $a_i = 1$ . Moreover, if instead we have $a_i = 0$ for some $i > 2$ , then $a_{i - 1} = a_{i - 2} = 1$
Proof: By the way $(a_n)$ is constructed in the problem statement, having two equal consecutive terms $a_i = a_{i + 1}$ implies that $a_i$ divides every term in the sequence. So $a_i | 999$ and $a_i | a_2$ , so $a_i | \gcd(999, a_2) = 1$ , so $a_i = 1$ . For the proof of the second result, note that if $a_i = 0$ , then $a_{i - 1} = a_{i - 2}$ , so by the first result we just proved, $a_{i - 2} = a_{i - 1} = 1$
Lemma: $(a_n)$ completes at $i$ for some $i \leq 2000$
Proof: Suppose $(a_n)$ completed at some $i > 2000$ or not at all. Then by the second lemma and the fact that neither $999$ nor $a_2$ are $0$ , none of the pairs $(a_1, a_2), ..., (a_{1999}, a_{2000})$ can have a $0$ or be equal to $(1, 1)$ . So the first lemma implies \[\max(a_1, a_2) > \max(a_3, a_4) > \cdots > \max(a_{1999}, a_{2000}) > 0,\] so $999 = \max(a_1, a_2) \geq 1000$ , a contradiction. Hence $(a_n)$ completes at $i$ for some $i \leq 2000$
Now we're ready to find exactly which values of $a_2$ we want to count.
Let's keep in mind that $2006 \equiv 2 \pmod 3$ and that $a_1 = 999$ is odd. We have two cases to consider.
Case 1: If $a_2$ is odd, then $a_3$ is even, so $a_4$ is odd, so $a_5$ is odd, so $a_6$ is even, and this pattern must repeat every three terms because of the recursive definition of $(a_n)$ , so the terms of $(a_n)$ reduced modulo 2 are \[1, 1, 0, 1, 1, 0, ...,\] so $a_{2006}$ is odd and hence $1$ (since if $(a_n)$ completes at $i$ , then $a_k$ must be $0$ or $1$ for all $k \geq i$ ).
Case 2: If $a_2$ is even, then $a_3$ is odd, so $a_4$ is odd, so $a_5$ is even, so $a_6$ is odd, and this pattern must repeat every three terms, so the terms of $(a_n)$ reduced modulo 2 are \[1, 0, 1, 1, 0, 1, ...,\] so $a_{2006}$ is even, and hence $0$
We have found that $a_{2006} = 1$ is true precisely when $\gcd(999, a_2) = 1$ and $a_2$ is odd. This tells us what we need to count.
There are $\phi(999) = 648$ numbers less than $999$ and relatively prime to it ( $\phi$ is the Euler totient function). We want to count how many of these are odd. Note that \[t \mapsto 999 - t\] is a 1-1 correspondence between the odd and even numbers less than and relatively prime to $999$ . So our final answer is $648/2 = 324$ , or $\boxed{324}$ | B | 324 |
731fbc455875635f207169d822fa7463 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_2 | The equations $2x + 7 = 3$ and $bx - 10 = - 2$ have the same solution. What is the value of $b$
$\textbf {(A)} -8 \qquad \textbf{(B)} -4 \qquad \textbf {(C) } 2 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 8$ | $2x + 7 = 3 \Longrightarrow x = -2, \quad -2b - 10 = -2 \Longrightarrow -2b = 8 \Longrightarrow b = \boxed{4}$ | B | 4 |
7aac10203524b2f2611933baa5c9b2bb | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_12 | line passes through $A\ (1,1)$ and $B\ (100,1000)$ . How many other points with integer coordinates are on the line and strictly between $A$ and $B$
$(\mathrm {A}) \ 0 \qquad (\mathrm {B}) \ 2 \qquad (\mathrm {C})\ 3 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 9$ | The slope of the line is $\frac{1000-1}{100-1}=\frac{111}{11},$ so all points on the line have the form $(1+11t, 1+111t)$ for some value of $t$ (the rise is 111 and the run is 11). Such a point has integer coordinates if and only if $t$ is an integer, and the point is strictly between $A$ and $B$ if and only if $0<t<9$ . Thus, there are $\boxed{8}$ points with the required property.
-Paixiao | null | 8 |
9ebb3b47e65e54f13ad06e090dd8df0f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_13 | In the five-sided star shown, the letters $A$ $B$ $C$ $D$ and $E$ are replaced by the
numbers 3, 5, 6, 7 and 9, although not necessarily in that order. The sums of the
numbers at the ends of the line segments $\overline{AB}$ $\overline{BC}$ $\overline{CD}$ $\overline{DE}$ , and $\overline{EA}$ form an
arithmetic sequence, although not necessarily in that order. What is the middle
term of the arithmetic sequence?
[asy] draw((0,0)--(0.5,1.54)--(1,0)--(-0.31,0.95)--(1.31,0.95)--cycle); label("$A$",(0.5,1.54),N); label("$B$",(1,0),SE); label("$C$",(-0.31,0.95),W); label("$D$",(1.31,0.95),E); label("$E$",(0,0),SW); [/asy]
$(\mathrm {A}) \ 9 \qquad (\mathrm {B}) \ 10 \qquad (\mathrm {C})\ 11 \qquad (\mathrm {D}) \ 12 \qquad (\mathrm {E})\ 13$ | Let the terms in the arithmetic sequence be $a$ $a + d$ $a + 2d$ $a + 3d$ , and $a + 4d$ . We seek the middle term $a + 2d$
These five terms are $A + B$ $B + C$ $C + D$ $D + E$ , and $E + A$ , in some order. The numbers $A$ $B$ $C$ $D$ , and $E$ are equal to 3, 5, 6, 7, and 9, in some order, so \[A + B + C + D + E = 3 + 5 + 6 + 7 + 9 = 30.\] Hence, the sum of the five terms is \[(A + B) + (B + C) + (C + D) + (D + E) + (E + A) = 2A + 2B + 2C + 2D + 2E = 60.\] But adding all five numbers, we also get $a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d$ , so \[5a + 10d = 60.\] Dividing both sides by 5, we get $a + 2d = \boxed{12}$ , which is the middle term. The answer is (D). | null | 12 |
66435d692b2b26a9f72c639a02455e96 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_19 | A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$ | Alternatively, consider that counting without the number $4$ is almost equivalent to counting in base $9$ ; only, in base $9$ , the number $9$ is not counted. Since $4$ is skipped, the symbol $5$ represents $4$ miles of travel, and we have traveled $2004_9$ miles. By basic conversion, $2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}$ | null | 1462 |
66435d692b2b26a9f72c639a02455e96 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_19 | A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005 , how many miles has the car actually traveled? $(\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804$ | This is very analogous to base $9$ . But, in base $9$ , we don't have a $9$ . So, this means that these are equal except for that base 9 will be one more than the operation here. $2005_9 = 5+0+0+1458 = 1463$ $1463 - 1 = 1462$
Therefore, our answer is $\boxed{1462}$ | null | 1462 |
d98886a57fb620435fa86d73f0a6ac15 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_22 | A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$ | Box P has dimensions $l$ $w$ , and $h$ .
Its surface area is \[2lw+2lh+2wh=384,\] and the sum of all its edges is \[l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.\]
The diameter of the sphere is the space diagonal of the prism, which is \[\sqrt{l^2 + w^2 +h^2}.\] Notice that \[(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,\] so the diameter is \[\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.\] The radius is half of the diameter, so \[r=\frac{20}{2} = \boxed{10}.\] | B | 10 |
d98886a57fb620435fa86d73f0a6ac15 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_22 | A rectangular box $P$ is inscribed in a sphere of radius $r$ . The surface area of $P$ is 384, and the sum of the lengths of its 12 edges is 112. What is $r$
$\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16$ | As in the previous solution, we have that $2lw+2lh+2wh=384$ and $l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28$ , and the diameter of the sphere is the space diagonal of the prism, $\sqrt{l^2 + w^2 + h^2}$
Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that $h=0$ . (This essentially means that we have an infinitesimally thin box.) We now have that $2lw = 384$ and $l + w = 28$ , and we are solving for $\sqrt{l^2 + w^2}$ . Because \[(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,\] this means that \[l^2 + w^2 = 28^2 - 384 = 400,\] so the space diagonal is $\sqrt{400} = 20$ . Since the diameter of the sphere is $20$ , the radius is $\boxed{10}$ . ~ emerald_block | B | 10 |
d93a957b5f4feae20329be4ab8b70744 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_24 | Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$ | We can write the problem as
Since $\deg P(x) = 3$ and $\deg R(x) = 3$ $\deg P(x)\cdot R(x) = 6$ . Thus, $\deg P(Q(x)) = 6$ , so $\deg Q(x) = 2$
Hence, we conclude $Q(1)$ $Q(2)$ , and $Q(3)$ must each be $1$ $2$ , or $3$ . Since a quadratic is uniquely determined by three points, there can be $3*3*3 = 27$ different quadratics $Q(x)$ after each of the values of $Q(1)$ $Q(2)$ , and $Q(3)$ are chosen.
However, we have included $Q(x)$ which are not quadratics: lines. Namely,
Clearly, we could not have included any other constant functions. For any linear function, we have $2\cdot Q(2) = Q(1) + Q(3)$ because $Q(2)$ is y-value of the midpoint of $(1, Q(1))$ and $(3, Q(3))$ . So we have not included any other linear functions. Therefore, the desired answer is $27 - 5 = \boxed{22}$ | B | 22 |
d93a957b5f4feae20329be4ab8b70744 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_24 | Let $P(x)=(x-1)(x-2)(x-3)$ . For how many polynomials $Q(x)$ does there exist a polynomial $R(x)$ of degree 3 such that $P(Q(x))=P(x) \cdot R(x)$
$\mathrm {(A) } 19 \qquad \mathrm {(B) } 22 \qquad \mathrm {(C) } 24 \qquad \mathrm {(D) } 27 \qquad \mathrm {(E) } 32$ | We see that \[P(Q(x))=(Q(x)-1)(Q(x)-2)(Q(x)-3)=P(x)\cdot R(x)=(x-1)(x-2)(x-3)\cdot R(x).\] Therefore, $P(x) | P(Q(x))$ . Since $\deg Q = 2,$ we must have $x-1, x-2, x-3$ divide $P(Q(x))$ . So, we pair them off with one of $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ to see that there are $3!+3 \cdot 2 \cdot \binom{3}{2} = 24$ without restrictions. (Note that this count was made by pairing off linear factors of $P(x)$ with $Q(x)-1, Q(x)-2,$ and $Q(x)-3$ , and also note that the degree of $Q$ is 2.) However, we have two functions which are constant, which are $Q(x) = x$ and $Q(x) = 4-x.$ So, we subtract $2$ to get a final answer of $\boxed{22}$ | B | 22 |
f4d3d6656de3078a339bb86429d5bbcf | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12A_Problems/Problem_25 | Let $S$ be the set of all points with coordinates $(x,y,z)$ , where $x$ $y$ , and $z$ are each chosen from the set $\{0,1,2\}$ . How many equilateral triangles all have their vertices in $S$
$(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88$ | For this solution, we will just find as many solutions as possible, until it becomes intuitive that there are no more size of triangles left.
First, try to make three of its vertices form an equilateral triangle. This we find is possible by taking any vertex , and connecting the three adjacent vertices into a triangle. This triangle will have a side length of $\sqrt{2}$ ; a quick further examination of this cube will show us that this is the only possible side length (red triangle in diagram). Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube (green triangle), giving us 9 cubes and $9 \cdot 8 = 72$ equilateral triangles.
NOTE: Connecting the centers of the faces will actually give an octahedron , not a cube, because it only has $6$ vertices.
Now, we look for any additional equilateral triangles. Connecting the midpoints of three non-adjacent, non-parallel edges also gives us more equilateral triangles (blue triangle). Notice that picking these three edges leaves two vertices alone (labelled A and B), and that picking any two opposite vertices determine two equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, for a total of $\boxed{80}$ | C | 80 |
dddb5ca3f68507e8e9d61fc5f6558fa4 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_1 | A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?
$\textbf{(A) }\ 100 \qquad \textbf{(B) }\ 200 \qquad \textbf{(C) }\ 300 \qquad \textbf{(D) }\ 400 \qquad \textbf{(E) }\ 500$ | \begin{align*} \mbox{Expenses} &= 1000 \cdot \frac25 = 400 \\ \mbox{Revenue} &= 1000 \cdot \frac12 = 500 \\ \mbox{Profit} &= \mbox{Revenue} - \mbox{Expenses} = 500-400 = \boxed{100} Note: Revenue is a gain. | A | 100 |
dddb5ca3f68507e8e9d61fc5f6558fa4 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_1 | A scout troop buys $1000$ candy bars at a price of five for $2$ dollars. They sell all the candy bars at the price of two for $1$ dollar. What was their profit, in dollars?
$\textbf{(A) }\ 100 \qquad \textbf{(B) }\ 200 \qquad \textbf{(C) }\ 300 \qquad \textbf{(D) }\ 400 \qquad \textbf{(E) }\ 500$ | Note that the troop buys $10$ candy bars at a price of $4$ dollars and sells $10$ bars at a price of $5$ dollars. So the troop gains $1$ dollar for every $10$ bars. So therefore we divide $1000 \div 10 = 100$ . So our answer is $\boxed{100}$ .
~HyperVoid | A | 100 |
553504b8886643993fa1386ac1143080 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_2 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Since $x\%$ means $0.01x$ , the statement " $x\% \text{ of } x \text{ is 4}$ " can be rewritten as " $0.01x \cdot x = 4$ ":
$0.01x \cdot x=4 \Rightarrow x^2 = 400 \Rightarrow x = \boxed{20}.$ | D | 20 |
553504b8886643993fa1386ac1143080 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_2 | A positive number $x$ has the property that $x\%$ of $x$ is $4$ . What is $x$
$\textbf{(A) }\ 2 \qquad \textbf{(B) }\ 4 \qquad \textbf{(C) }\ 10 \qquad \textbf{(D) }\ 20 \qquad \textbf{(E) }\ 40$ | Try the answer choices one by one. Upon examination, it is quite obvious that the answer is $\boxed{20}.$ Very fast. | D | 20 |
275ff2654b44fc16cc4242aa9038e283 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_3 | Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?
$\textbf{(A) }\ \frac15 \qquad\textbf{(B) }\ \frac13 \qquad\textbf{(C) }\ \frac25 \qquad\textbf{(D) }\ \frac23 \qquad\textbf{(E) }\ \frac45$ | Let $m =$ Brianna's money. We have $\frac15 m = \frac13 (\mbox{CDs}) \Rightarrow \frac35 m = (\mbox{CDs})$ . Thus, the money left over is $m-\frac35m = \frac25m$ , so the answer is $\boxed{25}$ | C | 25 |
7063bbb7bf2c1f63109516cd99a8ca7e | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_4 | At the beginning of the school year, Lisa's goal was to earn an $A$ on at least $80\%$ of her $50$ quizzes for the year. She earned an $A$ on $22$ of the first $30$ quizzes. If she is to achieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an $A$
$\textbf{(A) }\ 1 \qquad \textbf{(B) }\ 2 \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | Lisa's goal was to get an $A$ on $80\% \cdot 50 = 40$ quizzes. She already has $A$ 's on $22$ quizzes, so she needs to get $A$ 's on $40-22=18$ more. There are $50-30=20$ quizzes left, so she can afford to get less than an $A$ on $20-18=\boxed{2}$ of them. | B | 2 |
37222bfdd094fbbfa5badd090271b894 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Draw height $CH$ (Perpendicular line from point C to line AD). We have that $BH=1$ . By the Pythagorean Theorem $CH=\sqrt{48}$ . Since $CD=8$ $HD=\sqrt{8^2-48}=\sqrt{16}=4$ , and $BD=HD-1$ , so $BD=\boxed{3}$ | A | 3 |
37222bfdd094fbbfa5badd090271b894 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | After drawing out a diagram, let $\angle{ABC}=\theta$ . By the Law of Cosines, $7^2=2^2+7^2-2(7)(2)\cos{\theta} \rightarrow 0=4-28\cos{\theta} \rightarrow \cos{\theta}=\frac{1}{7}$ . In $\triangle CBD$ , we have $\angle{CBD}=(180-\theta)$ , and using the identity $\cos(180-\theta)=-\cos{\theta}$ and Law of Cosines one more time: $8^2=7^2+x^2-2(7)(x)\left( \frac{-1}{7} \right) \rightarrow 64=49+x^2+2x \rightarrow x^2+2x-15=0$ . The only positive value for $x$ is $3$ , which gives the length of $\overline{BD}$ . Thus the answer is $\boxed{3}$ | A | 3 |
37222bfdd094fbbfa5badd090271b894 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_6 | In $\triangle ABC$ , we have $AC=BC=7$ and $AB=2$ . Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$ . What is $BD$
$\textbf{(A) }\ 3 \qquad \textbf{(B) }\ 2\sqrt{3} \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 5 \qquad \textbf{(E) }\ 4\sqrt{2}$ | Let $BD=k$ . Then, by Stewart's Theorem
$2k(2+k)+7^2(2+k)=7^2k+8^2\cdot 2 \implies k^2+2k-15=0 \implies k=\boxed{3}$ | A | 3 |
79ffa12b6acbade1daa9e229121e0e15 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_7 | What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$ , then $a$ is either $b$ or $-b$ ):
\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}
We can then put these equations in slope-intercept form in order to graph them.
\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}
Now you can graph the lines to find the shape of the graph:
[asy] Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 4.0)); yaxis(-6,6,Ticks(f, 3.0)); fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); draw((-4,-6)--(8,3), Arrows(4)); draw((4,-6)--(-8,3), Arrows(4)); draw((-4,6)--(8,-3), Arrows(4)); draw((4,6)--(-8,-3), Arrows(4));[/asy]
We can easily see that it is a rhombus with diagonals of $6$ and $8$ . The area is $\dfrac{1}{2}\times 6\times8$ , or $\boxed{24}$ | D | 24 |
79ffa12b6acbade1daa9e229121e0e15 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_7 | What is the area enclosed by the graph of $|3x|+|4y|=12$
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is
\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{24}\] | D | 24 |
df5d798b823ce4cd31848833cd2dc1a4 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_8 | For how many values of $a$ is it true that the line $y = x + a$ passes through the
vertex of the parabola $y = x^2 + a^2$
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 2 \qquad \mathrm{(D)}\ 10 \qquad \mathrm{(E)}\ \text{infinitely many}$ | We see that the vertex of the quadratic function $y = x^2 + a^2$ is $(0,\,a^2)$ . The y-intercept of the line $y = x + a$ is $(0,\,a)$ . We want to find the values (if any) such that $a=a^2$ . Solving for $a$ , the only values that satisfy this are $0$ and $1$ , so the answer is $\boxed{2}$ | C | 2 |
c36e64c556ed41355d318c05d3b60aa8 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_9 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | To begin, we see that the remaining $30\%$ of the students got $95$ points. Assume that there are $20$ students; we see that $2$ students got $70$ points, $5$ students got $80$ points, $4$ students got $85$ points, $3$ students got $90$ points, and $6$ students got $95$ points. The median is $85$ , since the $10^{\text{th}}$ and $11^{\text{th}}$ terms are both $85$ . The mean is $\dfrac{70\,(2)+80\,(5)+85\,(4)+90\,(3)+95\,(6)}{20}=\dfrac{1720}{20}=86$ . The difference between the mean and median, therefore, is $\boxed{1}$ | B | 1 |
c36e64c556ed41355d318c05d3b60aa8 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_9 | On a certain math exam, $10\%$ of the students got $70$ points, $25\%$ got $80$ points, $20\%$ got $85$ points, $15\%$ got $90$ points, and the rest got $95$ points. What is the difference between the mean and the median score on this exam?
$\textbf{(A) }\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 4 \qquad \textbf{(E) }\ 5$ | The remaining $30\%$ of the students got $95$ points.
The mean is equal to $10\%\cdot70 + 25\%\cdot80 + 20\%\cdot85 + 15\%\cdot90 + 30\%\cdot95 = 86$ .
The score greater than or equal to $50\%$ of other scores is the median. Since $35\%$ scored $80$ or lower and the next $20\%$ scored $85$ , the median is $85$ . The difference between the mean and the median is $\boxed{1}$ | B | 1 |
0f8078a99f52bcb5de9a1ac4c7357156 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_10 | The first term of a sequence is $2005$ . Each succeeding term is the sum of the cubes of the digits of the previous term. What is the ${2005}^{\text{th}}$ term of the sequence?
$\textbf{(A) } 29 \qquad \textbf{(B) } 55 \qquad \textbf{(C) } 85 \qquad \textbf{(D) } 133 \qquad \textbf{(E) } 250$ | Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$ , a complete cycle. The cycle is, excluding the first term, the $2^{\text{nd}}$ $3^{\text{rd}}$ , and $4^{\text{th}}$ terms will equal $133$ $55$ , and $250$ , following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$ . It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$ , which leads us to the answer of $\boxed{250}$ | E | 250 |
8b12bf1aec82c699bbba2f4bc626be0c | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_12 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | Let $x^2 + px + m = 0$ have roots $a$ and $b$ . Then
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$ . Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$ , so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$ . Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{8}$ | D | 8 |
8b12bf1aec82c699bbba2f4bc626be0c | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_12 | The quadratic equation $x^2+mx+n$ has roots twice those of $x^2+px+m$ , and none of $m,n,$ and $p$ is zero. What is the value of $n/p$
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$ | If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$ , then using Vieta's formulas, \[2a + 2b = -m\] \[a + b = -p\] \[2a(2b) = n\] \[a(b) = m\] Therefore, substituting the second equation into the first equation gives \[m = 2(p)\] and substituting the fourth equation into the third equation gives \[n = 4(m)\] Therefore, $n = 8p$ , so $\frac{n}{p}= \boxed{8}$ | D | 8 |
3329a1a0c2b65b4bacb41acd2c9e9179 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | $221$ can be written as the sum of four two-digit numbers, let's say $\overline{ae}$ $\overline{bf}$ $\overline{cg}$ , and $\overline{dh}$ . Then $221= 10(a+b+c+d)+(e+f+g+h)$ . The last digit of $221$ is $1$ , and $10(a+b+c+d)$ won't affect the units digits, so $(e+f+g+h)$ must end with $1$ . The smallest value $(e+f+g+h)$ can have is $(1+2+3+4)=10$ , and the greatest value is $(6+7+8+9)=30$ . Therefore, $(e+f+g+h)$ must equal $11$ or $21$
Case 1: $(e+f+g+h)=11$
The only distinct positive integers that can add up to $11$ is $(1+2+3+5)$ . So, $a$ $b$ $c$ , and $d$ must include four of the five numbers $(4,6,7,8,9)$ . We have $10(a+b+c+d)=221-11=210$ , or $a+b+c+d=21$ . We can add all of $4+6+7+8+9=34$ , and try subtracting one number to get to $21$ , but none of them work. Therefore, $(e+f+g+h)$ cannot add up to $11$
Case 2: $(e+f+g+h)=21$
Checking all the values for $e$ $f$ $g$ ,and $h$ each individually may be time-consuming, instead of only having $1$ solution like Case 1. We can try a different approach by looking at $(a+b+c+d)$ first. If $(e+f+g+h)=21$ $10(a+b+c+d)=221-21=200$ , or $(a+b+c+d)=20$ . That means $(a+b+c+d)+(e+f+g+h)=21+20=41$ . We know $(1+2+3+4+5+6+7+8+9)=45$ , so the missing digit is $45-41=\boxed{4}$ | D | 4 |
3329a1a0c2b65b4bacb41acd2c9e9179 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_15 | The sum of four two-digit numbers is $221$ . None of the eight digits is $0$ and no two of them are the same. Which of the following is not included among the eight digits?
$\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5$ | Alternatively, we know that a number is congruent to the sum of its digits mod 9, so $221 \equiv 5 \equiv 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 - d \equiv -d$ , where $d$ is some digit. Clearly, $\boxed{4}$ | null | 4 |
37d210c73529676ea24a87574cd75cc2 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_17 | How many distinct four-tuples $(a,b,c,d)$ of rational numbers are there with
\[a\cdot\log_{10}2+b\cdot\log_{10}3+c\cdot\log_{10}5+d\cdot\log_{10}7=2005?\]
$\mathrm{(A)}\ 0 \qquad \mathrm{(B)}\ 1 \qquad \mathrm{(C)}\ 17 \qquad \mathrm{(D)}\ 2004 \qquad \mathrm{(E)}\ \text{infinitely many}$ | Using the laws of logarithms , the given equation becomes
\[\log_{10}2^{a}+\log_{10}3^{b}+\log_{10}5^{c}+\log_{10}7^{d}=2005\] \[\Rightarrow \log_{10}{2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}}=2005\] \[\Rightarrow 2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d} = 10^{2005}\]
As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$ $b=d=0$ . Then $2^{a}\cdot 5^{c}=2^{2005}\cdot 5^{2005} \Rightarrow a=c=2005$
Only the four-tuple $(2005,0,2005,0)$ satisfies the equation, so the answer is $\boxed{1}$ | B | 1 |
3dd77aa13d02f2f3e719679f06cf1263 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_18 | Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 39 \qquad \mathrm{(C)}\ 51 \qquad \mathrm{(D)}\ 60 \qquad \mathrm{(E)}\ 80$ | [asy] Label f; f.p=fontsize(6); xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0)); pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE); D(A--B); filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray); filldraw(CP(0.5(A+B),A),white); D(A); D(B); [/asy]
For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$ . For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\overline{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$ . Hence, the area of $R$ is the area of the large triangle minus the area of the small triangle minus the area of the circle, which is $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$ , which is approximately $51$ . The answer is $\boxed{51}$ | C | 51 |
dc179708fcad608b6535a75f799f5d82 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_20 | Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$ | The sum of the set is $-7-5-3-2+2+4+6+13=8$ , so if we could have the sum in each set of parenthesis be $4$ then the minimum value would be $2(4^2)=32$ . Considering the set of four terms containing $13$ , this sum could only be even if it had two or four odd terms. If it had all four odd terms then it would be $13-7-5-3=-2$ , and with two odd terms then its minimum value is $13-7+2-2=6$ , so we cannot achieve two sums of $4$ . The closest we could have to $4$ and $4$ is $3$ and $5$ , which can be achieved through $13-7-5+2$ and $6-3-2+4$ . So the minimum possible value is $3^2+5^2=34\Rightarrow\boxed{34}$ | C | 34 |
dc179708fcad608b6535a75f799f5d82 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_20 | Let $a,b,c,d,e,f,g$ and $h$ be distinct elements in the set $\{-7,-5,-3,-2,2,4,6,13\}.$
What is the minimum possible value of $(a+b+c+d)^{2}+(e+f+g+h)^{2}?$
$\mathrm{(A)}\ 30 \qquad \mathrm{(B)}\ 32 \qquad \mathrm{(C)}\ 34 \qquad \mathrm{(D)}\ 40 \qquad \mathrm{(E)}\ 50$ | Trying out the values for a bit leads to us getting $34$ with $(-3, -2, 2, 6)$ in 1 set for a total of 9 and the other numbers giving a total of 25.
We start off with trying to get a $2(4^2) = 32$ solution, so we only need to find one set with a sum of 4 (the other will automatically have a sum of 4).
We can add 7 to each number. We are now trying to get a set with 32.
\[(0, 2, 4, 5, 9, 11, 13, 20)\]
Observe that there are 4 odd and 4 even values. To get 32 from 4 numbers $a,b,c,d$ , the numbers either have to be 4 odds, 4 evens, or 2 odds and evens. The 4 odd numbers do not work (38) nor the 4 even numbers (26).
If $a,b,c,d$ has 2 odd numbers and 2 even numbers we can divide them into two cases: whether $a,b,c,d$ has 20 or doesn't have 20.
Case 1: $a,b,c,d$ doesn't have 20.
The maximum sum of the 2 even numbers are 2 + 4 = 6 and the maximum sum of the 2 odd numbers are 11 + 13 = 24. This is not enough to reach the 32 we need.
Case 2: $a,b,c,d$ has 20.
The minimum sum of the 2 even numbers are 20 + 0 = 20 and the minimum sum of the 2 odd numbers are 5 + 9 = 14. This is already greater than the 32 we need.
We have proven that we cannot partition the 8 numbers into two groups of 4 with the same sum, so the smallest value is $\boxed{34}$ | C | 34 |
c02b3a9978c426974348912ed4343238 | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_21 | A positive integer $n$ has $60$ divisors and $7n$ has $80$ divisors. What is the greatest integer $k$ such that $7^k$ divides $n$
$\mathrm{(A)}\ {{{0}}} \qquad \mathrm{(B)}\ {{{1}}} \qquad \mathrm{(C)}\ {{{2}}} \qquad \mathrm{(D)}\ {{{3}}} \qquad \mathrm{(E)}\ {{{4}}}$ | We may let $n = 7^k \cdot m$ , where $m$ is not divisible by 7. Using the fact that the number of divisors function $d(n)$ is multiplicative, we have $d(n) = d(7^k)d(m) = (k+1)d(m) = 60$ . Also, $d(7n) = d(7^{k+1})d(m) = (k+2)d(m) = 80$ . These numbers are in the ratio 3:4, so $\frac{k+1}{k+2} = \frac{3}{4} \implies k = 2 \Rightarrow \boxed{2}$ | C | 2 |
60e94bea236b973d4a3e9cbfa4781b2f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_24 | All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$ | Let the three points be at $A = (x_1, x_1^2)$ $B = (x_2, x_2^2)$ , and $C = (x_3, x_3^2)$ , such that the slope between the first two is $2$ , and $A$ is the point with the least $y$ -coordinate.
Then, we have $\textrm{Slope of }AC = \frac{x_1^2 - x_3^2}{x_1 - x_3} = x_1 + x_3$ . Similarly, the slope of $BC$ is $x_2 + x_3$ , and the slope of $AB$ is $x_1 + x_2 = 2$ . The desired sum is $x_1 + x_2 + x_3$ , which is equal to the sum of the slopes divided by $2$
To find the slope of $AC$ , we note that it comes at a $60^{\circ}$ angle with $AB$ . Thus, we can find the slope of $AC$ by multiplying the two complex numbers $1 + 2i$ and $1 + \sqrt{3}i$ . What this does is generate the complex number that is at a $60^{\circ}$ angle with the complex number $1 + 2i$ . Then, we can find the slope of the line between this new complex number and the origin: \[(1+2i)(1+\sqrt{3}i)\] \[= 1 - 2\sqrt{3} + 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}}\] \[= \frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}\] \[= \frac{8 + 5\sqrt{3}}{-11}\] \[= \frac{-8 - 5\sqrt{3}}{11}.\] The slope $BC$ can also be solved similarly, noting that it makes a $120^{\circ}$ angle with $AB$ \[(1+2i)(-1+\sqrt{3}i)\] \[= -1 - 2\sqrt{3} - 2i + \sqrt{3}i\] \[\textrm{Slope } = \frac{\sqrt{3} - 2}{-2\sqrt{3} - 1}\] \[= \frac{2 - \sqrt{3}}{1 + 2\sqrt{3}} \cdot \frac{1 - 2\sqrt{3}}{1 - 2\sqrt{3}}\]
At this point, we start to notice a pattern: This expression is equal to $\frac{2 + \sqrt{3}}{1 - 2\sqrt{3}} \cdot \frac{1 + 2\sqrt{3}}{1 + 2\sqrt{3}}$ , except the numerators of the first fractions are conjugates! Notice that this means that when we multiply out, the rational term will stay the same, but the coefficient of $\sqrt{3}$ will have its sign switched. This means that the two complex numbers are conjugates, so their irrational terms cancel out.
Our sum is simply $2 - 2\cdot\frac{8}{11} = \frac{6}{11}$ , and thus we can divide by $2$ to obtain $\frac{3}{11}$ , which gives the answer $\boxed{14}$ | A | 14 |
60e94bea236b973d4a3e9cbfa4781b2f | https://artofproblemsolving.com/wiki/index.php/2005_AMC_12B_Problems/Problem_24 | All three vertices of an equilateral triangle are on the parabola $y = x^2$ , and one of its sides has a slope of $2$ . The $x$ -coordinates of the three vertices have a sum of $m/n$ , where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$ | Using the slope formula and differences of squares, we find:
$a+b$ = the slope of $AB$
$b+c$ = the slope of $BC$
$a+c$ = the slope of $AC$
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$ . Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$ -axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$ . Translate the triangle so $A$ is at the origin. Then $\tan(BOJ) = 2$ . Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is $\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}$
Using $\tan(BOJ) = 2$ , and the tangent addition formula, this simplifies to $\dfrac{3}{11}$ , so the answer is $3 + 11 = \boxed{14}$ | A | 14 |
4613042c7185c4c8e3218a8966dc1aa3 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_1 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | $20$ dollars is the same as $2000$ cents, and $1.45\%$ of $2000$ is $0.0145\times2000=29$ cents. $\Rightarrow\boxed{29}$ | E | 29 |
4613042c7185c4c8e3218a8966dc1aa3 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_1 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes?
$\mathrm{(A) \ } 0.0029 \qquad \mathrm{(B) \ } 0.029 \qquad \mathrm{(C) \ } 0.29 \qquad \mathrm{(D) \ } 2.9 \qquad \mathrm{(E) \ } 29$ | Since there can't be decimal values of cents, the answer must be $\Rightarrow\boxed{29}$ | E | 29 |
32a572ee85b1fcb2ab91799ac31b07df | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_3 | For how many ordered pairs of positive integers $(x,y)$ is $x + 2y = 100$
$\text {(A)} 33 \qquad \text {(B)} 49 \qquad \text {(C)} 50 \qquad \text {(D)} 99 \qquad \text {(E)}100$ | If $x$ and $2y$ must each be positive integers, then we can say that $x$ is at least 1 and $2y$ is at least 1. From there, we want to find out how many ways there are to distribute the other 98 ones (the smallest positive integer addends of 100). 98 identical objects can be distributed to two distinct bins in 99 ways (think stars and bars), yet this 99 is an overcount. Because $y$ must be an integer, $2y$ must be even; thus only $\left\lfloor \frac{99}{2} \right\rfloor = \boxed{49}$ ways exist to distribute these ones. | B | 49 |
3341e128c70a1c63c6fb11978313a5ce | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Since Bertha has $6$ daughters, she has $30-6=24$ granddaughters, of which none have daughters. Of Bertha's daughters, $\frac{24}6=4$ have daughters, so $6-4=2$ do not have daughters. Therefore, of Bertha's daughters and granddaughters, $24+2=26$ do not have daughters. $\boxed{26}$ | E | 26 |
3341e128c70a1c63c6fb11978313a5ce | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Bertha has $30 - 6 = 24$ granddaughters, none of whom have any daughters. The granddaughters are the children of $24/6 = 4$ of Bertha's daughters, so the number of women having no daughters is $30 - 4 = \boxed{26}$ | null | 26 |
3341e128c70a1c63c6fb11978313a5ce | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_4 | Bertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no great-granddaughters. How many of Bertha's daughters and grand-daughters have no daughters?
$\mathrm{(A) \ } 22 \qquad \mathrm{(B) \ } 23 \qquad \mathrm{(C) \ } 24 \qquad \mathrm{(D) \ } 25 \qquad \mathrm{(E) \ } 26$ | Draw a tree diagram and see that the answer can be found in the sum of $6+6$ granddaughters, $5+5$ daughters, and $4$ more daughters. Adding them together gives the answer of $\boxed{26}$ | E | 26 |
57e6d869df3b6543a05b55028f6fab78 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | We look at a set of three rounds, where the players begin with $x+1$ $x$ , and $x-1$ tokens.
After three rounds, there will be a net loss of $1$ token per player (they receive two tokens and lose three). Therefore, after $36$ rounds -- or $12$ three-round sets, $A,B$ and $C$ will have $3$ $2$ , and $1$ tokens, respectively. After $1$ more round, player $A$ will give away $3$ tokens, leaving them empty-handed, and thus the game will end. We then have there are $36+1=\boxed{37}$ rounds until the game ends. | B | 37 |
57e6d869df3b6543a05b55028f6fab78 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | Let's bash a few rounds. The amounts are for players $1,2,$ and $3$ , respectively.
First round: $15,14,13$ (given)
Second round: $12,15,14$ Third round: $13,12,15$ Fourth round: $14,13,12$
We see that after $3$ rounds are played, we have the exact same scenario as the first round but with one token less per player. So, the sequence $1,4,7,10...$ where each of the next members are $3$ greater than the previous one corresponds with the sequence $15,14,13,12...$ where the first sequence represents the round and the second sequence represents the number of tokens player $1$ has. But we note that once player $1$ reaches $3$ coins, the game will end on his next turn as he must give away all his coins. Therefore, we want the $15-3+1=13$ th number in the sequence $1,4,7,10...$ which is $\boxed{37}$ | B | 37 |
57e6d869df3b6543a05b55028f6fab78 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_7 | A game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players $A$ $B$ , and $C$ start with $15$ $14$ , and $13$ tokens, respectively. How many rounds will there be in the game?
$\mathrm{(A) \ } 36 \qquad \mathrm{(B) \ } 37 \qquad \mathrm{(C) \ } 38 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ } 40$ | Looking at a set of five rounds, you'll see $A$ has $4$ fewer tokens than in the beginning. Looking at four more rounds, you'll notice $A$ has the same amount of tokens, namely $11$ , compared to round five. If you keep doing this process, you'll see a pattern: Every four rounds, the amount of tokens $A$ has either decreased by $4$ or stayed the same compared to the previous four rounds. For example, in round nine, $A$ had $11$ tokens, in round $13$ $A$ had $11$ tokens, and in round $17$ $A$ had $7$ tokens, etc. Using this weird pattern, you can find out that in round $37$ $A$ should have $3$ tokens, but since they would have given them away in that round, the game would end on $\boxed{37}$ | B | 37 |
5c45505d0b0e0790a63ea9078745b57a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Looking, we see that the area of $[\triangle EBA]$ is 16 and the area of $[\triangle ABC]$ is 12. Set the area of $[\triangle ADB]$ to be x. We want to find $[\triangle ADE]$ $[\triangle CDB]$ . So, that would be $[\triangle EBA]-[\triangle ADB]=16-x$ and $[\triangle ABC]-[\triangle ADB]=12-x$ . Therefore, $[\triangle ADE]-[\triangle DBC]=(16-x)-(12-x)=16-x-12+x= \boxed{4}$ | B | 4 |
5c45505d0b0e0790a63ea9078745b57a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Since $AE \perp AB$ and $BC \perp AB$ $AE \parallel BC$ . By alternate interior angles and $AA\sim$ , we find that $\triangle ADE \sim \triangle CDB$ , with side length ratio $\frac{4}{3}$ . Their heights also have the same ratio, and since the two heights add up to $4$ , we have that $h_{ADE} = 4 \cdot \frac{4}{7} = \frac{16}{7}$ and $h_{CDB} = 3 \cdot \frac 47 = \frac {12}7$ . Subtracting the areas, $\frac{1}{2} \cdot 8 \cdot \frac {16}7 - \frac 12 \cdot 6 \cdot \frac{12}7 = 4$ $\Rightarrow$ $\boxed{4}$ | B | 4 |
5c45505d0b0e0790a63ea9078745b57a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | Let $[X]$ represent the area of figure $X$ . Note that $[\triangle BEA]=[\triangle ABD]+[\triangle ADE]$ and $[\triangle BCA]=[\triangle ABD]+[\triangle BDC]$
$[\triangle ADE]-[\triangle BDC]=[\triangle BEA]-[\triangle BCA]=\frac{1}{2}\times8\times4-\frac{1}{2}\times6\times4= 16-12=4\Rightarrow\boxed{4}$ | B | 4 |
5c45505d0b0e0790a63ea9078745b57a | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_8 | In the overlapping triangles $\triangle{ABC}$ and $\triangle{ABE}$ sharing common side $AB$ $\angle{EAB}$ and $\angle{ABC}$ are right angles $AB=4$ $BC=6$ $AE=8$ , and $\overline{AC}$ and $\overline{BE}$ intersect at $D$ . What is the difference between the areas of $\triangle{ADE}$ and $\triangle{BDC}$
[asy] size(150); defaultpen(linewidth(0.4)); //Variable Declarations pair A, B, C, D, E; //Variable Definitions A=(0, 0); B=(4, 0); C=(4, 6); E=(0, 8); D=extension(A,C,B,E); //Initial Diagram draw(A--B--C--A--E--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,3N); label("$E$",E,NW); //Side labels label("$4$",A--B,S); label("$8$",A--E,W); label("$6$",B--C,ENE); [/asy]
$\mathrm {(A)}\ 2 \qquad \mathrm {(B)}\ 4 \qquad \mathrm {(C)}\ 5 \qquad \mathrm {(D)}\ 8 \qquad \mathrm {(E)}\ 9 \qquad$ | We want to figure out $Area(\triangle ADE) - Area(\triangle BDC)$ .
Notice that $\triangle ABC$ and $\triangle BAE$ "intersect" and form $\triangle ADB$
This means that $Area(\triangle BAE) - Area(\triangle ABC) = Area(\triangle ADE) - Area(\triangle BDC)$ because $Area(\triangle ADB)$ cancels out, which can be seen easily in the diagram.
$Area(\triangle BAE) = 0.5 * 4 * 8 = 16$
$Area(\triangle ABC) = 0.5 * 4 * 16 = 12$
$Area(\triangle BDC) - Area(\triangle ADE) = 16 - 12 =\boxed{4}$ | B | 4 |
852571302fb9cd8f2fde66fe520b9e48 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_9 | A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by $25\%$ without altering the volume , by what percent must the height be decreased?
$\mathrm{(A) \ } 10 \qquad \mathrm{(B) \ } 25 \qquad \mathrm{(C) \ } 36 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ } 60$ | When the diameter is increased by $25\%$ , it is increased by $\dfrac{5}{4}$ , so the area of the base is increased by $\left(\dfrac54\right)^2=\dfrac{25}{16}$
To keep the volume the same, the height must be $\dfrac{1}{\frac{25}{16}}=\dfrac{16}{25}$ of the original height, which is a $36\%$ reduction. $\boxed{36}$ | C | 36 |
a281e7015a126470f55c4c712c80c5d3 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_11 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | Let the total value, in cents, of the coins Paula has originally be $v$ , and the number of coins she has be $n$ . Then $\frac{v}{n}=20\Longrightarrow v=20n$ and $\frac{v+25}{n+1}=21$ . Substituting yields: $20n+25=21(n+1),$ so $n=4$ $v = 80.$ Then, we see that the only way Paula can satisfy this rule is if she had $3$ quarters and $1$ nickel in her purse. Thus, she has $\boxed{0}$ dimes. | A | 0 |
a281e7015a126470f55c4c712c80c5d3 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_11 | The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is $20$ cents. If she had one more quarter, the average value would be $21$ cents. How many dimes does she have in her purse?
$\text {(A)}\ 0 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 2 \qquad \text {(D)}\ 3\qquad \text {(E)}\ 4$ | If the new coin was worth $20$ cents, adding it would not change the mean. The additional $5$ cents raise the mean by $1$ , thus the new number of coins must be $5$ . Therefore there were $4$ coins worth a total of $4\times20=80$ cents. As in the previous solution, we conclude that the only way to get $80$ cents using $4$ coins is $25+25+25+5$ . Thus, having three quarters, one nickel, and no dimes $\boxed{0}.$ | A | 0 |
fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | Let $d$ be the common difference. Then $9$ $9+d+2=11+d$ $9+2d+20=29+2d$ are the terms of the geometric progression. Since the middle term is the geometric mean of the other two terms, $(11+d)^2 = 9(2d+29)$ $\Longrightarrow d^2 + 4d - 140$ $= (d+14)(d-10) = 0$ . The smallest possible value occurs when $d = -14$ , and the third term is $2(-14) + 29 = 1\Rightarrow\boxed{1}$ | A | 1 |
fe9158b8e93c11693e259fb0c7818076 | https://artofproblemsolving.com/wiki/index.php/2004_AMC_12A_Problems/Problem_14 | sequence of three real numbers forms an arithmetic progression with a first term of $9$ . If $2$ is added to the second term and $20$ is added to the third term, the three resulting numbers form a geometric progression . What is the smallest possible value for the third term in the geometric progression?
$\text {(A)}\ 1 \qquad \text {(B)}\ 4 \qquad \text {(C)}\ 36 \qquad \text {(D)}\ 49 \qquad \text {(E)}\ 81$ | Let $d$ be the common difference and $r$ be the common ratio. Then the arithmetic sequence is $9$ $9+d$ , and $9+2d$ . The geometric sequence (when expressed in terms of $d$ ) has the terms $9$ $11+d$ , and $29+2d$ . Thus, we get the following equations:
$9r=11+d\Rightarrow d=9r-11$
$9r^2=29+2d$
Plugging in the first equation into the second, our equation becomes $9r^2=29+18r-22\Longrightarrow9r^2-18r-7=0$ . By the quadratic formula, $r$ can either be $-\frac{1}{3}$ or $\frac{7}{3}$ . If $r$ is $-\frac{1}{3}$ , the third term (of the geometric sequence) would be $1$ , and if $r$ is $\frac{7}{3}$ , the third term would be $49$ . Clearly the minimum possible value for the third term of the geometric sequence is $\boxed{1}$ | A | 1 |
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