Answer: 4.0

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?
$\mathrm{(A)}\ 1      \qquad \mathrm{(B)}\ 2      \qquad \mathrm{(C)}\ 3      \qquad \mathrm{(D)}\ 4      \qquad \mathrm{(E)}\ 5$

Solution:
$221$ 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 $45-41=\boxed{4}$


Answer: 4.0

Solution: 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, $\boxed{4}$


Answer: 6.0

The sum of seven integers is $-1$ . What is the maximum number of the seven integers that can be larger than $13$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$

Solution:
If the first six integers are $14$ , the last number can be $(-14\cdot 6) - 1 = -85$ .  The sum of all seven integers will be $-1$
However, if all seven integers are over $13$ , the smallest possible sum is $14\cdot 7 = 98$
Thus, the answer is $6$ , which is option $\boxed{6}$


Answer: 338.0

The sum of six consecutive positive integers is 2013. What is the largest of these six integers?
$\textbf{(A)}\ 335 \qquad \textbf{(B)}\ 338 \qquad \textbf{(C)}\ 340 \qquad \textbf{(D)}\ 345 \qquad \textbf{(E)}\ 350$

Solution: The arithmetic mean of these numbers is $\frac{\frac{2013}{3}}{2}=\frac{671}{2}=335.5$ . Therefore the numbers are $333$ $334$ $335$ $336$ $337$ $338$ , so the answer is $\boxed{338}$

Solution:
Let the $4^{\text{th}}$ number be $x$ . Then our desired number is $x+2$
Our integers are $x-3,x-2,x-1,x,x+1,x+2$ , so we have that $6x-3 = 2013 \implies x = \frac{2016}{6} = 336 \implies x+2 = \boxed{338}$

Solution: Let the first term be $x$ . Our integers are $x,x+1,x+2,x+3,x+4,x+5$ . We have, $6x+15=2013\implies x=333\implies x+5=\boxed{338}$

Solution: Let the $6th$ number be $x$ . Then our list is: $x-6+x-5+x-4+x-3+x-x-1=2013$ . Simplifying this gets you $6x-21=2013\implies 6x=2034$ , which means that $x = \boxed{338}$


Answer: 338.0

Solution:
Since there are $6$ numbers, we divide $2013$ by $6$ to find the mean of the numbers. $\frac{2013}{6} = 335 \frac{1}{2}$ .
Then, $335 \frac{1}{2} + \frac{1}{2} = 336$ (the fourth number). Fifth: $337$ ; Sixth: $\boxed{338}$


Answer: 348.0

The sum of the areas of all triangles whose vertices are also vertices of a $1$ by $1$ by $1$ cube is $m + \sqrt{n} + \sqrt{p},$ where $m, n,$ and $p$ are integers . Find $m + n + p.$

Solution:
Since there are $8$ vertices of a cube , there are ${8 \choose 3} = 56$ total triangles to consider.  They fall into three categories: there are those which are entirely contained within a single face of the cube (whose sides are two edges and one face diagonal ), those which lie in a plane perpendicular to one face of the cube (whose sides are one edge, one face diagonal and one space diagonal of the cube) and those which lie in a plane oblique to the edges of the cube, whose sides are three face diagonals of the cube.
Each face of the cube contains ${4\choose 3} = 4$ triangles of the first type, and there are $6$ faces, so there are $24$ triangles of the first type. Each of these is a right triangle with legs of length $1$ , so each triangle of the first type has area $\frac 12$
Each edge of the cube is a side of exactly $2$ of the triangles of the second type, and there are $12$ edges, so there are $24$ triangles of the second type. Each of these is a right triangle with legs of length $1$ and $\sqrt 2$ , so each triangle of the second type has area $\frac{\sqrt{2}}{2}$
Each vertex of the cube is associated with exactly one triangle of the third type (whose vertices are its three neighbors), and there are $8$ vertices of the cube, so there are $8$ triangles of the third type. Each of the these is an equilateral triangle with sides of length $\sqrt 2$ , so each triangle of the third type has area $\frac{\sqrt 3}2$
Thus the total area of all these triangles is $24 \cdot \frac12 + 24\cdot\frac{\sqrt2}2 + 8\cdot\frac{\sqrt3}2 = 12 + 12\sqrt2 + 4\sqrt3 = 12 + \sqrt{288} + \sqrt{48}$ and the answer is $12 + 288 + 48 = \boxed{348}$


Answer: 11.0

The sum of the base- $10$ logarithms of the divisors of $10^n$ is $792$ . What is $n$
$\text{(A)}\ 11\qquad \text{(B)}\ 12\qquad \text{(C)}\ 13\qquad \text{(D)}\ 14\qquad \text{(E)}\ 15$

Solution:
Every factor of $10^n$ will be of the form $2^a \times 5^b , a\leq n , b\leq n$ . Not all of these base ten logarithms will be rational, but we can add them together in a certain way to make it rational. Recall the logarithmic property $\log(a \times b) = \log(a)+\log(b)$ . For any factor $2^a \times 5^b$ , there will be another factor $2^{n-a} \times 5^{n-b}$ . Note this is not true if $10^n$ is a perfect square. When these are multiplied, they equal $2^{a+n-a} \times 5^{b+n-b} = 10^n$ $\log 10^n=n$ so the number of factors divided by 2 times n equals the sum of all the factors, 792.
There are $n+1$ choices for the exponent of 5 in each factor, and for each of those choices, there are $n+1$ factors (each corresponding to a different exponent of 2), yielding $(n+1)^2$ total factors. $\frac{(n+1)^2}{2}*n = 792$ . We then plug in answer choices and arrive at the answer $\boxed{11}$


Answer: 4.0

The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$ , where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad  \textbf{(B)}\ 4n \qquad  \textbf{(C)}\ 2+2n \qquad  \textbf{(D)}\ 4n^2 \qquad  \textbf{(E)}\ n^2+n+2$

Solution: We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: \[110\text{ some number of zeros }011.\] This works because of the way they are multiplied. Therefore, the answer is $\boxed{4}$


Answer: 10.0

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$ . How many two-digit numbers have this property?
$\textbf{(A) } 5\qquad \textbf{(B) } 7\qquad \textbf{(C) } 9\qquad \textbf{(D) } 10\qquad \textbf{(E) } 19$

Solution:
Let the number be $10a+b$ where $a$ and $b$ are the tens and units digits of the number.
So $(10a+b)-(a+b)=9a$ must have a units digit of $6$
This is only possible if $9a=36$ , so $a=4$ is the only way this can be true.
So the numbers that have this property are $40, 41, 42, 43, 44, 45, 46, 47, 48, 49$
Therefore the answer is $\boxed{10}$

Solution:
Let a two-digit number equal $10a+b$ , where $a$ and $b$ are the tens and units digits of the number.
From the problem, we have $10a+b-(a+b)=9a$
Now let $9a=10x+y$ , where $x$ and $y$ are the tens and units digits of the number. Then it must be that $y=6$ as stated in the problem.
Note that $10a$ ends in $0$ , but $9a$ ends in $6$ , so $a=4$ . We need not to care about $b$ , since it cancels out in the calculation.
So the answer is $\boxed{10}$ , since there are $10$ numbers that have $a=4$


Answer: 542.0

The sum of the first $2011$ terms of a geometric sequence is $200$ . The sum of the first $4022$ terms is $380$ . Find the sum of the first $6033$ terms.

Solution:
Since the sum of the first $2011$ terms is $200$ , and the sum of the first $4022$ terms is $380$ , the sum of the second $2011$ terms is $180$ .
This is decreasing from the first 2011, so the common ratio is less than one.
Because it is a geometric sequence and the sum of the first 2011 terms is $200$ , second $2011$ is $180$ , the ratio of the second $2011$ terms to the first $2011$ terms is $\frac{9}{10}$ . Following the same pattern, the sum of the third $2011$ terms is $\frac{9}{10}*180 = 162$
Thus, $200+180+162=542$ , so the sum of the first $6033$ terms is $\boxed{542}$

Solution:
Solution by e_power_pi_times_i
The sum of the first $2011$ terms can be written as $\dfrac{a_1(1-k^{2011})}{1-k}$ , and the first $4022$ terms can be written as $\dfrac{a_1(1-k^{4022})}{1-k}$ . Dividing these equations, we get $\dfrac{1-k^{2011}}{1-k^{4022}} = \dfrac{10}{19}$ . Noticing that $k^{4022}$ is just the square of $k^{2011}$ , we substitute $x = k^{2011}$ , so $\dfrac{1}{x+1} = \dfrac{10}{19}$ . That means that $k^{2011} = \dfrac{9}{10}$ . Since the sum of the first $6033$ terms can be written as $\dfrac{a_1(1-k^{6033})}{1-k}$ , dividing gives $\dfrac{1-k^{2011}}{1-k^{6033}}$ . Since $k^{6033} = \dfrac{729}{1000}$ , plugging all the values in gives $\boxed{542}$

Solution: The sum of the first 2011 terms of the sequence is expressible as $a_1 + a_1r + a_1r^2 + a_1r^3$ .... until $a_1r^{2010}$ . The sum of the 2011 terms following the first 2011 is expressible as $a_1r^{2011} + a_1r^{2012} + a_1r^{2013}$ .... until $a_1r^{4021}$ . Notice that the latter sum of terms can be expressed as $(r^{2011})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ . We also know that the latter sum of terms can be obtained by subtracting 200 from 380, which then means that $r^{2011} = 9/10$ . The terms from 4023 to 6033 can be expressed as $(r^{4022})(a_1 + a_1r + a_1r^2 + a_1r^3...a_1r^{2010})$ , which is equivalent to $((9/10)^2)(200) = 162$ . Adding 380 and 162 gives the answer of $\boxed{542}$


Answer: 255.0

The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n$
$\textbf{(A)}\ 255\qquad\textbf{(B)}\ 256\qquad\textbf{(C)}\ 257\qquad\textbf{(D)}\ 258\qquad\textbf{(E)}\ 259$

Solution:
The sum of the first $m$ odd integers is given by $m^2$ . The sum of the first $n$ even integers is given by $n(n+1)$
Thus, $m^2 = n^2 + n + 212$ . Since we want to solve for n, rearrange as a quadratic equation: $n^2 + n + (212 - m^2) = 0$
Use the quadratic formula: $n = \frac{-1 + \sqrt{1 - 4(212 - m^2)}}{2}$ . Since $n$ is clearly an integer, $1 - 4(212 - m^2) = 4m^2 - 847$ must be not only a perfect square, but also an odd perfect square for $n$ to be an integer.
Let $x = \sqrt{4m^2 - 847}$ ; note that this means $n = \frac{-1 + x}{2}$ . It can be rewritten as $x^2 = 4m^2 - 847$ , so $4m^2 - x^2 = 847$ . Factoring the left side by using the difference of squares, we get $(2m + x)(2m - x) = 847 = 7\cdot11^2$
Our goal is to find possible values for $x$ , then use the equation above to find $n$ . The difference between the factors is $(2m + x) - (2m - x) = 2m + x - 2m + x = 2x.$ We have three pairs of factors, $847\cdot1, 121\cdot 7,$ and $77\cdot 11$ . The differences between these factors are $846$ $114$ , and $66$ - those are all possible values for $2x$ . Thus the possibilities for $x$ are $423$ $57$ , and $33$
Now plug in these values into the equation $n = \frac{-1 + x}{2}$ , so $n$ can equal $211$ $28$ , or $16$ , hence the answer is $\boxed{255}$

Solution:
As above, start off by noting that the sum of the first $m$ odd integers $= m^2$ and the sum of the first $n$ even integers $= n(n+1)$ . Clearly $m > n$ , so let $m = n + a$ , where $a$ is some positive integer. We have:
$(n+a)^2 = n(n+1) + 212$ .
Expanding, grouping like terms and factoring, we get: $n = \frac{(212 - a^2)}{(2a - 1)}$
We know that $n$ and $a$ are both positive integers, so we need only check values of $a$ from $1$ to $14$ $14^2 = 196 < 212 < 15^2 = 225$ ). Plugging in, the only values of $a$ that give integral solutions are $1, 4,$ and $6$ . These gives $n$ values of $211, 28,$ and $16$ , respectively. $211 + 28 + 16 = 255$ . Hence, the answer is $\boxed{255}$

Solution: Using the closed forms for the sums, we get $m^2=n(n+1)+212$ , or $m^2=n^2+n+212$ . We would like to factor this equation, but the current expressions don't allow for this. So we multiply both sides by 4 to let us complete the square. Our equation is now $4m^2=4n^2+4n+848$ . Complete the square on the right hand side: $4m^2=(4n^2+4n+1)+848-1=(2n+1)^2+847$ . Move over the $(2n+1)^2$ and factor to get $(2m-2n-1)(2m+2n+1)=847=7\cdot11\cdot11$ . The second factor is clearly greater than the first, and the only possible factor pairs are $1$ and $847$ $7$ and $121$ $11$ and $77$ . In each of these cases, solve for $m$ and $n$ and we find the solutions $(m,n)=(212,211), (32,28), (22,16)$ . The sum of all possible values of $n$ is $211+28+16=\boxed{255}$


Answer: 80.0

The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 40 \qquad  \textbf{(D)}\ 60 \qquad  \textbf{(E)}\ 80$

Solution:
Solution by e_power_pi_times_i
When the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{80}$


Answer: 61.0

The sum of the following seven numbers is exactly 19: $a_1 = 2.56$ $a_2 = 2.61$ $a_3 = 2.65$ $a_4 = 2.71$ $a_5 = 2.79$ $a_6 = 2.82$ $a_7 = 2.86$ . It is desired to replace each $a_i$ by an integer approximation $A_i$ $1\le i \le 7$ , so that the sum of the $A_i$ 's is also 19 and so that $M$ , the maximum of the "errors" $\| A_i-a_i\|$ , the maximum absolute value of the difference, is as small as possible. For this minimum $M$ , what is $100M$

Solution: If any of the approximations $A_i$ is less than 2 or more than 3, the error associated with that term will be larger than 1, so the largest error will be larger than 1.  However, if all of the $A_i$ are 2 or 3, the largest error will be less than 1.  So in the best case, we write 19 as a sum of 7 numbers, each of which is 2 or 3.  Then there must be five 3s and two 2s.  It is clear that in the best appoximation, the two 2s will be used to approximate the two smallest of the $a_i$ , so our approximations are $A_1 = A_2 = 2$ and $A_3 = A_4 = A_5 = A_6 = A_7 = 3$ and the largest error is $|A_2 - a_2| = 0.61$ , so the answer is $\boxed{061}$


Answer: 784.0

The sum of the lengths of the twelve edges of a rectangular box is $140$ , and
the distance from one corner of the box to the farthest corner is $21$ . The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$

Solution:
Let $x, y$ , and $z$ be the unique lengths of the edges of the box.  Each box has $4$ edges of each length, so: \[4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.\] The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$ .  Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$
Our target expression is the surface area of the box:
\[S = 2xy + 2xz + 2yz.\]
Since $S$ is a symmetric polynomial of degree $2$ , we try squaring the first equation to get:
\[35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.\]
Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$ , which is option $\boxed{784}$


Answer: 1.0

The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$

Solution:
For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$ ) is found by setting all variables equal to $1$ .  Note that we don't have to worry about whether a constant is a coefficient of an "invisible" $x^0y^0$ term, because there is no such term here.
Setting $x=y=1$ gives $(-1)^{18}$ , which is equal to $1$ , which is answer $\boxed{1}$


Answer: 14.0

The sum of the two 5-digit numbers $AMC10$ and $AMC12$ is $123422$ . What is $A+M+C$
$\mathrm{(A) \ } 10\qquad \mathrm{(B) \ } 11\qquad \mathrm{(C) \ } 12\qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 14$

Solution:
$AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2\cdot AMC=1234$
$AMC=\frac{1234}{2}=617$
Since $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$
Therefore, $A+M+C = 6+1+7 = \boxed{14}$

Solution: We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \boxed{14}$

Solution:
Consider the place values of the digits of $AMC10$ and $AMC12$
When we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \boxed{14}$


Answer: 14.0

Solution:
$AMC10+AMC12=123422$
$AMC00+AMC00=123400$
$AMC+AMC=1234$
$2\cdot AMC=1234$
$AMC=\frac{1234}{2}=617$
Since $A$ $M$ , and $C$ are digits, $A=6$ $M=1$ $C=7$
Therefore, $A+M+C = 6+1+7 = \boxed{14}$

Solution: We know that $AMC12$ is $2$ more than $AMC10$ . We set up $AMC10=x$ and $AMC12=x+2$ . We have $x+x+2=123422$ . Solving for $x$ , we get $x=61710$ . Therefore, the sum $A+M+C= \boxed{14}$

Solution:
Consider the place values of the digits of $AMC10$ and $AMC12$
When we add $AMC10$ and $AMC12$ $C + C$ must result in a units digit of $4$ , meaning $C$ is either $2$ or $7$ . Since $M$ is odd, this means a ten was carried over to the next place value from $C + C$ , and thus $C = 7$ (as $7 + 7 = 14$ and the ten is carried over). Now, we know 3 is the units digit of $M + M + 1$ , so $M$ is either $1$ or $6$ . Again, we must look at the digit before $M$ , or $A$ $A$ is even, so $M$ must be less than $5$ , or else the ten would be carried over. Ergo, $M$ is $1$ . Nothing is carried over, so we have $A + A = 12$ , and $A = 6$ . Therefore, the sum of $A$ $M$ , and $C$ is $6 + 1 + 7 = \boxed{14}$


Answer: 28.0

The sum of three numbers is $20$ . The first is four times the sum of the other two. The second is seven times the third. What is the product of all three?
$\mathrm{(A) \ } 28\qquad \mathrm{(B) \ } 40\qquad \mathrm{(C) \ } 100\qquad \mathrm{(D) \ } 400\qquad \mathrm{(E) \ } 800$

Solution:
Let the numbers be $x$ $y$ , and $z$ in that order. The given tells us that
\begin{eqnarray*}y&=&7z\\ x&=&4(y+z)=4(7z+z)=4(8z)=32z\\ x+y+z&=&32z+7z+z=40z=20\\ z&=&\frac{20}{40}=\frac{1}{2}\\ y&=&7z=7\cdot\frac{1}{2}=\frac{7}{2}\\ x&=&32z=32\cdot\frac{1}{2}=16 \end{eqnarray*}
Therefore, the product of all three numbers is $xyz=16\cdot\frac{7}{2}\cdot\frac{1}{2}=28 \Rightarrow \boxed{28}$

Solution:
Alternatively, we can set up the system in equation form:
\begin{eqnarray*}1x+1y+1z&=&20\\ 1x-4y-4z&=&0\\ 0x+1y-7z&=&0\\ \end{eqnarray*}
Or, in matrix form $\begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} =\begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
To solve this matrix equation, we can rearrange it thus:
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -4 & -4 \\ 0 & 1 & -7 \end{bmatrix} ^{-1} \begin{bmatrix} 20 \\ 0 \\ 0 \\ \end{bmatrix}$
Solving this matrix equation by using inverse matrices and matrix multiplication yields
$\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix} = \begin{bmatrix} \frac{1}{2} \\ \frac{7}{2} \\ 16 \\ \end{bmatrix}$
Which means that $x = \frac{1}{2}$ $y = \frac{7}{2}$ , and $z = 16$ . Therefore, $xyz = \frac{1}{2}\cdot\frac{7}{2}\cdot16 = 28 \Rightarrow \boxed{28}$


Answer: 5.0

The sum of three numbers is $96.$ The first number is $6$ times the third number, and the third number is $40$ less than the second number. What is the absolute value of the difference between the first and second numbers?
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 5$

Solution:
Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$

Solution:
Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$
So, the answer is $|x-y|=|42-47|=\boxed{5}.$

Solution: In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 5.0

Solution:
Let $x$ be the third number. It follows that the first number is $6x,$ and the second number is $x+40.$
We have \[6x+(x+40)+x=8x+40=96,\] from which $x=7.$
Therefore, the first number is $42,$ and the second number is $47.$ Their absolute value of the difference is $|42-47|=\boxed{5}.$

Solution:
Solve this using a system of equations. Let $x,y,$ and $z$ be the three numbers, respectively. We get three equations: \begin{align*} x+y+z&=96, \\ x&=6z, \\ z&=y-40. \end{align*} Rewriting the third equation gives us $y=z+40,$ so we can substitute $x$ as $6z$ and $y$ as $z+40.$
Therefore, we get \begin{align*} 6z+(z+40)+z&=96 \\ 8z+40&=96 \\ 8z&=56 \\ z&=7. \end{align*} Substituting 7 in for $z$ gives us $x=6z=6(7)=42$ and $y=z+40=7+40=47.$
So, the answer is $|x-y|=|42-47|=\boxed{5}.$

Solution: In accordance with Solution 2, \[y = z+40, x = 6z \implies |x-y| = |6z - z - 40| = 5|z - 8| \implies \boxed{5}.\] vladimir.shelomovskii@gmail.com, vvsss


Answer: 30.0

The sum of three numbers is $98$ . The ratio of the first to the second is $\frac {2}{3}$ ,
and the ratio of the second to the third is $\frac {5}{8}$ . The second number is:
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$

Solution:
Let the $3$ numbers be $a,$ $b,$ and $c$ . We see that \[a+b+c = 98\] and \[\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b\] \[\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c\] Writing $a$ and $c$ in terms of $b$ we have $a = \frac{2}{3} b$ and $c = \frac{8}{5} b$ .
Substituting in the sum, we have \[\frac{2}{3} b + b + \frac{8}{5} b = 98\] \[\frac{49}{15} b = 98\] \[b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30\] $\boxed{30}$


Answer: 72.0

The sum of two angles of a triangle is $\frac{6}{5}$ of a right angle, and one of these two angles is $30^{\circ}$ larger than the other. What is the degree measure of the largest angle in the triangle?
$\textbf{(A)}\ 69 \qquad\textbf{(B)}\ 72 \qquad\textbf{(C)}\ 90 \qquad\textbf{(D)}\ 102 \qquad\textbf{(E)}\ 108$

Solution:
The sum of two angles in a triangle is $\frac{6}{5}$ of a right angle $\longrightarrow \frac{6}{5} \times 90 = 108$
If $x$ is the measure of the first angle, then the measure of the second angle is $x+30$ \[x + x + 30 = 108 \longrightarrow 2x = 78 \longrightarrow x = 39\]
Now we know the measure of two angles are $39^{\circ}$ and $69^{\circ}$ . By the Triangle Sum Theorem, the sum of all angles in a triangle is $180^{\circ},$ so the final angle is $72^{\circ}$ . Therefore, the largest angle in the triangle is $\boxed{72}$


Answer: 14238.0

The sum of two natural numbers is $17402$ . One of the two numbers is divisible by $10$ . If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?
$\textbf{(A)} ~10272\qquad\textbf{(B)} ~11700\qquad\textbf{(C)} ~13362\qquad\textbf{(D)} ~14238\qquad\textbf{(E)} ~15426$

Solution:
The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$
Let the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$
We know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \boxed{14238}$

Solution: Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$

Solution: Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc}   & A & B & C & D & 0 \\ +\quad &   & A & B & C & D \\ \hline   &   &   &   &   &   \\ [-2.5ex]   & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$

Solution:
We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$


Answer: 14238.0

Solution:
The units digit of a multiple of $10$ will always be $0$ . We add a $0$ whenever we multiply by $10$ . So, removing the units digit is equal to dividing by $10$
Let the smaller number (the one we get after removing the units digit) be $a$ . This means the bigger number would be $10a$
We know the sum is $10a+a = 11a$ so $11a=17402$ . So $a=1582$ . The difference is $10a-a = 9a$ . So, the answer is $9(1582) = \boxed{14238}$

Solution: Since the unit's place of a multiple of $10$ is $0$ , the other integer must end with a $2$ , for both integers sum up to a number ending in a $2$ . Thus, the unit's place of the difference must be $10-2=8$ , and the only answer choice that ends with an $8$ is $\boxed{14238}$

Solution: Let the larger number be $\underline{ABCD0}.$ It follows that the smaller number is $\underline{ABCD}.$ Adding vertically, we have \[\begin{array}{cccccc}   & A & B & C & D & 0 \\ +\quad &   & A & B & C & D \\ \hline   &   &   &   &   &   \\ [-2.5ex]   & 1 & 7 & 4 & 0 & 2 \\ \end{array}\] Working from right to left, we get \[D=2\implies C=8 \implies B=5 \implies A=1.\] The larger number is $15820$ and the smaller number is $1582.$ Their difference is $15820-1582=\boxed{14238}.$

Solution:
We know that the larger number has a units digit of $0$ since it is divisible by 10. If $D$ is the ten's digit of the larger number, then $D$ is the units digit of the smaller number. Since the sum of the natural numbers has a unit's digit of $2$ $D=2$
The units digit of the larger number is $0$ and the units digit of the smaller number is $2$ , so the positive difference between the numbers is 8. There is only one answer choice that has this units digit, and that is $\boxed{14238}.$


Answer: 4.0

The sum of two nonzero real numbers is $4$ times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution:
Let the two real numbers be $x,y$ . We are given that $x+y=4xy,$ and dividing both sides by $xy$ $\frac{x}{xy}+\frac{y}{xy}=4.$
\[\frac{1}{y}+\frac{1}{x}=\boxed{4}.\]

Solution:
Instead of using algebra, another approach at this problem would be to notice the fact that one of the nonzero numbers has to be a fraction.
See for yourself. And by looking into fractions, we immediately see that $\frac{1}{3}$ and $1$ would fit the rule. $\boxed{4}.$

Solution:
Notice that from the information given above, $x+y=4xy$
Because the sum of the reciprocals of two numbers is just the sum of the two numbers over the product of the two numbers or $\frac{x+y}{xy}$
We can solve this by substituting $x+y\implies 4xy$
Our answer is simply $\frac{4xy}{xy}\implies4$
Therefore, the answer is $\boxed{4}$


Answer: 4.0

The sum of two nonzero real numbers is 4 times their product. What is the sum of the reciprocals of the two numbers?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 12$

Solution:
Let $x, y$ be our two numbers. Then $x+y = 4xy$ . Thus,
$\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy} = 4$
$\boxed{4}$

Solution: We can let $x=y.$ Then, we have that $2x=4x^2$ making $x=\tfrac{1}{2}.$ The answer is $\dfrac{2}{x}=4=\boxed{4}.$ Solasky talk


Answer: 12.0

The sum of two numbers is $10$ ; their product is $20$ . The sum of their reciprocals is:
$\textbf{(A)}\ \frac{1}{10}\qquad  \textbf{(B)}\ \frac{1}{2}\qquad  \textbf{(C)}\ 1\qquad  \textbf{(D)}\ 2\qquad  \textbf{(E)}\ 4$

Solution:
$x+y=10$
$xy=20$
$\frac1x+\frac1y=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}=\frac{10}{20}=\boxed{12}$


Answer: 32.0

The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$

Solution:
Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$

Solution: Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$


Answer: 32.0

Solution:
Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$

Solution: Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$


Answer: 166.0

The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$

Solution: Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\cdot2=\boxed{166}$


Answer: 7.0

The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$

Solution:
Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:
$a+b=12$
$b+c=17$
$a+c=19$
Adding these equations together, we get that
$2(a+b+c)=48$ and
$a+b+c=24$
Substituting the original equations into this one, we find
$c+12=24$
$a+17=24$
$b+19=24$
Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{7}$

Solution:
Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:
$a+b=12$
$a+c=17$
$b+c=19$
To isolate $b$ , We add the first and last equations and then subtract the second one.
$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$
Because $b$ is the middle number, the middle number is $\boxed{7}$


Answer: 7.0

Solution:
Let the three numbers be equal to $a$ $b$ , and $c$ . We can now write three equations:
$a+b=12$
$b+c=17$
$a+c=19$
Adding these equations together, we get that
$2(a+b+c)=48$ and
$a+b+c=24$
Substituting the original equations into this one, we find
$c+12=24$
$a+17=24$
$b+19=24$
Therefore, our numbers are 12, 7, and 5. The middle number is $\boxed{7}$

Solution:
Let the three numbers be $a$ $b$ and $c$ and $a<b<c$ . We get the three equations:
$a+b=12$
$a+c=17$
$b+c=19$
To isolate $b$ , We add the first and last equations and then subtract the second one.
$(a+b)+(b+c)-(a+c) = 12+19-17 \Rightarrow 2b=14 \Rightarrow b = 7$
Because $b$ is the middle number, the middle number is $\boxed{7}$


Answer: 38.0

The symbolism $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . For example, $\lfloor 3 \rfloor = 3,$ and $\lfloor 9/2 \rfloor = 4$ . Compute \[\lfloor \sqrt{1} \rfloor + \lfloor \sqrt{2} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{16} \rfloor.\]
$\textbf{(A) } 35 \qquad\textbf{(B) } 38 \qquad\textbf{(C) } 40 \qquad\textbf{(D) } 42 \qquad\textbf{(E) } 136$

Solution:
The first three values in the sum are equal to $1,$ the next five equal to $2,$ the next seven equal to $3,$ and the last one equal to $4.$ For example, since $2^2=4$ any square root of a number less than $4$ must be less than $2.$ Sum them all together to get
\[3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{38}\]


Answer: 25.0

Solution:
Since $\log ab = \log a + \log b$ , we can reduce the equations to a more recognizable form:
\begin{eqnarray*} -\log x \log y + \log x + \log y - 1 &=& 3 - \log 2000\\ -\log y \log z + \log y + \log z - 1 &=& - \log 2\\ -\log x \log z + \log x + \log z - 1 &=& -1\\ \end{eqnarray*}
Let $a,b,c$ be $\log x, \log y, \log z$ respectively. Using SFFT , the above equations become (*)
\begin{eqnarray*}(a - 1)(b - 1) &=& \log 2 \\ (b-1)(c-1) &=& \log 2 \\ (a-1)(c-1) &=& 1  \end{eqnarray*}
Small note from different author: $-(3 - \log 2000) = \log 2000 - 3 = \log 2000 - \log 1000 = \log 2.$
From here, multiplying the three equations gives
\begin{eqnarray*}(a-1)^2(b-1)^2(c-1)^2 &=& (\log 2)^2\\ (a-1)(b-1)(c-1) &=& \pm\log 2\end{eqnarray*}
Dividing the third equation of (*) from this equation, $b-1 = \log y - 1 = \pm\log 2 \Longrightarrow \log y = \pm \log 2 + 1$ . (Note from different author if you are confused on this step: if $\pm$ is positive then $\log y = \log 2 + 1 = \log 2 + \log 10 = \log 20,$ so $y=20.$ if $\pm$ is negative then $\log y = 1 - \log 2 = \log 10 - \log 2 = \log 5,$ so $y=5.$ ) This gives $y_1 = 20, y_2 = 5$ , and the answer is $y_1 + y_2 = \boxed{025}$

Solution:
Subtracting the second equation from the first equation yields \begin{align*} \log 2000xy-\log 2yz-((\log x)(\log y)-(\log y)(\log z)) &= 3 \\ \log\frac{2000xy}{2yz}-\log y(\log x-\log z) &= 3 \\ \log1000+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ 3+\log\frac{x}{z}-\log y(\log\frac{x}{z}) &= 3 \\ \log\frac{x}{z}(1-\log y) &= 0 \\ \end{align*} If $1-\log y=0$ then $y=10$ . Substituting into the first equation yields $\log20000=4$ which is not possible.
If $\log\frac{x}{z}=0$ then $\frac{x}{z}=1\Longrightarrow x=z$ . Substituting into the third equation gets \begin{align*} \log x^2-(\log x)(\log x) &= 0 \\ \log x^2-\log x^x &= 0 \\ \log x^{2-x} &= 0 \\ x^{2-x} &= 1 \\ \end{align*} Thus either $x=1$ or $2-x=0\Longrightarrow x=2$ . (Note that here $x\neq-1$ since logarithm isn't defined for negative number.)
Substituting $x=1$ and $x=2$ into the first equation will obtain $y=5$ and $y=20$ , respectively. Thus $y_1+y_2=\boxed{025}$

Solution:
Let $a = \log x$ $b = \log y$ and $c = \log z$ . Then the given equations become:
\begin{align*} \log 2 + a + b - ab = 1 \\ \log 2 + b + c - bc = 1 \\ a+c = ac \\ \end{align*}
Equating the first and second equations, solving, and factoring, we get $a(1-b) = c(1-b) \implies{a = c}$ . Plugging this result into the third equation, we get $c = 0$ or $2$ . Substituting each of these values of $c$ into the second equation, we get $b = 1 - \log 2$ and $b = 1 + \log 2$ . Substituting backwards from our original substitution, we get $y = 5$ and $y = 20$ , respectively, so our answer is $\boxed{025}$


Answer: 943.0

The table below displays some of the results of last summer's Frostbite Falls Fishing Festival, showing how many contestants caught $n\,$ fish for various values of $n\,$
In the newspaper story covering the event, it was reported that
What was the total number of fish caught during the festival?

Solution:
Suppose that the number of fish is $x$ and the number of contestants is $y$ . The $y-(9+5+7)=y-21$ fishers that caught $3$ or more fish caught a total of $x - \left(0\cdot(9) + 1\cdot(5) + 2\cdot(7)\right) = x - 19$ fish. Since they averaged $6$ fish,
Similarily, those who caught $12$ or fewer fish averaged $5$ fish per person, so
Solving the two equation system, we find that $y = 175$ and $x = \boxed{943}$ , the answer.

Solution: Let $f$ be the total number of fish caught by the contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish and let $a$ be the number of contestants who didn't catch $0, 1, 2, 3, 13, 14$ , or $15$ fish. From $\text{(b)}$ , we know that $\frac{69+f+65+28+15}{a+31}=6\implies f=6a+9$ . From $\text{(c)}$ we have $\frac{f+69+14+5}{a+44}=5\implies f=5a+132$ . Using these two equations gets us $a=123$ . Plug this back into the equation to get $f=747$ . Thus, the total number of fish caught is $5+14+69+f+65+28+15=\boxed{943}$ - Heavytoothpaste


Answer: 15.0

The table below gives the percent of students in each grade at Annville and Cleona elementary schools:
\[\begin{tabular}{rccccccc}&\textbf{\underline{K}}&\textbf{\underline{1}}&\textbf{\underline{2}}&\textbf{\underline{3}}&\textbf{\underline{4}}&\textbf{\underline{5}}&\textbf{\underline{6}}\\ \textbf{Annville:}& 16\% & 15\% & 15\% & 14\% & 13\% & 16\% & 11\%\\ \textbf{Cleona:}& 12\% & 15\% & 14\% & 13\% & 15\% & 14\% & 17\%\end{tabular}\]
Annville has 100 students and Cleona has 200 students.  In the two schools combined, what percent of the students are in grade 6?
$\text{(A)}\ 12\% \qquad \text{(B)}\ 13\% \qquad \text{(C)}\ 14\% \qquad \text{(D)}\ 15\% \qquad \text{(E)}\ 28\%$

Solution: By the tables, Annville has $11$ 6th graders and Cleona has $34$ . Together they have $45$ 6th graders and $300$ total students, so the percent is $\frac{45}{300}=\frac{15}{100}= \boxed{15}$


Answer: 62.5

The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.
\[\begin{tabular}{|c|c|c|c|c|c|c|}\hline t & 0 & 1 & 2 & 3 & 4 & 5\\ \hline s & 0 & 10 & 40 & 90 & 160 & 250\\ \hline\end{tabular}\] The distance $s$ for $t = 2.5$ is:
$\textbf{(A)}\ 45\qquad \textbf{(B)}\ 62.5\qquad \textbf{(C)}\ 70\qquad \textbf{(D)}\ 75\qquad \textbf{(E)}\ 82.5$

Solution: Looking at the pattern, we can determine that $t=10s^2$ . Applying the relationship, we can see that $s = \boxed{62.5}$ when $t=2.5$


Answer: 75.0

The table shows some of the results of a survey by radiostation KACL. What percentage of the males surveyed listen to the station?
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & ? & 26 & ?\\ \hline Females & 58 & ? & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
$\textbf{(A)}\ 39\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 52\qquad\textbf{(D)}\ 55\qquad\textbf{(E)}\ 75$

Solution:
Filling out the chart, it becomes
$\begin{tabular}{|c|c|c|c|}\hline & Listen & Don't Listen & Total\\ \hline Males & 78 & 26 & 104\\ \hline Females & 58 & 38 & 96\\ \hline Total & 136 & 64 & 200\\ \hline\end{tabular}$
Thus, the percentage of males surveyed that listen to the station is $100 \cdot \frac{78}{104} \%= \boxed{75}$


Answer: 3.3

The taxi fare in Gotham City is $2.40 for the first $\frac12$ mile and additional mileage charged at the rate $0.20 for each additional 0.1 mile. You plan to give the driver a $2 tip. How many miles can you ride for $10?
$\textbf{(A) }3.0\qquad\textbf{(B) }3.25\qquad\textbf{(C) }3.3\qquad\textbf{(D) }3.5\qquad\textbf{(E) }3.75$

Solution:
Let $x$ be the number of miles you ride. The number of miles you ride after the first half mile is $x-0.5.$ We can write this equation:
\begin{align*} 10 &= 2.4 + 0.2 \times \frac{x-0.5}{0.1} + 2\\ 5.6 &= 2(x-0.5)\\ 2.8 &= x-0.5\\ x &= \boxed{3.3}

Solution:
\begin{array}{|c|c|c|}
Miles & Money & Remark\\
\hline
0 & 8 & \text{2  for  tip} \\
\hline
0.5 & 5.6 & \text{-2.4  for  first  0.5  mi} \\
\hline
3.3 & 0 & \text{-5.6  for  2.8  mi}
\end{array}
Miles travelled = $\boxed{3.3}$


Answer: 781.0

The teams $T_1$ $T_2$ $T_3$ , and $T_4$ are in the playoffs. In the semifinal matches, $T_1$ plays $T_4$ , and $T_2$ plays $T_3$ . The winners of those two matches will play each other in the final match to determine the champion. When $T_i$ plays $T_j$ , the probability that $T_i$ wins is $\frac{i}{i+j}$ , and the outcomes of all the matches are independent. The probability that $T_4$ will be the champion is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: There are two scenarios in which $T_4$ wins. The first scenario is where $T_4$ beats $T_1$ $T_3$ beats $T_2$ , and $T_4$ beats $T_3$ , and the second scenario is where $T_4$ beats $T_1$ $T_2$ beats $T_3$ , and $T_4$ beats $T_2$ . Consider the first scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{4+1}$ , the probability $T_3$ beats $T_2$ is $\frac{3}{3+2}$ , and the probability $T_4$ beats $T_3$ is $\frac{4}{4+3}$ . Therefore the first scenario happens with probability $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}$ . Consider the second scenario. The probability $T_4$ beats $T_1$ is $\frac{4}{1+4}$ , the probability $T_2$ beats $T_3$ is $\frac{2}{2+3}$ , and the probability $T_4$ beats $T_2$ is $\frac{4}{4+2}$ . Therefore the second scenario happens with probability $\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . By summing these two probabilities, the probability that $T_4$ wins is $\frac{4}{4+1}\cdot\frac{3}{3+2}\cdot\frac{4}{4+3}+\frac{4}{1+4}\cdot\frac{2}{2+3}\cdot\frac{4}{4+2}$ . Because this expression is equal to $\frac{256}{525}$ , the answer is $256+525=\boxed{781}$


Answer: 8671.0

The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$ , in order. What 4-digit number is represented by the code word $\text{CLUE}$
$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$

Solution: We can derive that $C=8$ $L=6$ $U=7$ , and $E=1$ . Therefore, the answer is $\boxed{8671}$ ~edited by Owencheng


Answer: 195.0

The terms of an arithmetic sequence add to $715$ . The first term of the sequence is increased by $1$ , the second term is increased by $3$ , the third term is increased by $5$ , and in general, the $k$ th term is increased by the $k$ th odd positive integer. The terms of the new sequence add to $836$ . Find the sum of the first, last, and middle terms of the original sequence.

Solution: If the sum of the original sequence is $\sum_{i=1}^{n} a_i$ then the sum of the new sequence can be expressed as $\sum_{i=1}^{n} a_i + (2i - 1) = n^2 + \sum_{i=1}^{n} a_i.$ Therefore, $836 = n^2 + 715 \rightarrow n=11.$ Now the middle term of the original sequence is simply the average of all the terms, or $\frac{715}{11} = 65,$ and the first and last terms average to this middle term, so the desired sum is simply three times the middle term, or $\boxed{195}.$

Solution: After the adding of the odd numbers, the total of the sequence increases by $836 - 715 = 121 = 11^2$ . Since the sum of the first $n$ positive odd numbers is $n^2$ , there must be $11$ terms in the sequence, so the mean of the sequence is $\dfrac{715}{11} = 65$ . Since the first, last, and middle terms are centered around the mean, our final answer is $65 \cdot 3 = \boxed{195}$

Solution: Proceed as in Solution 2 until it is noted that there are 11 terms in the sequence. Since the sum of the terms of the original arithmetic sequence is 715, we note that $\frac{2a_1 + 10d}{2} \cdot 11 = 715$ or $2a_1 + 10d = 130$ for all sets of first terms and common differences that fit the conditions given in the problem. Assume WLOG that $a_1 = 60$ and $d = 1$ . Then the first term of the corresponding arithmetic sequence will be $60$ , the sixth (middle) term will be $65$ , and the eleventh (largest) term will be $70$ . Thus, our final answer is $60 + 65 + 70 = \boxed{195}$


Answer: 90.0

The terms of the sequence $(a_i)$ defined by $a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}$ for $n \ge 1$ are positive integers. Find the minimum possible value of $a_1 + a_2$

Solution:
This question is guessable but let's prove our answer
\[a_{n + 2} = \frac {a_n + 2009} {1 + a_{n + 1}}\]
\[a_{n + 2}(1 + a_{n + 1})= a_n + 2009\]
\[a_{n + 2}+a_{n + 2} a_{n + 1}-a_n= 2009\]
lets put $n+1$ into $n$ now
\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= 2009\]
and set them equal now
\[a_{n + 3}+a_{n + 3} a_{n + 2}-a_{n+1}= a_{n + 2}+a_{n + 2} a_{n + 1}-a_n\]
\[a_{n + 3}-a_{n+1}+a_{n + 3} a_{n + 2}-a_{n + 2} a_{n + 1}= a_{n + 2}-a_n\]
let's rewrite it
\[(a_{n + 3}-a_{n+1})(a_{n + 2}+1)= a_{n + 2}-a_n\]
Let's make it look nice and let $b_n=a_{n + 2}-a_n$
\[(b_{n+1})(a_{n + 2}+1)= b_n\]
Since $b_n$ and $b_{n+1}$ are integers, we can see $b_n$ is divisible by $b_{n+1}$
But we can't have an infinite sequence of proper factors, unless $b_n=0$
Thus, $a_{n + 2}-a_n=0$
\[a_{n + 2}=a_n\]
So now, we know $a_3=a_1$
\[a_{3} = \frac {a_1 + 2009} {1 + a_{2}}\]
\[a_{1} = \frac {a_1 + 2009} {1 + a_{2}}\]
\[a_{1}+a_{1}a_{2} = a_1 + 2009\]
\[a_{1}a_{2} = 2009\]
To minimize $a_{1}+a_{2}$ , we need $41$ and $49$
Thus, our answer $= 41+49=\boxed{090}$

Solution:
If $a_{n} \ne \frac {2009}{a_{n+1}}$ , then either \[a_{n} = \frac {a_{n}}{1} < \frac {a_{n} + 2009}{1 + a_{n+1}} < \frac {2009}{a_{n+1}}\]
or
\[\frac {2009}{a_{n+1}} < \frac {2009 + a_{n}}{a_{n+1} + 1} < \frac {a_{n}}{1} = a_{n}\]
All the integers between $a_{n}$ and $\frac {2009}{a_{n+1}}$ would be included in the sequence. However the sequence is infinite, so eventually there will be a non-integral term.
So $a_{n} = \frac {2009}{a_{n+1}}$ , which $a_{n} \cdot a_{n+1} = 2009$ . When $n = 1$ $a_{1} \cdot a_{2} = 2009$ . The smallest sum of two factors which have a product of $2009$ is $41 + 49=\boxed{090}$

Solution:
Essentially you see that it must be an integer for infinite numbers, which doesn't quite seem probable. The most logical explanation is that the sequence repeats, and the numbers in the sequence that repeat are integers. We list out some terms. \begin{align*} a_{1} &= a \\ a_{2} &= b \\ a_{3} &=\frac{a+2009}{1+b} \\ a_{4} &=\frac{(b+1)(b+2009)}{a+b+2010} \\ \end{align*} The terms get more and more wacky, so we just solve for $a,b$ such that $a_{1}=a_{3}$ and $a_{2}=a_{4}.$
Solving we find both equations end up to the equation $ab=2009$ in which we see to minimize we see that $a = 49$ and $b=41$ or vice versa for an answer of $\boxed{90}.$ This solution is VERY non rigorous and not recommended.


Answer: 130.0

The third exit on a highway is located at milepost 40 and the tenth exit is at milepost 160. There is a service center on the highway located three-fourths of the way from the third exit to the tenth exit. At what milepost would you expect to find this   service center?
$\text{(A)}\ 90 \qquad \text{(B)}\ 100 \qquad \text{(C)}\ 110 \qquad \text{(D)}\ 120 \qquad \text{(E)}\ 130$

Solution: There are $160-40=120$ miles between the third and tenth exits, so the service center is at milepost $40+(3/4)(120) = 40+90=\boxed{130}$


Answer: 4.0

The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution:
If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):
\[\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]\]
Or you leave at least one row and one column unchanged:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]\]
In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums.  (In the example given, the sum in row $x$ is the same as in column $x$ .)
In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal.  (In the second example given, row $3$ and column $3$ are untouched.)
Either way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]\]
Letting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\boxed{4}$


Answer: 6.0

The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ .  If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution:
If $5b9$ is divisible by $9$ , this must mean that $5 + b + 9$ is a multiple of $9$ . So, \[5 + b + 9 = 9, 18, 27, 36...\]
Because $5 + 9 = 14$ and $b$ is in between 0 and 9,
\[5 + b + 9 = 18\] \[b = 4\]
The question states that \[2a3 + 326 = 549\] so \[2a3 = 549 - 326\] \[2a3 = 223\] \[a = 2\]
\[a + b = 6\] which is answer choice $\boxed{6}$


Answer: 64.0

The top of one tree is $16$ feet higher than the top of another tree. The heights of the two trees are in the ratio $3:4$ . In feet, how tall is the taller tree?
$\textbf{(A)}\ 48 \qquad\textbf{(B)}\ 64 \qquad\textbf{(C)}\ 80 \qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 112$

Solution: Let the height of the taller tree be $h$ and let the height of the smaller tree be $h-16$ . Since the ratio of the smaller tree to the larger tree is $\frac{3}{4}$ , we have $\frac{h-16}{h}=\frac{3}{4}$ . Solving for $h$ gives us $h=64 \Rightarrow \boxed{64}$

Solution: To answer this problem, you have to make it so that we have the same proportion as 3:4, but the difference between them is 16. Since the two numbers are consecutive, if we multiply both of them by 16, we would get a difference of 16 between them. So, it would be 48:64 and since we need to find the height of the taller tree, we get $h=64 \Rightarrow \boxed{64}$


Answer: 2013.0

The tower function of twos is defined recursively as follows: $T(1) = 2$ and $T(n + 1) = 2^{T(n)}$ for $n\ge1$ . Let $A = (T(2009))^{T(2009)}$ and $B = (T(2009))^A$ . What is the largest integer $k$ for which \[\underbrace{\log_2\log_2\log_2\ldots\log_2B}_{k\text{ times}}\] is defined?
$\textbf{(A)}\ 2009\qquad \textbf{(B)}\ 2010\qquad \textbf{(C)}\ 2011\qquad \textbf{(D)}\ 2012\qquad \textbf{(E)}\ 2013$

Solution:
(Note: This for the people who need clear, concise reasoning in the form of mostly words, instead of a compact proof written in set theory symbols and the like. I figured if I had trouble understanding the above solutions, others would too.)
We begin by contemplating what $B$ actually is. Calculating the first few values of $T(n)$ , we see that it quickly becomes absolutely, massively, well, MASSIVE. So actually calculating $T(2009)$ , is sadly, not an option (this is a final five problem, after all). Next we consider if it is possible to express $B$ differently. We have:
$B = T(2009)^{T(2009)^{T(2009)}} \Rightarrow \text{calculating.... ERROR}$
(I didn't actually try plug it into a computer program; the point is, it's not that easy)
So, the only thing left to do is jump right into the problem. We start by applying the $\log$ s one at a time, and seeing where it goes. We have:
$\log_2{\left(T(2009)^{\left(T(2009)^{T(2009)}\right)}\right)}$
(Uhhhhhh....)
But wait! We can use the rule $\log_b{x^y} = y*\log_b{x}$ . Then the expression becomes
$(T(2009)^{T(2009)}*\log_2{T(2009)} \Rightarrow (T(2009)^{T(2009)}*T(2008)$
(because $\log_2{T(n)} = {T(n-1)}$
Now we try $\log$ ing it again:
$\log_2{(T(2009)^{T(2009)}*T(2008)}$
This time, we can make use of the $\log$ addition rule $\log_b{x*y} = \log_b{x}+\log_b{y}$ . The expression becomes
$\log_2{(T(2009)^{T(2009)})} + \log_2{T(2008)}$
Pulling out the exponent again and further simplifying, we have
$\log_2{(\log_2{(B)})} = T(2009)*T(2008)+T(2007)$
Now we $\log$ it a third time... and we're stuck 😐. There is no pretty formula for taking the $\log$ of two things added together. So we rack our brains, sigh in frustration... and give up.
JUST KIDDING! (Don't give up)
Whenever you get stuck, you want to "zoom out" and re-gain perspective of what it is you're actually trying to find. The problem asks us to find the number of times we can keep taking the $\log$ of this expression, until we no longer can (once you reach a negative value, you can't take the log of it anymore; i.e. $\log$ s are only defined for positive inputs). So, for the context of this problem, what we want to know is:
"How many times can I take the $\log$ of $B$ and get a positive value?"
Let's think about what $\log$ actually means. We want a number $x$ , such that when $2$ is raised to the power of $x$ , we get the expression in the $\log$ . So, when we think about
$\log_2{T(2009)*T(2008)+T(2007)}$
What actually matters? Well, even if you have a big number, say, $124986068$ , what happens when you raise $2$ to the power of that number? We would end up getting another one of these:
Calculating... ERROR
(lol 🤣)
But in all seriousness, whenever we consider tower functions, the next tower number is always much, much bigger. So, back to the problem:
$\log_2{(T(2009)*T(2008)+T(2007))}$
$T(2007)$ is huge... but when you put it next to $T(2008)$ , it looks like an ant. Even more so when it's next to $T(2009)*T(2008)$ . So... what does that mean? Well, if we could get rid of the $T(2007)$ , we could at least go another step. So... let's get rid of it! (At this point I can almost see the audience: *surprised pikachu face)
\[\text{WARNING: THIS SECTION IS FOR CALCULUS USERS ONLY}\]
(Just kidding, but yes, the following part of the explanation does incorporate some calculus-related ideas)
If we were to consider just
$\log_2{(T(2009)*T(2008))}$
we would get
$\log_2{T(2009)} + \log_2{T(2008)} \Rightarrow T(2008) + T(2007)$
But what if we wanted to get
$T(2008) + T(2007)+1$
as our answer? What would the argument have to be? (In this context, "argument" refers to the value in the parentheses). Well, we can work backwards:
$T(2008) + T(2007)+1$
$\Rightarrow \log_2{T(2009)} + \log_2{T(2008)} + \log_2{2}$
$\Rightarrow \log_2{(T(2009)*T(2008)*2)}$
So the argument would have to be at least
$2*T(2009)*T(2008) = T(2009)*T(2008)+T(2009)*T(2008)$
But when we compare
$T(2009)*T(2008)+T(2009)*T(2008)$
to
$T(2009)*T(2008)+T(2007)$
Clearly
$T(2009)*T(2008) > T(2007)$
So,
$\log_2{(T(2009)*T(2008)+T(2007))}$
equals some value between
$T(2008) + T(2007)$
and
$T(2008) + T(2007) + 1$
Let's call that small excess value $\varepsilon$ (lowercase epsilon, literally "infinitely small but not 0"). We want to keep taking the $\log$ of the expression, and $\varepsilon$ is going to be pitifully small compared to $T(2008)$ or $T(2007)$ . So, within the context of this problem,
$T(2008) + T(2007) + \varepsilon \approx T(2008) + T(2007)$
Now we take the $\log$ again:
$\log_2{(T(2008)+T(2007))}$
Look familiar? It should, because this is almost exactly the same scenario as the one I just made a big fuss about (Take that, you no-know-calculus noobs! On an off note, it's completely fine if you don't know calculus; AMC problems are supposed to solvable by students who don't know calculus.) So, applying the same idea as earlier we have
$\log_2{(T(2008)+T(2007))} \approx \log_2{T(2008)}$
From here on out, it's calm waters. To recap,
$\log_2{(\log_2{(\log_2{(B)})})} \approx T(2008)$
Taking the $\log$ $2007$ more times(for a total of $2010$ ), we get
$\log_2\log_2... T(2008) = T(1) = 2$
Taking the log again, we get
$\log_2{2} = 1$
(Now we're at $2011$ times)
Once more and we have
$\log_2{1} = 0$
$2012$ times)
But wait, there's more! (IYKYK lol) Remember how we ignored the $\varepsilon$ ? Well, It's time to bring it back. Every time we took the log, we would get $T(\text{something})$ $\varepsilon$ because we defined $\varepsilon$ to be some very, very small (but finite) number. Let's go back a few steps:
$\log_2{2} = 1$
should really be
$\log_2{(2+\varepsilon)} = 1+\epsilon$
and
$\log_2{1} = 0$
should really be
$\log_2{(1+\varepsilon)} = \varepsilon$
Since $\varepsilon$ is still a positive number, we can take the $\log$ of it one last time before it becomes negative. So, after a long and very tiring journey (for both you, the reader, and me, the writer) we have $k$ (the number of times we took the $\log$ $= 2013 \Rightarrow \boxed{2013}$ .
~TheAsian


Answer: 47.0

The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$

Solution:
Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ .
We know \[3h=p\] \[4c=s\] \[3p=d\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3h+h+4c+c+(3\times3h)$ . This is equivalent to $13h+5c$ . Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$ . So the answer is $\boxed{47}$

Solution: As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$ . However, instead of going through each of the solutions and testing the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$ \[13\cdot 5-13-5=47,\] so our answer is $\boxed{47}$


Answer: 4.5

The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad.  Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles.  The width of the railroad track can be ignored.  How many square miles are in the plot of land ACD?
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0);  fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); }  label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Solution: The area of a triangle is $\frac12 bh$ . If we let $CD$ be the base of the triangle, then the height is $AB$ , and the area is $\frac12 \cdot 3 \cdot 3 = \boxed{4.5}$

Solution: We can see that there is a big triangle encasing $ACD$ . The area of that triangle is $\frac12 \cdot 3 \cdot 6 = 9$ . We can easily see that triangle $ABC$ is $\frac12 \cdot 3 \cdot 3$ , which is $4.5$ $9-4.5$ is $4.5$ , so the answer is $\boxed{4.5}$


Answer: 13.0

The twelve-sided figure shown has been drawn on $1 \text{ cm}\times 1 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$
[asy] unitsize(8mm); for (int i=0; i<7; ++i) {   draw((i,0)--(i,7),gray);   draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
$\textbf{(A) } 12 \qquad \textbf{(B) } 12.5 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 13.5 \qquad \textbf{(E) } 14$

Solution: We count $3 \cdot 3=9$ unit squares in the middle, and $8$ small triangles, which gives 4 rectangles each with an area of $1$ . Thus, the answer is $9+4=\boxed{13}$

Solution: We can see here that there are $9$ total squares in the middle. We also see that the triangles that make the corners of the shape have an area half the squares' area. Then, we can easily find that each corner has an area of one square and there are $4$ corners so we add that to the original 9 squares to get $9+4=\boxed{13}$ . That is how I did it.

Solution:
We can apply Pick's Theorem here. There are $8$ lattice points, and $12$ lattice points on the boundary. Then,
\[8 + 12 \div 2 - 1 = \boxed{13}.\]


Answer: 64.0

Solution: Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{64}$


Answer: 18.0

The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?
$\mathrm{(A) \ } 9 \qquad \mathrm{(B) \ } 18 \qquad \mathrm{(C) \ } 27 \qquad \mathrm{(D) \ } 36\qquad \mathrm{(E) \ } 45$

Solution:
If Jack's current age is $\overline{ab}=10a+b$ , then Bill's current age is $\overline{ba}=10b+a$
In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$
We are given that $10a+b+5=2(10b+a+5)$
Thus $8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}$
For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\boxed{18}$

Solution:
Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.
The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$
Thus Bill's age is in the set $\{13,22,31,40,49,58,67,76,85,94\}$
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\{13,49,58,67\}$
Checking each of them, we see that only $13$ works, and gives the solution $31-13=\boxed{18}$


Answer: 18.0

Solution:
If Jack's current age is $\overline{ab}=10a+b$ , then Bill's current age is $\overline{ba}=10b+a$
In five years, Jack's age will be $10a+b+5$ and Bill's age will be $10b+a+5$
We are given that $10a+b+5=2(10b+a+5)$
Thus $8a=19b+5 \Rightarrow a=\dfrac{19b+5}{8}$
For $b=1$ we get $a=3$ . For $b=2$ and $b=3$ the value $\frac{19b+5}8$ is not an integer, and for $b\geq 4$ $a$ is more than $9$ . Thus the only solution is $(a,b)=(3,1)$ , and the difference in ages is $31-13=\boxed{18}$

Solution:
Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.
The age difference is $(10a+b)-(10b+a)=9(a-b)$ , hence it is a multiple of $9$ . Thus Bill's current age modulo $9$ must be $4$
Thus Bill's age is in the set $\{13,22,31,40,49,58,67,76,85,94\}$
As Jack is older, we only need to consider the cases where the tens digit of Bill's age is smaller than the ones digit. This leaves us with the options $\{13,49,58,67\}$
Checking each of them, we see that only $13$ works, and gives the solution $31-13=\boxed{18}$


Answer: 3.0

The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$ . How long is the third altitude of the triangle?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution:
We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$ . By the Pythagorean Theorem , we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.
Let $h$ be the third height of the triangle. We have $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{3}$

Solution: By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$ . Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$ . Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is $\boxed{3}$ (We can also check from the other side).


Answer: 79.0

The two spinners shown are spun once and each lands on one of the numbered sectors. What is the probability that the sum of the numbers in the two sectors is prime?
[asy] unitsize(30);  draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$1$",(0,.5)); label("$3$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$5$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy] [asy] unitsize(30);  draw(unitcircle); draw((0,0)--(0,-1)); draw((0,0)--(cos(pi/6),sin(pi/6))); draw((0,0)--(-cos(pi/6),sin(pi/6))); label("$2$",(0,.5)); label("$4$",((cos(pi/6))/2,(-sin(pi/6))/2)); label("$6$",(-(cos(pi/6))/2,(-sin(pi/6))/2));[/asy]
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{2}{3}\qquad\textbf{(C)}\ \frac{3}{4}\qquad\textbf{(D)}\ \frac{7}{9}\qquad\textbf{(E)}\ \frac{5}{6}$

Solution:
The possible sums are \[\begin{tabular}{c|ccc} & 1 & 3 & 5 \\ \hline 2 & 3 & 5 & 7 \\ 4 & 5 & 7 & 9 \\ 6 & 7 & 9 & 11 \end{tabular}\]
Only $9$ is not prime, so there are $7$ prime numbers and $9$ total numbers for a probability of $\boxed{79}$


Answer: 185.0

The two squares shown share the same center $O_{}$ and have sides of length 1. The length of $\overline{AB}$ is $43/99$ and the area of octagon $ABCDEFGH$ is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers .  Find $m+n.$
[asy] //code taken from thread for problem real alpha = 25; pair W=dir(225), X=dir(315), Y=dir(45), Z=dir(135), O=origin; pair w=dir(alpha)*W, x=dir(alpha)*X, y=dir(alpha)*Y, z=dir(alpha)*Z; draw(W--X--Y--Z--cycle^^w--x--y--z--cycle); pair A=intersectionpoint(Y--Z, y--z),  C=intersectionpoint(Y--X, y--x),  E=intersectionpoint(W--X, w--x),  G=intersectionpoint(W--Z, w--z),  B=intersectionpoint(Y--Z, y--x),  D=intersectionpoint(Y--X, w--x),  F=intersectionpoint(W--X, w--z),  H=intersectionpoint(W--Z, y--z); dot(O); label("$O$", O, SE); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$E$", E, dir(O--E)); label("$F$", F, dir(O--F)); label("$G$", G, dir(O--G)); label("$H$", H, dir(O--H));[/asy]

Solution: Triangles $AOB$ $BOC$ $COD$ , etc. are congruent by symmetry (you can prove it rigorously by using the power of a point to argue that exactly two chords of length $1$ in the circumcircle of the squares pass through $B$ , etc.), and each area is $\frac{\frac{43}{99}\cdot\frac{1}{2}}{2}$ . Since the area of a triangle is $bh/2$ , the area of all $8$ of them is $\frac{86}{99}$ and the answer is $\boxed{185}$

Solution:
Define the two possible distances from one of the labeled points and the corners of the square upon which the point lies as $x$ and $y$ . The area of the octagon (by subtraction of areas) is $1 - 4\left(\frac{1}{2}xy\right) = 1 - 2xy$
By the Pythagorean theorem \[x^2 + y^2 = \left(\frac{43}{99}\right)^2\]
Also, \begin{align*}x + y + \frac{43}{99} &= 1\\ x^2 + 2xy + y^2 &= \left(\frac{56}{99}\right)^2\end{align*}
Substituting, \begin{align*}\left(\frac{43}{99}\right)^2 + 2xy &= \left(\frac{56}{99}\right)^2 \\ 2xy = \frac{(56 + 43)(56 - 43)}{99^2} &= \frac{13}{99} \end{align*}
Thus, the area of the octagon is $1 - \frac{13}{99} = \frac{86}{99}$ , so $m + n = \boxed{185}$


Answer: 0.0

The unit's digit (one's digit) of the product of any six consecutive positive whole numbers is
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$

Solution: Within six consecutive integers, there must be a number with a factor of $5$ and an even integer with a factor of $2$ . Multiplied together, these would produce a number that is a multiple of $10$ and has a units digit of $\boxed{0}$

Solution: We can easily compute the product of the first 6 positive integers: $(1*2*3*4*5*6)=6!=720$ Therefore the units digit must be $\boxed{0}$


Answer: 9.0

The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution:
First, we notice that $3^0$ is congruent to $1 \ \text{(mod 10)}$ $3^1$ is $3 \ \text{(mod 10)}$ $3^2$ is $9 \ \text{(mod 10)}$ $3^3$ is $7 \ \text{(mod 10)}$ $3^4$ is $1 \ \text{(mod 10)}$ , and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \ \text{(mod 10)}$
The number $7$ has a similar cycle, going $1, 7, 9, 3, 1, ...$ Hence we have that $7^{1002}$ is congruent to $9 \ \text{(mod 10)}$ . Finally, $13^{1003}$ is congruent to $3^{1003} \equiv 7 \ \text{(mod 10)}$ . Thus the required units digit is $3\cdot 9\cdot 7 \equiv 9 \ \text{(mod 10)}$ , so the answer is $\boxed{9}$


Answer: 10000000000.0

Solution:
We can make the approximations \begin{align*} 487,000 &\approx 500,000 \\  12,027,300 &\approx 12,000,000 \\ 9,621,001 &\approx 10,000,000 \\ 19,367 &\approx 20,000. \end{align*}
Using these instead of the original numbers for an estimate, we have \begin{align*} \frac{(500,000)(12,000,000)+(10,000,000)(500,000)}{(20000)(.05)} &= \frac{500,000\times 22,000,000}{1000} \\ &= 500,000 \times 22,000 \\ &= 1.1\times 10^{10} \\ &\approx 10,000,000,000 \rightarrow \boxed{10,000,000,000}


Answer: 4.0

Solution: When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \cdot 256^{0.09}$ \[256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\] Now we can convert the decimal exponent to a fraction: \[256^{0.25} = 256^{\frac{1}{4}}.\] Now, let us convert the expression into radical form. Since $4$ is the denominator of the fractional exponent, it will be the index exponent: \[256^{\frac{1}{4}}=\sqrt[4]{256}.\] Since $256 =16^2=(4^2)^2=4^4$ , we can solve for the fourth root of $256$ \[\sqrt[4]{256}=\sqrt[4]{4^4}=4.\] Therefore, $(256)^{.16} \cdot (256)^{.09}=\boxed{4}.$


Answer: 5.0

The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$

Solution: $\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{5}$

Solution: $\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{5}$


Answer: 14.0

The value of $x - y^{x - y}$ when $x = 2$ and $y = -2$ is:
$\textbf{(A)}\ -18 \qquad \textbf{(B)}\ -14\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 256$

Solution: Just plug in the numbers and follow the order of operations: \[2-(-2)^{2-(-2)}\] \[2-(-2)^4\] \[2-16\] \[\boxed{14}\]


Answer: 103.0

The value of $x$ that satisfies $\log_{2^x} 3^{20} = \log_{2^{x+3}} 3^{2020}$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Let $\log _{2^x}3^{20}=\log _{2^{x+3}}3^{2020}=n$ . Based on the equation, we get $(2^x)^n=3^{20}$ and $(2^{x+3})^n=3^{2020}$ . Expanding the second equation, we get $8^n\cdot2^{xn}=3^{2020}$ . Substituting the first equation in, we get $8^n\cdot3^{20}=3^{2020}$ , so $8^n=3^{2000}$ . Taking the 100th root, we get $8^{\frac{n}{100}}=3^{20}$ . Therefore, $(2^{\frac{3}{100}})^n=3^{20}$ , and using the our first equation( $2^{xn}=3^{20}$ ), we get $x=\frac{3}{100}$ and the answer is $\boxed{103}$ .
~rayfish

Solution:
Recall the identity $\log_{a^n} b^{m} = \frac{m}{n}\log_{a} b$ (which is easily proven using exponents or change of base).
Then this problem turns into \[\frac{20}{x}\log_{2} 3 = \frac{2020}{x+3}\log_{2} 3\] Divide $\log_{2} 3$ from both sides. And we are left with $\frac{20}{x}=\frac{2020}{x+3}$ .Solving this simple equation we get \[x = \tfrac{3}{100} \Rightarrow \boxed{103}\] ~mlgjeffdoge21

Solution:
$\log_{2^x} 3^{20} = 2^{xy} = 3^{20}$
$\log_{2^{x+3}} 3^{2020} = (2^{x+3})^y = 3^{2020}$
$3^{2020} = (3^{20})^{101}$
$(2^{xy})^{101} = (2^{x+3})^y = 3^{2020}$
$(2^{xy})^{101} = (2^{x+3})^y \Rightarrow 2^{101xy} = 2^{xy+3y} \Rightarrow 101xy = xy + 3y \Rightarrow 101xy = y(x+3)$
$101x = x + 3$
$100x = 3$
$x = \frac{3}{100}$
$100 + 3 = \boxed{103}$ ~Airplanes2007


Answer: 3.0

The value of the expression $\frac{(304)^5}{(29.7)(399)^4}$ is closest to
$\text{(A)}\ .003 \qquad \text{(B)}\ .03 \qquad \text{(C)}\ .3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 30$

Solution: \[\frac{(304)^5}{(29.7)(399)^4} \approx \frac{300^5}{30\cdot400^4} = \frac{3^5 \cdot 10^{10}}{3\cdot 4^4 \cdot 10^9} = \frac{3^4\cdot 10}{4^4} = \frac{810}{256}\] Which is closest to $3\rightarrow\boxed{3}$


Answer: 16.0

The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad  \textbf{(B) }\ - 4 \qquad  \textbf{(C) }\ 4 \qquad  \textbf{(D) }\ 8 \qquad  \textbf{(E) }\ 16$

Solution:
Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,
\begin{align*}     0 &= b^2-4ac\\     0 &= (-8)^2-4c\\     c &=  64\\     c &= 16\\ \end{align*}
Therefore $c=16$ , and our answer is $\boxed{16}$


Answer: 344.0

The vertices of $\triangle ABC$ are $A = (0,0)\,$ $B = (0,420)\,$ , and $C = (560,0)\,$ .  The six faces of a die are labeled with two $A\,$ 's, two $B\,$ 's, and two $C\,$ 's.  Point $P_1 = (k,m)\,$ is chosen in the interior of $\triangle ABC$ , and points $P_2\,$ $P_3\,$ $P_4, \dots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L\,$ , where $L \in \{A, B, C\}$ , and $P_n\,$ is the most recently obtained point, then $P_{n + 1}^{}$ is the midpoint of $\overline{P_n L}$ .  Given that $P_7 = (14,92)\,$ , what is $k + m\,$

Solution:
If we have points $(p,q)$ and $(r,s)$ and we want to find $(u,v)$ so $(r,s)$ is the midpoint of $(u,v)$ and $(p,q)$ , then $u=2r-p$ and $v=2s-q$ . So we start with the point they gave us and work backwards. We make sure all the coordinates stay within the triangle. We have \[P_{n-1}=(x_{n-1},y_{n-1}) = (2x_n\bmod{560},\ 2y_n\bmod{420})\] Then $P_7=(14,92)$ , so $x_7=14$ and $y_7=92$ , and we get \[\begin{array}{c||ccccccc} n & 7 & 6 & 5 & 4 & 3 & 2 & 1 \\ \hline\hline x_n & 14 & 28 & 56 & 112 & 224 & 448 & 336 \\ \hline y_n & 92 & 184 & 368 & 316 & 212 & 4 & 8  \end{array}\]
So the answer is $\boxed{344}$

Solution:
Let $L_1$ be the $n^{th}$ roll that directly influences $P_{n + 1}$
Note that $P_7 = \cfrac{\cfrac{\cfrac{P_1 + L_1}2 + L_2}2 + \cdots}{2\ldots} = \frac {(k,m)}{64} + \frac {L_1}{64} + \frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 + \frac {L_5}4 + \frac {L_6}2 = (14,92)$
Then quickly checking each addend from the right to the left, we have the following information (remembering that if a point must be $(0,0)$ , we can just ignore it!):
for $\frac {L_6}2,\frac {L_5}4$ , since all addends are nonnegative, a non- $(0,0)$ value will result in a $x$ or $y$ value greater than $14$ or $92$ , respectively, and we can ignore them,
for $\frac {L_4}8,\frac {L_3}{16},\frac {L_2}{32}$ in a similar way, $(0,0)$ and $(0,420)$ are the only possibilities,
and for $\frac {L_1}{64}$ , all three work.
Also, to be in the triangle, $0\le k\le560$ and $0\le m\le420$
Since $L_1$ is the only point that can possibly influence the $x$ coordinate other than $P_1$ , we look at that first.
If $L_1 = (0,0)$ , then $k = 2^6\cdot14 = 64\cdot14 > 40\cdot14 = 560$
so it can only be that $L_1 = (560,0)$ , and $k + 560 = 2^6\cdot14$
$\implies k = 64\cdot14 - 40\cdot14 = 24\cdot14 = 6\cdot56 = 336$
Now, considering the $y$ coordinate, note that if any of $L_2,L_3,L_4$ are $(0,0)$ $L_2$ would influence the least, so we test that),
then $\frac {L_2}{32} + \frac {L_3}{16} + \frac {L_4}8 < \frac {420}{16} + \frac {420}8 = 79\pm\epsilon < 80$
which would mean that $P_1 > 2^6\cdot(92 - 80) = 64\cdot12 > 42\cdot10 = 420\ge m$ , so $L_2,L_3,L_4 = (0,420)$
and now $\frac {P_1}{64} + \frac {420}{2^5} + \frac {420}{2^4} + \frac {420}{2^3} = 92$
$\implies P_1$
$= 64\cdot92 - 420(2 + 4 + 8)$
$= 64\cdot92 - 420\cdot14= 64(100 - 8) - 14^2\cdot30$
$= 6400 - 512 - (200 - 4)\cdot30$
$= 6400 - 512 - 6000 + 120$
$= - 112 + 120$
$= 8$
and finally, $k + m = 336 + 8 = \boxed{344}$


Answer: 12.0

The vertices of a quadrilateral lie on the graph of $y=\ln{x}$ , and the $x$ -coordinates of these vertices are consecutive positive integers. The area of the quadrilateral is $\ln{\frac{91}{90}}$ . What is the $x$ -coordinate of the leftmost vertex?
$\textbf{(A) } 6 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 13$

Solution: Let the coordinates of the quadrilateral be $(n,\ln(n)),(n+1,\ln(n+1)),(n+2,\ln(n+2)),(n+3,\ln(n+3))$ . We have by shoelace's theorem, that the area is \begin{align*} &\frac{\ln(n)(n+1) + \ln(n+1)(n+2) + \ln(n+2)(n+3)+n\ln(n+3)}{2} - \frac{\ln(n+1)(n) + \ln(n+2)(n+1) + \ln(n+3)(n+2)+\ln(n)(n+3)}{2} \\ &=\frac{\ln \left( \frac{n^{n+1}(n+1)^{n+2}(n+2)^{n+3}(n+3)^n}{(n+1)^n(n+2)^{n+1}(n+3)^{n+2}n^{n+3}}\right)}{2} \\ &= \ln \left( \sqrt{\frac{(n+1)^2(n+2)^2}{n^2(n+3)^2}} \right)  \\ &= \ln \left(\frac{(n+1)(n+2)}{n(n+3)}\right) \\ &= \ln \left( \frac{91}{90} \right). \end{align*} We know that the numerator must have a factor of $13$ , so given the answer choices, $n$ is either $12$ or $11$ . If $n=11$ , the expression $\frac{(n+1)(n+2)}{n(n+3)}$ does not evaluate to $\frac{91}{90}$ , but if $n=12$ , the expression evaluates to $\frac{91}{90}$ . Hence, our answer is $\boxed{12}$

Solution: Like above, use the shoelace formula to find that the area of the quadrilateral is equal to $\ln\frac{(n+1)(n+2)}{n(n+3)}$ . Because the final area we are looking for is $\ln\frac{91}{90}$ , the numerator factors into $13$ and $7$ , which one of $n+1$ and $n+2$ has to be a multiple of $13$ and the other has to be a multiple of $7$ . Clearly, the only choice for that is $\boxed{12}$


Answer: 144.0

The vertices of a regular nonagon (9-sided polygon) are to be labeled with the digits 1 through 9 in such a way that the sum of the numbers on every three consecutive vertices is a multiple of 3.  Two acceptable arrangements are considered to be indistinguishable if one can be obtained from the other by rotating the nonagon in the plane.  Find the number of distinguishable acceptable arrangements.

Solution:
First, we determine which possible combinations of digits $1$ through $9$ will yield sums that are multiples of $3$ . It is simplest to do this by looking at each of the digits $\bmod{3}$
We see that the numbers $1, 4,$ and $7$ are congruent to $1 \pmod{3}$ , that the numbers $2, 5,$ and $8$ are congruent to $2 \pmod{3}$ , and that the numbers $3, 6,$ and $9$ are congruent to $0 \pmod{3}$ . In order for a sum of three of these numbers to be a multiple of three, the mod $3$ sum must be congruent to $0$ . Quick inspection reveals that the only possible combinations are $0+0+0, 1+1+1, 2+2+2,$ and $0+1+2$ . However, every set of three consecutive vertices must sum to a multiple of three, so using any of $0+0+0, 1+1+1$ , or $2+2+2$ would cause an adjacent sum to include exactly $2$ digits with the same $\bmod{3}$ value, and this is an unacceptable arrangement. Thus the only possible groupings are composed of three digits congruent to three different $\bmod{3}$ values.
We see also that there are two possible arrangements for these trios on the nonagon: a digit congruent to $1 \pmod{3}$ can be located counterclockwise of a digit congruent to $0$ and clockwise of a digit congruent to $2 \pmod{3}$ , or the reverse can be true.
We set the first digit as $3$ avoid overcounting rotations, so we have one option as a choice for the first digit.  The other two $0 \pmod{3}$ numbers can be arranged in $2!=2$ ways.  The three $1 \pmod{3}$ and three $2 \pmod{3}$ can both be arranged in $3!=6$ ways.  Therefore, the desired result is $2(2 \times 6 \times 6)=\boxed{144}$

Solution: Notice that there are three triplets of congruent integers $\mod 3$ $(1,4,7),$ $(2,5,8),$ and $(3,6,9).$ There are $3!$ ways to order each of the triplets individually and $3!$ ways to order the triplets as a group (see solution 1). Rotations are indistinguishable, so there are $(3!)^4/9=\boxed{144}$ total arrangements.


Answer: 108.0

The vertices of an equilateral triangle lie on the hyperbola $xy=1$ , and a vertex of this hyperbola is the centroid of the triangle. What is the square of the area of the triangle?
$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 60\qquad\textbf{(C)}\ 108\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 169$

Solution: WLOG, let the centroid of $\triangle ABC$ be $I = (-1,-1)$ . The centroid of an equilateral triangle is the same as the circumcenter. It follows that the circumcircle must intersect the graph exactly three times. Therefore, $A = (1,1)$ , so $AI = BI = CI = 2\sqrt{2}$ , so since $\triangle AIB$ is isosceles and $\angle AIB = 120^{\circ}$ , then by the Law of Cosines $AB = 2\sqrt{6}$ . Alternatively, we can use the fact that the circumradius of an equilateral triangle is equal to $\frac {s}{\sqrt{3}}$ . Therefore, the area of the triangle is $\frac{(2\sqrt{6})^2\sqrt{3}}4 = 6\sqrt{3}$ , so the square of the area of the triangle is $\boxed{108}$

Solution: Without loss of generality, let the centroid of $\triangle ABC$ be $G = (-1,-1)$ . Then, one of the vertices must be the other curve of the hyperbola. Without loss of generality, let $A = (1,1)$ . Then, point $B$ must be the reflection of $C$ across the line $y=x$ , so let $B = \left(a,\frac{1}{a}\right)$ and $C=\left(\frac{1}{a},a\right)$ , where $a <-1$ . Because $G$ is the centroid, the average of the $x$ -coordinates of the vertices of the triangle is $-1$ . So we know that $a + 1/a+ 1 = -3$ . Multiplying by $a$ and solving gives us $a=-2-\sqrt{3}$ . So $B=(-2-\sqrt{3},-2+\sqrt{3})$ and $C=(-2+\sqrt{3},-2-\sqrt{3})$ . So $BC=2\sqrt{6}$ , and finding the square of the area gives us $\boxed{108}$

Solution: Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ and let point $A$ be $(-1, -1)$ . It is known that the centroid is equidistant from the three vertices of $\triangle ABC$ . Because we have the coordinates of both $A$ and $G$ , we know that the distance from $G$ to any vertice of $\triangle ABC$ is $\sqrt{(1-(-1))^2+(1-(-1))^2} = 2\sqrt{2}$ . Therefore, $AG=BG=CG=2\sqrt{2}$ . It follows that from $\triangle ABG$ , where $AG=BG=2\sqrt{2}$ and $\angle AGB = \dfrac{360^{\circ}}{3} = 120^{\circ}$ $[\triangle ABG]= \dfrac{(2\sqrt{2})^2 \cdot \sin(120)}{2} = 4 \cdot \dfrac{\sqrt{3}}{2} = 2\sqrt{3}$ using the formula for the area of a triangle with sine $\left([\triangle ABC]= \dfrac{1}{2} AB \cdot BC \sin(\angle ABC)\right)$ . Because $\triangle ACG$ and $\triangle BCG$ are congruent to $\triangle ABG$ , they also have an area of $2\sqrt{3}$ . Therefore, $[\triangle ABC] = 3(2\sqrt{3}) = 6\sqrt{3}$ . Squaring that gives us the answer of $\boxed{108}$


Answer: 108.0

Solution:
Without loss of generality, let the centroid of $\triangle ABC$ be $G = (1, 1)$ . Assuming we don't know one vertex is $(-1, -1)$ we let the vertices be $A\left(x_1, \frac{1}{x_1}\right), B\left(x_2, \frac{1}{x_2}\right), C\left(x_3, \frac{1}{x_3}\right).$
Since the centroid coordinates are the average of the vertex coordinates, we have that $\frac{x_1+x_2+x_3}{3}=1$ and $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1.$
We also know that the centroid is the orthocenter in an equilateral triangle, so $CG \perp AB.$ Examining slopes, we simplify the equation to $x_1x_2x_3 = -1$ . From the equation $\frac{\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}}{3}=1,$ we get that $x_1x_2+x_1x_3+x_2x_3 = -3$ . These equations are starting to resemble Vieta's:
$x_1+x_2+x_3=3$
$x_1x_2+x_1x_3+x_2x_3 = -3$
$x_1x_2x_3=-1$
$x_1,x_2,x_3$ are the roots of the equation $x^3 - 3x^2 - 3x + 1 = 0$ . This factors as $(x+1)(x^2-4x+1)=0 \implies x = -1, 2 \pm \sqrt3,$ for the points $(-1, -1), (2+\sqrt3, 2-\sqrt3), (2-\sqrt3, 2+\sqrt3)$ . The side length is clearly $\sqrt{24}$ , so the square of the area is $\boxed{108}.$


Answer: 32.0

The volume of a certain rectangular solid is $8$ cm , its total surface area is $32$ cm , and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \  } 36 \qquad \mathrm{(D) \  } 40 \qquad \mathrm{(E) \  }44$

Solution:
As the dimensions are in geometric progression, let them be $\frac{b}{r}$ $b$ , and $br$ cm, so the volume is $\left(\frac{b}{r}\right)(b)(br) = b^3$ , giving $b^3 = 8$ and thus $b = 2$ . The surface area condition now yields \begin{align*}2\left(\frac{2}{r}\right)(2)+2(2)(2r)+2(2r)\left(\frac{2}{r}\right) = 32 &\iff \frac{8}{r}+8+8r = 32 \\ &\iff r+\frac{1}{r} = 3 \\ &\iff r^2-3r+1 = 0 \\ &\iff r = \frac{3 \pm \sqrt{5}}{2}.\end{align*}
Since \[\frac{3-\sqrt{5}}{2} = \frac{1}{\left(\frac{3+\sqrt{5}}{2}\right)},\] the two possible values of $r$ do not actually give different dimensions, but merely determine whether they are in increasing or decreasing order. Therefore, without loss of generality, we take $r = \frac{3+\sqrt{5}}{2}$ , giving the dimensions (in cm) as \begin{align*}&\frac{2}{\left(\frac{3+\sqrt{5}}{2}\right)}, 2, \text{ and } 2\left(\frac{3+\sqrt{5}}{2}\right) \\ &= \frac{4}{3+\sqrt{5}}, 2, \text{ and } 3+\sqrt{5} \\ &= 3-\sqrt{5}, 2, \text{ and } 3+\sqrt{5}.\end{align*}
As there are $4$ edges with each of these distinct lengths, it follows that the sum of all the edge lengths (in cm) is \begin{align*}4\left(3-\sqrt{5}+2+3+\sqrt{5}\right) &= 4(8) \\ &= \boxed{32}

Solution: Similarly to in Solution 1, we let the dimensions (in cm) be $b$ $br$ , and $br^2$ , so that the volume condition gives $8 = b^3r^3 = (br)^3$ and thus $br = 2$ . The surface area condition now becomes \begin{align*}2(b)(br)+2(br)\left(br^2\right)+2\left(br^2\right)(b) = 32 &\iff b^2r+b^2r^2+b^2r^3 = 16 \\&\iff br\left(b+br+br^2\right) = 16,\end{align*} so substituting $br = 2$ from above immediately gives \[b+br+br^2 = \frac{16}{2} = 8,\] and hence, without needing to actually compute the dimensions, we deduce that the sum of the edge lengths (in cm) is \[4b + 4br + 4br^2 = 4(8) = \boxed{32}.\]


Answer: 9.0

The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$

Solution:
Draw an altitude towards the equilateral triangle base.  By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle.  This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\sqrt{3}$

Using the Pythagorean Theorem , the length of the altitude is $\sqrt{3}$ , so the volume of the triangular pyramid is $\tfrac13 \cdot \tfrac{6^2 \cdot \sqrt{3}}{4} \cdot \sqrt{3} = \boxed{9}$


Answer: 24.0

The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$

Solution:
If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2 &= 576\\ abc &= \boxed{24}


Answer: 4.0

The wheel shown below consists of two circles and five spokes, with a label at each point where a spoke meets a circle. A bug walks along the wheel, starting at point $A$ . At every step of the process, the bug walks from one labeled point to an adjacent labeled point. Along the inner circle the bug only walks in a counterclockwise direction, and along the outer circle the bug only walks in a clockwise direction. For example, the bug could travel along the path $AJABCHCHIJA$ , which has $10$ steps. Let $n$ be the number of paths with $15$ steps that begin and end at point $A$ . Find the remainder when $n$ is divided by $1000.$

Solution:
We divide this up into casework.  The "directions" the bug can go are $\text{Clockwise}$ $\text{Counter-Clockwise}$ , and $\text{Switching}$ .  Let an $I$ signal going clockwise (because it has to be in the inner circle), an $O$ signal going counter-clockwise, and an $S$ switching between inner and outer circles.  An example string of length fifteen that gets the bug back to $A$ would be $ISSIIISOOSISSII$ .
For the bug to end up back at $A$ , the difference between the number of $I$ 's and $O$ 's must be a multiple of $5$
So, the total number of ways is $1+2002+1001=3004$ which gives $\boxed{004}$ as the answer.

Solution: Define $A_n$ to be the number of sequences of length $n$ that ends at $A$ and similarly for the other spokes. Also let \[S_n=A_n+B_n+C_n+D_n+E_n+F_n+G_n+H_n+I_n+J_n\] Apparently everytime the bug has $2$ choices for its next move, thus we have $S_n=2^n$ . Now we attempt to find a recursive formula for $A_n$ \begin{align*} A_n&=J_{n-1}+E_{n-1} \\ &=(I_{n-2}+A_{n-2})+(D_{n-2}+F_{n-2}) \\ &=(H_{n-3}+B_{n-3})+(J_{n-3}+E_{n-3})+(G_{n-3}+C_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(B_{n-3}+C_{n-3}+E_{n-3}+G_{n-3}+H_{n-3}+J_{n-3})+(J_{n-3}+E_{n-3}) \\ &=(S_{n-3}-(I_{n-3}+A_{n-3}+D_{n-3}+F_{n-3}))+A_{n-2} \\ &=2^{n-3}-(J_{n-2}+E_{n-2})+A_{n-2} \\ &=2^{n-3}-A_{n-1}+A_{n-2} \\ \end{align*} Computing a few easy terms we have $A_0=0$ $A_1=0$ $A_2=1$ $A_3=0$ $A_4=3$ . Continuing the process yields $A_{15}=3\boxed{004}$


Answer: 36.0

Solution: Count the number of black squares in each letter. H has 11, E has 11, L has 7, and P has 10, giving the number of black squares to be $11+11+7+10=39$ . The total number of squares is $(15)(5)=75$ and the number of white squares is $75-39=\boxed{36}$


Answer: 450.0

The workers in a factory produce widgets and whoosits. For each product, production time is constant and identical for all workers, but not necessarily equal for the two products. In one hour, $100$ workers can produce $300$ widgets and $200$ whoosits. In two hours, $60$ workers can produce $240$ widgets and $300$ whoosits. In three hours, $50$ workers can produce $150$ widgets and $m$ whoosits. Find $m$

Solution:
Suppose that it takes $x$ hours for one worker to create one widget, and $y$ hours for one worker to create one whoosit.
Therefore, we can write that (note that two hours is similar to having twice the number of workers, and so on):
Solve the system of equations with the first two equations to find that $(x,y) = \left(\frac{1}{7}, \frac{2}{7}\right)$ . Substitute this into the third equation to find that $1050 = 150 + 2m$ , so $m = \boxed{450}$


Answer: 4.0

The year 2002 is a palindrome (a number that reads the same from left to right as it does from right to left). What is the product of the digits of the next year after 2002 that is a palindrome?
$\text{(A)}\ 0 \qquad \text{(B)}\ 4 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 16 \qquad \text{(E)}\ 25$

Solution: The palindrome right after 2002 is 2112. The product of the digits of 2112 is $\boxed{4}$

Solution: The palindrome formula is to add 110 to the number in order to get the next palindrome, a palindrome needs to be in the form as ABBA . We can use this in this case to get 2112. 2*1*1*2=4. Therefore, the answer is $\boxed{4}$


Answer: 12.0

The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$

Solution:
A 25% increase means the new population is $\frac{5}{4}$ of the original population.  A 25% decrease means the new population is $\frac{3}{4}$ of the original population.
Thus, after four years, the population is $1 \cdot \frac{5}{4} \cdot \frac{5}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{225}{256}$ times the original population.
Thus, the net change is -12%, so the answer is $\boxed{12}$


Answer: 16.0

The zeroes of the function $f(x)=x^2-ax+2a$ are integers. What is the sum of the possible values of $a?$
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18$

Solution:
By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.
Because the zeros are integral, the discriminant of the function, $a^2 - 8a$ , is a perfect square, say $k^2$ . Then adding 16 to both sides and completing the square yields \[(a - 4)^2 = k^2 + 16.\] Therefore $(a-4)^2 - k^2 = 16$ and \[((a-4) - k)((a-4) + k) = 16.\] Let $(a-4) - k = u$ and $(a-4) + k = v$ ; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$ . Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ( $u + v$ ), $(2, 8), (4, 4), (-2, -8), (-4, -4)$ , yields $a = 9, 8, -1, 0$ . These $a$ sum to $16$ , so our answer is $\boxed{16}$

Solution:
Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. Since the coefficient of the $x^2$ term is $1$ , the quadratic can be written as \[(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2\]
By comparing this with $x^2 - ax + 2a$ \[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]
Plugging the first equation in the second, \[r_1r_2 = 2 (r_1 + r_2).\] Rearranging gives \[r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.\] These factors can be $(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),$ or $(-2, -2).$
We want the number of distinct $a = r_1 + r_2$ , and these factors gives $a = -1, 0, 8, 9$ . So the answer is $-1 + 0 + 8 + 9 = \boxed{16}$