Answer: 2.0

Suppose that $f(x+3)=3x^2 + 7x + 4$ and $f(x)=ax^2 + bx + c$ . What is $a+b+c$
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ 3$

Solution:
As $f(x)=ax^2 + bx + c$ , we have $f(1)=a\cdot 1^2 + b\cdot 1 + c = a+b+c$
To compute $f(1)$ , set $x=-2$ in the first formula. We get $f(1) = f(-2+3) = 3(-2)^2 + 7(-2) + 4 = 12 - 14 + 4 = \boxed{2}$

Solution:
Combining the two formulas, we know that $f(x+3) = a(x+3)^2 + b(x+3) + c$
We can rearrange the right hand side to $ax^2 + (6a+b)x + (9a+3b+c)$
Comparing coefficients we have $a=3$ $6a+b=7$ , and $9a+3b+c = 4$ . From the second equation we get $b=-11$ , and then from the third we get $c=10$ . Hence $a+b+c = 3-11+10 = \boxed{2}$


Answer: 60.0

Suppose that $m$ and $n$ are positive integers such that $75m = n^{3}$ . What is the minimum possible value of $m + n$
$\text{(A)}\ 15 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 60 \qquad \text{(E)}\ 5700$

Solution: $3 \cdot 5^2m$ must be a perfect cube, so each power of a prime in the factorization for $3 \cdot 5^2m$ must be divisible by $3$ . Thus the minimum value of $m$ is $3^2 \cdot 5 = 45$ , which makes $n = \sqrt[3]{3^3 \cdot 5^3} = 15$ . The minimum possible value for the sum of $m$ and $n$ is $\boxed{60}.$

Solution: First, we need to prime factorize $75$ $75$ $5^2 \cdot 3$ . We need $75m$ to be in the form $x^3y^3$ . Therefore, the smallest $m$ is $5 \cdot 3^2$ $m$ = 45, and since $5^3 \cdot 3^3 = 15^3$ , our answer is $45 + 15$ $\boxed{60}$


Answer: 28.0

Suppose that $n$ is the product of three consecutive integers and that $n$ is divisible by $7$ . Which of the following is not necessarily a divisor of $n$
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 28 \qquad \textbf{(E)}\ 42$

Solution:
Whenever $n$ is the product of three consecutive integers, $n$ is divisible by $3!$ , meaning it is divisible by $6$
It also mentions that it is divisible by $7$ , so the number is definitely divisible by all the factors of $42$
In our answer choices, the one that is not a factor of $42$ is $\boxed{28}$

Solution: We can look for counterexamples. For example, letting $n = 13 \cdot 14 \cdot 15$ , we see that $n$ is not divisible by 28, so $\boxed{28}$ is our answer.


Answer: 74.0

Suppose that $x$ $y$ , and $z$ are complex numbers such that $xy = -80 - 320i$ $yz = 60$ , and $zx = -96 + 24i$ , where $i$ $=$ $\sqrt{-1}$ . Then there are real numbers $a$ and $b$ such that $x + y + z = a + bi$ . Find $a^2 + b^2$

Solution: The First (pun intended) thing to notice is that $xy$ and $zx$ have a similar structure, but not exactly conjugates, but instead once you take out the magnitudes of both, simply multiples of a root of unity. It turns out that root of unity is $e^{\frac{3\pi i}{2}}$ . Anyway this results in getting that $\left(\frac{-3i}{10}\right)y=z$ . Then substitute this into $yz$ to get, after some calculation, that $y=10e^{\frac{5\pi i}{4}}\sqrt{2}$ and $z=-3e^{\frac{7\pi i}{4}}\sqrt{2}$ . Then plug $z$ into $zx$ , you could do the same thing with $xy$ but $zx$ looks like it's easier due to it being smaller. Anyway you get $x=20+12i$ . Then add all three up, it turns out easier than it seems because for $z$ and $y$ the $\sqrt{2}$ disappears after you expand the root of unity (e raised to a specific power). Long story short, you get $x=20+12i, y=-3+3i, z=-10-10i \implies x+y+z=7+5i \implies a^2+b^2=\boxed{074}$

Solution: First we evaluate the magnitudes. $|xy|=80\sqrt{17}$ $|yz|=60$ , and $|zx|=24\sqrt{17}$ . Therefore, $|x^2y^2z^2|=17\cdot80\cdot60\cdot24$ , or $|xyz|=240\sqrt{34}$ . Divide to find that $|z|=3\sqrt{2}$ $|x|=4\sqrt{34}$ , and $|y|=10\sqrt{2}$  This allows us to see that the argument of $y$ is $\frac{\pi}{4}$ , and the argument of $z$ is $-\frac{\pi}{4}$ . We need to convert the polar form to a standard form. Simple trig identities show $y=10+10i$ and $z=3-3i$ . More division is needed to find what $x$ is. \[x=-20-12i\] \[x+y+z=-7-5i\] \[(-7)^2+(-5)^2=\boxed{74}\] \[QED\blacksquare\] Written by a1b2

Solution:
Solve this system the way you would if the RHS of all equations were real. Multiply the first and 3rd equations out and then factor out $60$ to find $x^2$ , then use standard techniques that are used to evaluate square roots of irrationals. let \[x = c+di\] , then you get \[c^2 - d^2 = 256\] and \[2cd = 480\] Solve to get $x$ as $20+12i$ and $-20-12i$ . Both will give us the same answer, so use the positive one. Divide $-80-320i$ by $x$ , and you get $10+10i$ as $y$ . This means that $z$ is a multiple of $1-i$ to get a real product, so you find $z$ is $3-3i$ . Now, add the real and imaginary parts separately to get $-7-5i$ , and calculate $a^2 + b^2$ to get $\boxed{74}$ .
~minor latex improvements done by jske25 and jdong2006

Solution:
Dividing the first equation by the second equation given, we find that $\frac{xy}{yz}=\frac{x}{z}=\frac{-80-320i}{60}=-\frac{4}{3}-\frac{16}{3}i \implies x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)$ . Substituting this into the third equation, we get $z^2=\frac{-96+24i}{-\frac{4}{3}-\frac{16}{3}i}=3\cdot \frac{-24+6i}{-1-4i}=3\cdot \frac{(-24+6i)(-1+4i)}{1+16}=3\cdot \frac{-102i}{17}=-18i$ . Taking the square root of this is equivalent to halving the argument and taking the square root of the magnitude. Furthermore, the second equation given tells us that the argument of $y$ is the negative of that of $z$ , and their magnitudes multiply to $60$ . Thus, we have $z=\sqrt{-18i}=3-3i$ and $3\sqrt{2}\cdot |y|=60 \implies |y|=10\sqrt{2} \implies y=10+10i$ . To find $x$ , we can use the previous substitution we made to find that $x=z\left(-\frac{4}{3}-\frac{16}{3}i\right)=-\frac{4}{3}\cdot (3-3i)(1+4i)=-4(1-i)(1+4i)=-4(5+3i)=-20-12i$ .
Therefore, $x+y+z=(-20+10+3)+(-12+10-3)i=-7-5i \implies a^2+b^2=(-7)^2+(-5)^2=49+25=\boxed{74}$

Solution:
We observe that by multiplying $xy,$ $yz,$ and $zx,$ we get $(xyz)^2=(-80-320i)(60)(-96+24i).$ Next, we divide $(xyz)^2$ by $(yz)^2$ to
get $x^2.$ We have $x^2=\frac{(-80-320i)(60)(-96+24i)}{3600}=256+480i.$ We can write $x$ in the form of $a+bi,$ so we get $(a+bi)^2=256+480i.$ Then, $a^2-b^2+2abi=256+480i,$ $a^2-b^2=256,$ and $2ab=480.$ Solving this system of equations is relatively
simple. We have two cases, $a=20, b=12,$ and $a=-20, b=-12.$ Case 1: $a=20, b=12,$ so $x=20+12i.$ We solve for $y$ and $z$ by plugging in $x$ to the two equations. We see $y=\frac{-80-320i}{20+12i}=-10-10i$ and $z=\frac{-96+24i}{20+12i}=-3+3i.$ $x+y+z=7+5i,$ so $a=7$ and $b=5.$ Solving, we end up with $7^2+5^2=\boxed{074}$ as our answer.
Case 2: $a=-20, b=-12,$ so $x=-20-12i.$ Again, we solve for $y$ and $z.$ We find $y=\frac{-80-320i}{-20-12i}=10+10i,$ $z=\frac{-96+24i}{-20-12i}=3-3i,$ so $x+y+z=-7-5i.$ We again have $(-7)^2+(-5)^2=\boxed{074}.$ Solution by Airplane50

Solution:
According to the Euler's Theory, we can rewrite $x$ $y$ and $z$ as \[x=r_{1}e^{i{\theta}_1}\] \[y=r_{2}e^{i{\theta}_2}\] \[x=r_{3}e^{i{\theta}_3}\] As a result, \[|xy|=r_{1}r_{2}=\sqrt{80^2+320^2}=80\sqrt{17}\] \[|yz|=r_{2}r_{3}=60\] \[|xz|=r_{1}r_{3}=\sqrt{96^2+24^2}=24\sqrt{17}\] Also, it is clear that \[yz=r_{2}e^{i{\theta}_2}r_{3}e^{i{\theta}_3}=|yz|e^{i({\theta}_2+{\theta}_3)}=|yz|=60\] So ${\theta}_2+{\theta}_3=0$ , or \[{\theta}_2=-{\theta}_3\] Also, we have \[xy=-80\sqrt{17}e^{i\arctan{4}}\] \[yz=60\] \[xz=-24\sqrt{17}e^{i\arctan{-\frac{1}{4}}}\] So now we have $r_{1}r_{2}=80\sqrt{17}$ $r_{2}r_{3}=60$ $r_{1}r_{3}=24\sqrt{17}$ ${\theta}_1+{\theta}_2=\arctan{4}$ and ${\theta}_1-{\theta}_2=\arctan {-\frac{1}{4}}$ . Solve these above, we get \[r_{1}=4\sqrt{34}\] \[r_{2}=10\sqrt{2}\] \[r_{3}=3\sqrt{2}\] \[{\theta}_2=\frac{\arctan{4}-\arctan{-\frac{1}{4}}}{2}=\frac{\frac{\pi}{2}}{2}=\frac{\pi}{4}\] So we can get \[y=r_{2}e^{i{\theta}_2}=10\sqrt{2}e^{i\frac{\pi}{4}}=10+10i\] \[z=r_{3}e^{i{\theta}_3}=r_{3}e^{-i{\theta}_2}=3\sqrt{2}e^{-i\frac{\pi}{4}}=3-3i\] Use $xy=-80-320i$ we can find that \[x=-20-12i\] So \[x+y+z=-20-12i+10+10i+3-3i=-7-5i\] So we have $a=-7$ and $b=-5$
As a result, we finally get \[a^2+b^2=(-7)^2+(-5)^2=\boxed{074}\]

Solution:
We can turn the expression $x+y+z$ into $\sqrt{x^2+y^2+z^2+2xy+2yz+2xz}$ , and this would allow us to plug in the values after some computations.
Based off of the given products, we have
$xy^2z=60(-80-320i)$
$xyz^2=60(-96+24i)$
$x^2yz=(-96+24i)(-80-320i)$
Dividing by the given products, we have
$y^2=\frac{60(-80-320i)}{-96+24i}$
$z^2=\frac{60(-96+24i)}{-80-320i}$
$x^2=\frac{(-96+24i)(-80-320i)}{60}$
Simplifying, we get that this expression becomes $\sqrt{24+70i}$ . This equals $\pm{(7+5i)}$ , so the answer is $7^2+5^2=\boxed{74}$

Solution: Multiplying $xy \cdot yz \cdot zx = (xyz)^2$ we obtain $60 \cdot 960(16+30i)$ (too lazy to do $60 \cdot 960$ , you don't need to). Taking the square root, we get $240\sqrt{16+30i}$ . Letting $(a+bi)^2=16+30i,$ we have $a^2+2abi-b^2=16+30i.$ Thus, $(a+b)(a-b)=16,$ and $2ab=30.$ Guessing and checking, we get $a+bi=5+3i$ . Therefore, $xyz=240(5+3i).$ Dividing this by each of the equations provided in the original problem, we get $x=20+12i,y=-10-10i,$ and $z=-3+3i$ $20+12i-10-10i-3+3i=7+5i$ . Finally, $7^2+5^2=\boxed{074}.$


Answer: 5.0

Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$
note: this is the type of problem that makes you think symmetry, but actually can be solved easily with substitution, and other normal technniques

Solution:
We can rewrite $xyz=1$ as $\frac{1}{z}=xy$
Substituting into one of the given equations, we have \[x+xy=5\] \[x(1+y)=5\] \[\frac{1}{x}=\frac{1+y}{5}.\]
We can substitute back into $y+\frac{1}{x}=29$ to obtain \[y+\frac{1+y}{5}=29\] \[5y+1+y=145\] \[y=24.\]
We can then substitute once again to get \[x=\frac15\] \[z=\frac{5}{24}.\] Thus, $z+\frac1y=\frac{5}{24}+\frac{1}{24}=\frac{1}{4}$ , so $m+n=\boxed{005}$

Solution:
Let $r = \frac{m}{n} = z + \frac {1}{y}$
\begin{align*} (5)(29)(r)&=\left(x + \frac {1}{z}\right)\left(y + \frac {1}{x}\right)\left(z + \frac {1}{y}\right)\\ &=xyz + \frac{xy}{y} + \frac{xz}{x} + \frac{yz}{z} + \frac{x}{xy} + \frac{y}{yz} + \frac{z}{xz} + \frac{1}{xyz}\\ &=1 + x + z + y + \frac{1}{y} + \frac{1}{z} + \frac{1}{x} + \frac{1}{1}\\ &=2 + \left(x + \frac {1}{z}\right) + \left(y + \frac {1}{x}\right) + \left(z + \frac {1}{y}\right)\\ &=2 + 5 + 29 + r\\ &=36 + r \end{align*}
Thus $145r = 36+r \Rightarrow 144r = 36 \Rightarrow r = \frac{36}{144} = \frac{1}{4}$ . So $m + n = 1 + 4 = \boxed{5}$

Solution:
(Hybrid between 1/2)
Because $xyz = 1, \hspace{0.15cm} \frac{1}{x} = yz, \hspace{0.15cm} \frac{1}{y} = xz,$ and $\hspace{0.05cm}\frac{1}{z} = xy$ . Substituting and factoring, we get $x(y+1) = 5$ $\hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . Multiplying them all together, we get, $xyz(x+1)(y+1)(z+1) = 145k$ , but $xyz$ is $1$ , and by the Identity property of multiplication, we can take it out. So, in the end, we get $(x+1)(y+1)(z+1) = 145k$ . And, we can expand this to get $xyz+xy+yz+xz+x+y+z+1 = 145k$ , and if we make a substitution for $xyz$ , and rearrange the terms, we get $xy+yz+xz+x+y+z = 145k-2$ This will be important.
Now, lets add the 3 equations $x(y+1) = 5, \hspace{0.15cm}y(z+1) = 29$ , and $\hspace{0.05cm}z(x+1) = k$ . We use the expand the Left hand sides, then, we add the equations to get $xy+yz+xz+x+y+z = k+34$ Notice that the LHS of this equation matches the LHS equation that I said was important. So, the RHS of both equations are equal, and thus $145k-2 = k+34$ We move all constant terms to the right, and all linear terms to the left, to get $144k = 36$ , so $k = \frac{1}{4}$ which gives an answer of $1+4 = \boxed{005}$

Solution: Get rid of the denominators in the second and third equations to get $xz-5z=-1$ and $xy-29x=-1$ . Then, since $xyz=1$ , we have $\tfrac 1y-5z=-1$ and $\tfrac 1z-29x=-1$ . Then, since we know that $\tfrac 1z+x=5$ , we can subtract these two equations to get that $30x=6\implies x=5$ . The result follows that $z=\tfrac 5{24}$ and $y=24$ , so $z+\tfrac 1y=\tfrac 1{24}+\tfrac 5{24}=\tfrac 14$ , and the requested answer is $1+4=\boxed{005}.$

Solution:
Rewrite the equations in terms of x.
$x+\frac{1}{z}=5$ becomes $z=\frac{1}{x+5}$
$y+\frac{1}{x}=29$ becomes $y=29-\frac{1}{x}$
Now express $xyz=1$ in terms of x.
$\frac{1}{5-x}\cdot(29-\frac{1}{x})\cdot x=1$
This evaluates to $29x-1=5-x$ , giving us $x=\frac{1}{5}$ . We can now plug x into the other equations to get $y=24$ and $z=\frac{5}{24}$
Therefore, $z+\frac{1}{y}=\frac{5}{24}+\frac{1}{24}=\frac{6}{24}=\frac{1}{4}$
$1+4=\boxed{5}$ , and we are done.
~MC413551


Answer: 529.0

Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$

Solution: Substitute $y = \frac34x$ into $x^y = y^x$ and solve. \[x^{\frac34x} = \left(\frac34x\right)^x\] \[x^{\frac34x} = \left(\frac34\right)^x \cdot x^x\] \[x^{-\frac14x} = \left(\frac34\right)^x\] \[x^{-\frac14} = \frac34\] \[x = \frac{256}{81}\] \[y = \frac34x = \frac{192}{81}\] \[x + y = \frac{448}{81}\] \[448 + 81 = \boxed{529}\]

Solution:
We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:
Dividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$
Taking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$
In the case $c = \frac34$ $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$ $y = \frac {64}{27}$ $x + y = \frac {256 + 192}{81} = \frac {448}{81}$ , yielding an answer of $448 + 81 = \boxed{529}$

Solution: Taking the logarithm base $x$ of both sides, we arrive with: \[y = \log_x y^x \Longrightarrow \frac{y}{x} = \log_{x} y = \log_x \left(\frac{3}{4}x\right) = \frac{3}{4}\] Now we proceed by the logarithm rule $\log(ab)=\log a + \log b$ . The equation becomes: \[\log_x \frac{3}{4} + \log_x x = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} + 1 = \frac{3}{4}\] \[\Longleftrightarrow \log_x \frac{3}{4} = -\frac{1}{4}\] \[\Longleftrightarrow x^{-\frac{1}{4}} = \frac{3}{4}\] \[\Longleftrightarrow \frac{1}{x^{\frac{1}{4}}} = \frac{3}{4}\] \[\Longleftrightarrow x^{\frac{1}{4}} = \frac{4}{3}\] \[\Longleftrightarrow \sqrt[4]{x} = \frac{4}{3}\] \[\Longleftrightarrow x = \left(\frac{4}{3}\right)^4=\frac{256}{81}\] Then find $y$ as in solution 3, and we get $\boxed{529}$


Answer: 20.0

Suppose that $|x_i| < 1$ for $i = 1, 2, \dots, n$ .  Suppose further that $|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n|.$ What is the smallest possible value of $n$

Solution:
Since $|x_i| < 1$ then
\[|x_1| + |x_2| + \dots + |x_n| = 19 + |x_1 + x_2 + \dots + x_n| < n.\]
So $n \ge 20$ . We now just need to find an example where $n = 20$ : suppose $x_{2k-1} = \frac{19}{20}$ and $x_{2k} = -\frac{19}{20}$ ; then on the left hand side we have $\left|\frac{19}{20}\right| + \left|-\frac{19}{20}\right| + \dots + \left|-\frac{19}{20}\right| = 20\left(\frac{19}{20}\right) = 19$ . On the right hand side, we have $19 + \left|\frac{19}{20} - \frac{19}{20} + \dots - \frac{19}{20}\right| = 19 + 0 = 19$ , and so the equation can hold for $n = \boxed{020}$

Solution: Straight off, we notice that the RHS must be greater than or equal to 19, because an absolute value only gives nonnegative values. It then becomes clear that $|x_1|+|x_2|+...+|x_n|\ge19$ . If each $x_n$ were equal to 1, then $n=19$ . However, $x_n<1$ , so there must be at least one extra term to satisfy the inequality. Therefore, $n=\boxed{020}$


Answer: 9.0

Suppose that 7 boys and 13 girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$ . The average value of $S$ (if all possible orders of these 20 people are considered) is closest to
$\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13$

Solution:
We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\frac7{20}\cdot\frac{13}{19}$ . Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also $\frac{7\cdot 13}{20\cdot 19}$ . Thus, the total probability of the two people being one boy and one girl is $\frac{91}{190}$
There are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of $S$ is $\frac{91}{10} \to \boxed{9}$

Solution: Suppose that the class tried every configuration. Boy $i$ and girl $j$ would stand next to each other in $2$ different orders, in $19$ different positions, $18!$ times each. Summing over all $i,j$ gives $7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!$ , so the average value of $S$ is $\boxed{9}$


Answer: 1.0

Solution:
The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$
Solving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$
Since we have to add the 2 triangles the final answer is $1$ , which is $\boxed{1}$


Answer: 11.0

Suppose that a parabola has vertex $\left(\frac{1}{4},-\frac{9}{8}\right)$ and equation $y = ax^2 + bx + c$ , where $a > 0$ and $a + b + c$ is an integer.  The minimum possible value of $a$ can be written in the form $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers.  Find $p + q$

Solution: If the vertex is at $\left(\frac{1}{4}, -\frac{9}{8}\right)$ , the equation of the parabola can be expressed in the form \[y=a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}.\] Expanding, we find that \[y=a\left(x^2-\frac{x}{2}+\frac{1}{16}\right)-\frac{9}{8},\] and \[y=ax^2-\frac{ax}{2}+\frac{a}{16}-\frac{9}{8}.\] From the problem, we know that the parabola can be expressed in the form $y=ax^2+bx+c$ , where $a+b+c$ is an integer. From the above equation, we can conclude that $a=a$ $-\frac{a}{2}=b$ , and $\frac{a}{16}-\frac{9}{8}=c$ . Adding up all of these gives us \[\frac{9a-18}{16}=a+b+c.\] We know that $a+b+c$ is an integer, so $9a-18$ must be divisible by $16$ .  Let $9a=z$ . If ${z-18}\equiv {0} \pmod{16}$ , then ${z}\equiv {2} \pmod{16}$ . Therefore, if $9a=2$ $a=\frac{2}{9}$ . Adding up gives us $2+9=\boxed{011}$

Solution: Complete the square. Since $a>0$ , the parabola must be facing upwards. $a+b+c=\text{integer}$ means that $f(1)$ must be an integer. The function can be recasted into $a\left(x-\frac{1}{4}\right)^2-\frac{9}{8}$ because the vertex determines the axis of symmetry and the critical value of the parabola. The least integer greater than $-\frac{9}{8}$ is $-1$ . So the $y$ -coordinate must change by $\frac{1}{8}$ and the $x$ -coordinate must change by $1-\frac{1}{4}=\frac{3}{4}$ . Thus, $a\left(\frac{3}{4}\right)^2=\frac{1}{8}\implies \frac{9a}{16}=\frac{1}{8}\implies a=\frac{2}{9}$ . So $2+9=\boxed{011}$

Solution:
To do this, we can use the formula for the minimum (or maximum) value of the $x$ coordinate at a vertex of a parabola, $-\frac{b}{2a}$ and equate this to $\frac{1}{4}$ . Solving, we get $-\frac{a}{2}=b$ . Enter $x=\frac{1}{4}$ to get $-\frac{9}{8}=\frac{a}{16}+\frac{b}{4}+c=-\frac{a}{16}+c$ so $c=\frac{a-18}{16}$ . This means that $\frac{9a-18}{16}\in \mathbb{Z}$ so the minimum of $a>0$ is when the fraction equals -1, so $a=\frac{2}{9}$ . Therefore, $p+q=2+9=\boxed{011}$ .
-Gideontz


Answer: 399.0

Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$

Solution:
WLOG, let C be the largest angle in the triangle.
As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$
Expanding, we get
$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$
$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$
$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$
CASE 1: If $\sin 3A = 0$ or $\sin 3B = 0$
This implies one or both of A or B are 60 or 120.
If one of A or B is 120, we have a contradiction, since C must be the largest angle.
Otherwise, if one of A or B is 60, WLOG, assume A = 60, we would have $\cos(3B) + \cos(3C) = 2$ , and thus, cos(3B) and cos(3C) both equal 1, implying $B = C = 120$ , a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees.
CASE 2: If $\sin 3A \neq 0$ and $\sin 3B \neq 0$
$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$
$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$
Note that $\tan{x}=\frac{1}{\tan(90-x)}$ , or $\tan{x}\tan(90-x)=1$
Thus $\frac{3A}{2}+\frac{3B}{2}=90$ , or $A+B=60$
Now we know that $C=120$ , so we can just use the Law of Cosines to get $\boxed{399}$

Solution: \[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$ , then $\frac{3}{2}(A+B)=90$ $A+B=60$ , so $C=120$ ; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A=0$ or $\sin\frac{3}{2}B=0$ , i.e., either $A=120$ or $B=120$ . In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $\boxed{399}$

Solution: Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \sqrt{269-260\cos{A}}$ . We notice that $\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$ , so $269-260\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\cos{A} = \frac{-1}{2}$ . We do this because $120$ degrees is a "common" value and it makes the value inside the square root an integer. Plugging this into $269-260\cos{A}$ for $m$ we get that it is $\boxed{399}$


Answer: 4.0

Suppose that the euro is worth 1.3 dollars. If Diana has 500 dollars and Etienne has 400 euros, by what percent is the value of Etienne's money greater that the value of Diana's money?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6.5\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 13$

Solution: The ratio $\frac{400 \text{ euros}}{500 \text{ dollars}}$ can be simplified using conversion factors: \[\frac{400 \text{ euros}}{500 \text{ dollars}} \cdot \frac{1.3 \text{ dollars}}{1 \text{ euro}} = \frac{520}{500} = 1.04\] which means the money is greater by $\boxed{4}$ percent.

Solution: If we divide each of Etienne's and Diana's values by $100$ , the problem stays the same. Then, Etienne has $1.3$ times the amount of money Diana has, so Etienne has $5.2$ dollars. Since $\dfrac{5.2}{5} = 1.04$ , Etienne has $\boxed{4}$ percent more money than Diana. ~Extremelysupercooldude


Answer: 275.0

Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$

Solution: There are $24 \cdot 60=1440$ normal minutes in a day , and $10 \cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$ , which simplifies to $\frac{36}{25}$ . This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ AM, $6 \cdot 60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \cdot 11=275$ to $\text{0:00}$ gives $\text{2:75}$ , so the answer is $\boxed{275}$

Solution: First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\frac{1}{24}$ of a day, and 36 minutes is $\frac{36}{60}=\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\frac{11}{40}$ of a day is $\frac{11}{40}\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\boxed{275}$ - mathleticguyyy


Answer: 194.0

Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$ . What is the value of $a^{4} + a^{ - 4}$
$\textbf{(A)}\ 164 \qquad \textbf{(B)}\ 172 \qquad \textbf{(C)}\ 192 \qquad \textbf{(D)}\ 194 \qquad \textbf{(E)}\ 212$

Solution: Note that for all real numbers $k,$ we have $a^{2k} + a^{-2k} + 2 = (a^{k} + a^{-k})^2,$ from which \[a^{2k} + a^{-2k} = (a^{k} + a^{-k})^2-2.\] We apply this result twice to get the answer: \begin{align*} a^4 + a^{-4} &= (a^2 + a^{-2})^2 - 2 \\ &= [(a + a^{-1})^2 - 2]^2 - 2 \\ &= \boxed{194} ~Azjps (Fundamental Logic)

Solution:
Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Squaring both sides of $a^2+a^{-2}=14$ gives $a^4+a^{-4}+2=196,$ from which $a^4+a^{-4}=\boxed{194}.$

Solution: The detailed explanation of Solution 2 is as follows: \begin{alignat*}{8} a+a^{-1}&=4 \\ (a+a^{-1})^2&=4^2 \\ a^2+2aa^{-1}+a^{-2}&=16 \\ a^2+a^{-2}&=16-2&&=14 \\ (a^2+a^{-2})^2&=14^2 \\ a^4+2a^2a^{-2}+a^{-4}&=196 \\ a^4+a^{-4}&=196-2&&=\boxed{194} ~MathFun1000 (Solution)

Solution:
Squaring both sides of $a+a^{-1}=4$ gives $a^2+a^{-2}+2=16,$ from which $a^2+a^{-2}=14.$
Applying the Binomial Theorem, we raise both sides of $a+a^{-1}=4$ to the fourth power: \begin{align*} \binom40a^4a^0+\binom41a^3a^{-1}+\binom42a^2a^{-2}+\binom43a^1a^{-3}+\binom44a^0a^{-4}&=256 \\ a^4+4a^2+6+4a^{-2}+a^{-4}&=256 \\ \left(a^4+a^{-4}\right)+4\left(a^2+a^{-2}\right)&=250 \\ \left(a^4+a^{-4}\right)+4(14)&=250 \\ a^4+a^{-4}&=\boxed{194} ~MRENTHUSIASM

Solution:
We multiply both sides of $4=a+a^{-1}$ by $a,$ then rearrange: \[a^2-4a+1=0.\]
We apply the Quadratic Formula to get $a=2\pm\sqrt3.$
By Vieta's Formulas, note that the roots are reciprocals of each other. Therefore, both values of $a$ produce the same value of $a^4+a^{-4}:$ \begin{align*} a^4+a^{-4}&=\left(2+\sqrt{3}\right)^4 + \left(2+\sqrt{3}\right)^{-4} \\ &=\left(2+\sqrt{3}\right)^4+\left(2-\sqrt{3}\right)^4 &&(*) \\ &=\boxed{194}$

Solution: From the first sentence of Solution 4, we conclude that $a$ and $a^{-1}$ are the roots of $x^2-4x+1=0.$ Let \begin{align*} P_1&=a+a^{-1}, \\ P_2&=a^2+a^{-2}, \\ P_3&=a^3+a^{-3}, \\ P_4&=a^4+a^{-4}. \end{align*} By Newton's Sums, we have \begin{alignat*}{12} &1\cdot P_1-4\cdot 1&&=0 &&\qquad\implies\qquad P_1&&=4, \\ &1\cdot P_2-4\cdot P_1+1\cdot2 &&=0 &&\qquad\implies\qquad P_2&&=14, \\ &1\cdot P_3-4\cdot P_2+1\cdot P_1&&=0 &&\qquad\implies\qquad P_3&&=52, \\ &1\cdot P_4-4\cdot P_3+1\cdot P_2&&=0 &&\qquad\implies\qquad P_4&&=\boxed{194} ~Albert1993 (Fundamental Logic)

Solution: Note that \[a^{4} + a^{-4} = (a^{2} + a^{-2})^{2} - 2.\] We guess that $a^{2} + a^{-2}$ is an integer, so the answer must be $2$ less than a perfect square. The only possibility is $\boxed{194}.$


Answer: 4.0

Suppose that the sum of the squares of two complex numbers $x$ and $y$ is $7$ and the sum of the cubes is $10$ . What is the largest real value that $x + y$ can have?

Solution:
One way to solve this problem is by substitution . We have
$x^2+y^2=(x+y)^2-2xy=7$ and $x^3+y^3=(x+y)(x^2-xy+y^2)=(7-xy)(x+y)=10$
Hence observe that we can write $w=x+y$ and $z=xy$
This reduces the equations to $w^2-2z=7$ and $w(7-z)=10$
Because we want the largest possible $w$ , let's find an expression for $z$ in terms of $w$
$w^2-7=2z \implies z=\frac{w^2-7}{2}$
Substituting, $w^3-21w+20=0$ , which factorizes as $(w-1)(w+5)(w-4)=0$ (the Rational Root Theorem may be used here, along with synthetic division).
The largest possible solution is therefore $x+y=w=\boxed{004}$

Solution:
An alternate way to solve this is to let $x=a+bi$ and $y=c+di$
Because we are looking for a value of $x+y$ that is real, we know that $d=-b$ , and thus $y=c-bi$
Expanding $x^2+y^2=7+0i$ will give two equations, since the real and imaginary parts must match up.
$(a+bi)^2+(c-bi)^2=7+0i$
$(a^2+c^2-2b^2)+(2ab-2cb)i=7+0i$
Looking at the imaginary part of that equation, $2ab-2cb=0$ , so $a=c$ , and $x$ and $y$ are actually complex conjugates.
Looking at the real part of the equation and plugging in $a=c$ $2a^2-2b^2=7$ , or $2b^2=2a^2-7$
Now, evaluating the real part of $(a+bi)^3+(a-bi)^3$ , which equals $10$ (ignoring the odd powers of $i$ , since they would not result in something in the form of $10+0i$ ):
$a^3+3a(bi)^2+a^3+3a(-bi)^2=10$
$2a^3-6ab^2=10$
Since we know that $2b^2=2a^2-7$ , it can be plugged in for $b^2$ in the above equation to yield:
$2a^3-3a(2a^2-7)=10$
$-4a^3+21a=10$
$4a^3-21a+10=0$
Since the problem is looking for $x+y=2a$ to be a positive integer, only positive half-integers (and whole-integers) need to be tested.  From the Rational Roots theorem, $a=10, a=5, a=\frac{5}{2}$ all fail, but $a=2$ does work.  Thus, the real part of both numbers is $2$ , and their sum is $\boxed{004}$

Solution:
Begin by assuming that $x$ and $y$ are roots of some polynomial of the form $w^2+bw+c$ , such that by Vieta's Formulæ and some algebra (left as an exercise to the reader), $b^2-2c=7$ and $3bc-b^3=10$ .
Substituting $c=\frac{b^2-7}{2}$ , we deduce that $b^3-21b-20=0$ , whose roots are $-4$ $-1$ , and $5$ .
Since $-b$ is the sum of the roots and is maximized when $b=-4$ , the answer is $-(-4)=\boxed{004}$

Solution: $x^3 + y^3 = 10 = (x+y)(x^2-xy+y^2) = (x+y)(7-xy) \implies xy = 7 - \frac{10}{x+y}.$ Also, $(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 = 10 + 3xy(x+y).$ Substituting our above into this, we get $10 + 3(7-\frac{10}{x+y})(x+y) = 21x+21y-20 = (x+y)^3$ .  Letting $p = x+y$ , we have that $p^3 - 21p + 20 = 0$ .  Testing $p = 1$ , we find that this is a root, to get $(p-1)(p^2+p-20) = 0 \implies p = -5, 1, 4 \implies \boxed{4}$


Answer: 80.0

Suppose the estimated $20$ billion dollar cost to send a person to the planet Mars is shared equally by the $250$ million people in the U.S. Then each person's share is
$\text{(A)}\ 40\text{ dollars} \qquad \text{(B)}\ 50\text{ dollars} \qquad \text{(C)}\ 80\text{ dollars} \qquad \text{(D)}\ 100\text{ dollars} \qquad \text{(E)}\ 125\text{ dollars}$

Solution: We want the cost per person, which is \begin{align*} \frac{20\text{ billion}}{250\text{ million}} &= \frac{20000\text{ million}}{250\text{ million}} \\ &= 80 \rightarrow \boxed{80}


Answer: 0.25

Suppose there is a special key on a calculator that replaces the number $x$ currently displayed with the number given by the formula $1/(1-x)$ .  For example, if the calculator is displaying 2 and the special key is pressed, then the calculator will display -1 since $1/(1-2)=-1$ .  Now suppose that the calculator is displaying 5.  After the special key is pressed 100 times in a row, the calculator will display
$\text{(A)}\ -0.25 \qquad \text{(B)}\ 0 \qquad \text{(C)}\ 0.8 \qquad \text{(D)}\ 1.25 \qquad \text{(E)}\ 5$

Solution:
We look for a pattern, hoping this sequence either settles down to one number, or that it forms a cycle that repeats.
After $1$ press, the calculator displays $\frac{1}{1 - 5} = -\frac{1}{4}$
After $2$ presses, the calculator displays $\frac{1}{1 - (-\frac{1}{4})} = \frac{1}{\frac{5}{4}} = \frac{4}{5}$
After $3$ presses, the calculator displays $\frac{1}{1 - \frac{4}{5}} = \frac{1}{\frac{1}{5}} = 5$
Thus, every three presses, the display will be $5$ .  On press $3\cdot 33 = 99$ , the display will be $5$ .  One more press will give $-\frac{1}{4}$ , which is answer $\boxed{0.25}$


Answer: 2.0

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$ . What is the value of $\frac{x+3y}{3x-y}$
$\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solution:
Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\implies x=y$ .
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{2}$

Solution: Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\frac{x+3y}{3x-y}=\boxed{2}$

Solution: Let $y=ax$ . The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{2}$ - mathleticguyyy

Solution: Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\boxed{2}$


Answer: 2.0

Solution:
Rearranging, we find $3x+y=-2x+6y$ , or $5x=5y\implies x=y$ .
Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{2}$

Solution: Substituting each $x$ and $y$ with $1$ , we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$ . Thus, $\frac{x+3y}{3x-y}=\boxed{2}$

Solution: Let $y=ax$ . The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$ , which simplifies to $3+a=-2(1-3a)$ . After a bit of algebra we found out $a=1$ , which means that $x=y$ . Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{2}$ - mathleticguyyy

Solution: Let $x=1$ . Then $y=1$ . So the desired result is $2$ . Select $\boxed{2}$


Answer: 14.0

Susan had 50 dollars to spend at the carnival. She spent 12 dollars on food and twice as much on rides. How many dollars did she have left to spend?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 26 \qquad \textbf{(D)}\ 38 \qquad \textbf{(E)}\ 50$

Solution: If Susan spent 12 dollars, then twice that much on rides, then she spent $12+12 \times 2=36$ dollars in total. We subtract $36$ from $50$ to get $\boxed{14}$


Answer: 21.0

Suzanne went to the bank and withdrew $$800$ . The teller gave her this amount using $$20$ bills, $$50$ bills, and $$100$ bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?
$\textbf{(A) } 45 \qquad \textbf{(B) } 21 \qquad \text{(C) } 36 \qquad \text{(D) } 28 \qquad \text{(E) } 32$

Solution:
Denote by $x$ $y$ $z$ the amount of $20 bills, $50 bills and $100 bills, respectively.
Thus, we need to find the number of tuples $\left( x , y, z \right)$ with $x, y, z \in \Bbb N$ that satisfy \[ 20 x + 50 y + 100 z = 800.  \]
First, this equation can be simplified as \[ 2 x + 5 y + 10 z = 80. \]
Second, we must have $5 |x$ . Denote $x = 5 x'$ .
The above equation can be converted to \[ 2 x' + y + 2 z = 16 . \]
Third, we must have $2 | y$ . Denote $y = 2 y'$ .
The above equation can be converted to \[ x' + y' + z = 8 . \]
Denote $x'' = x' - 1$ $y'' = y' - 1$ and $z'' = z - 1$ .
Thus, the above equation can be written as \[ x'' + y'' + z'' = 5 . \]
Therefore, the number of non-negative integer solutions $\left( x'', y'', z'' \right)$ is $\binom{5 + 3 - 1}{3 - 1} = \boxed{21}$

Solution:
We notice that each $100 can be split 3 ways: 5 $20 dollar bills, 2 $50 dollar bills, or 1 $100 dollar bill.
There are 8 of these $100 chunks in total--take away 3 as each split must be used at least once.
Now there are five left--so we use stars and bars.
5 chunks, 3 categories or 2 bars. This gives us $\binom{5+2}{2}=\boxed{21}$


Answer: 78.0

Solution:
Finding the area of the shaded walkway can be achieved by computing the total area of Tamara's garden and then subtracting the combined area of her six flower beds.
Since the width of Tamara's garden contains three margins, the total width is $2\cdot 6+3\cdot 1 = 15$ feet.
Similarly, the height of Tamara's garden is $3\cdot 2+4\cdot 1 = 10$ feet.
Therefore, the total area of the garden is $15\cdot 10 =150$ square feet.
Finally, since the six flower beds each have an area of $2\cdot 6 = 12$ square feet, the area we seek is $150 - 6\cdot 12$ , and our answer is $\boxed{78}$

Solution:
The long horizontal walkways have an area of $15 ft^{2}$ each, and there are $4$ of them $\longrightarrow 60 ft^{2}$
The short vertical walkways have an area of $2 ft^{2}$ each, and there are $9$ of them $\longrightarrow 18 ft^{2}$
Adding the areas together, we have $60 ft^{2} + 18 ft^{2} = 78 ft^{2} \Longrightarrow \boxed{78}$ .
~JH. L


Answer: 42.0

Teams $A$ and $B$ are playing in a basketball league where each game results in a win for one team and a loss for the other team. Team $A$ has won $\tfrac{2}{3}$ of its games and team $B$ has won $\tfrac{5}{8}$ of its games. Also, team $B$ has won $7$ more games and lost $7$ more games than team $A.$ How many games has team $A$ played?
$\textbf{(A) } 21 \qquad \textbf{(B) } 27 \qquad \textbf{(C) } 42 \qquad \textbf{(D) } 48 \qquad \textbf{(E) } 63$

Solution:
Suppose team $A$ has played $g$ games in total so that it has won $\frac23g$ games.
It follows that team $B$ has played $g+14$ games in total so that it has won $\frac23g+7$ games.
We set up and solve an equation for team $B$ 's win ratio: \begin{align*} \frac{\frac23g+7}{g+14}&=\frac58 \\ \frac{16}{3}g+56&=5g+70 \\ \frac13g&=14 \\ g&=\boxed{42} ~MRENTHUSIASM

Solution: If we consider the number of games team $B$ has played as $x$ and the number of games that team $A$ has played as $y$ , then we can set up the following system of equations: \begin{align*} \frac{5}{8}x &= \frac{2}{3}y+7, \\ \frac{3}{8}x &= \frac{1}{3}y+7. \end{align*} The first system equated the number of wins of each team, while the second system equates the number of losses by each team. By multiplying the second equation by $2$ and solving the system, we get $y = 42$ or answer choice $\boxed{42}.$

Solution: First, let us assign some variables. Let \[A_w=2x, \ A_l=x, \ A_g=3x,\] \[B_w=5y, \ B_l=3y, \ B_g=8y,\] where $X_w$ denotes number of games won, $X_l$ denotes number of games lost, and $X_g$ denotes total games played for $X\in \{A, B\}$ . Using the given information, we can set up the following two equations: \begin{align*} B_w=A_w+7&\implies 5y=2x+7, \\ B_l=A_l+7&\implies 3y=x+7. \end{align*} We can solve through substitution, as the second equation can be written as $x=3y-7$ , and plugging this into the first equation gives $5y=6y-7\implies y=7$ , which means $x=3(7)-7=14$ . Finally, we want the total number of games team $A$ has played, which is $A_g=3(14)=\boxed{42}$

Solution: Using the information from the problem, we can note that team $A$ has lost $\frac{1}{3}$ of their matches. Using the answer choices, we can construct the following list of possible win-lose scenarios for $A,$ represented in the form $(w, l)$ for convenience: \begin{align*} \textbf{(A)} &\implies (14, 7) \\ \textbf{(B)} &\implies (18, 9) \\ \textbf{(C)} &\implies (28, 14) \\ \textbf{(D)} &\implies (32, 16) \\ \textbf{(E)} &\implies (42, 21) \end{align*} Thus, we have $5$ matching $B$ scenarios, simply adding $7$ to $w$ and $l.$ We can then test each of the five $B$ scenarios for $\frac{w}{w+l} = \frac{5}{8}$ and find that $(35, 21)$ fits this description. Then working backwards and subtracting $7$ from $w$ and $l$ gives us the point $(28, 14),$ making the answer $\boxed{42}.$

Solution: Let's say that team $A$ plays $n$ games in total. Therefore, team $B$ must play $n + 14$ games in total (7 wins, 7 losses) Since the ratio of $A$ is \[\frac{2}{3} \implies n \equiv 0 \pmod{3}\] Similarly, since the ratio of $B$ is \[\frac{5}{8} \implies n + 14 \equiv 0 \pmod{8}\] Now, we can go through the answer choices and see which ones work: \begin{align*} \textbf{(A) } 21 &\implies 21 + 14 = 35 \not \equiv 0\pmod{8} \\ \textbf{(B) } 27 &\implies 27 + 14 = 41 \not \equiv 0\pmod{8} \\ \textbf{(C) } 42 &\implies 42 + 14 = 56  \equiv 0\pmod{8} \\ \textbf{(D) } 48 &\implies 48 + 14 = 62 \not \equiv 0\pmod{8} \\ \textbf{(E) } 63 &\implies 63 + 14 = 77 \not \equiv 0\pmod{8} \\ \end{align*} So we can see $\boxed{42}$ is the only valid answer.


Answer: 7.0

Ted mistakenly wrote $2^m\cdot\sqrt{\frac{1}{4096}}$ as $2\cdot\sqrt[m]{\frac{1}{4096}}.$ What is the sum of all real numbers $m$ for which these two expressions have the same value?
$\textbf{(A) } 5 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 8 \qquad \textbf{(E) } 9$

Solution: We are given that \[2^m\cdot\sqrt{\frac{1}{4096}} = 2\cdot\sqrt[m]{\frac{1}{4096}}.\] Converting everything into powers of $2,$ we have \begin{align*} 2^m\cdot(2^{-12})^{\frac12} &= 2\cdot (2^{-12})^{\frac1m} \\ 2^{m-6} &= 2^{1-\frac{12}{m}} \\ m-6 &= 1-\frac{12}{m}. \end{align*} We multiply both sides by $m$ , then rearrange as \[m^2-7m+12=0.\] By Vieta's Formulas, the sum of such values of $m$ is $\boxed{7}.$

Solution: We can rewrite the equation using fractional exponents and take logarithms of both sides: \[\log_2{(2^{m}\cdot4096^{-1/2}}) = \log_2{(2\cdot4096^{-1/m})}.\] We can then use the additive properties of logarithms to split them up: \[\log_2{(2^{m})} + \log_2{(4096^{-1/2})} = \log_2{2} + \log_2{(4096^{-1/m})}.\] Using the power rule, the fact that $4096 = 2^{12},$ and bringing the exponents down, we get \begin{align*} m - 6 &= 1 - \frac{12}{m} \\ m + \frac{12}{m} &= 7 \\ m^{2} + 12 &= 7m \\ m^{2} - 7m + 12 &= 0 \\ (m-3)(m-4) &= 0, \end{align*} from which $m = 3$ or $m = 4$ . Therefore, the answer is $3+4 = \boxed{7}.$

Solution: Since surd roots are conventionally positive integers, assume $m$ is an integer, so $m$ can only be $1$ $2$ $3$ $4$ $6$ , and $12$ $\sqrt{\frac{1}{4096}}=\frac{1}{64}$ . Testing out $m$ , we see that only $3$ and $4$ work. Hence, $3+4=\boxed{7}$


Answer: 4.0

Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$

Solution:
We use that fact that $d=rt$ . Let d= distance, r= rate or speed, and t=time. In this case, let $x$ represent the time.
On Monday, he was at a rate of $5 \text{ m.p.h}$ . So, $5x = 2 \text{ miles}\implies x = \frac{2}{5} \text { hours}$
For Wednesday, he walked at a rate of $3 \text{ m.p.h}$ . Therefore, $3x = 2 \text{ miles}\implies x = \frac{2}{3}  \text { hours}$
On Friday, he walked at a rate of $4 \text{ m.p.h}$ . So, $4x = 2 \text{ miles}\implies x=\frac{2}{4}=\frac{1}{2}  \text {hours}$
Adding up the hours yields $\frac{2}{5}  \text { hours}$ $\frac{2}{3}  \text { hours}$ $\frac{1}{2}  \text { hours}$ $\frac{47}{30}  \text { hours}$
We now find the amount of time Grandfather would have taken if he walked at $4 \text{ m.p.h}$ per day. Set up the equation, $4x = 2 \text{ miles} \times  3 \text{ days}\implies x = \frac{3}{2}  \text { hours}$
To find the amount of time saved, subtract the two amounts: $\frac{47}{30}  \text { hours}$ $\frac{3}{2}  \text { hours}$ $\frac{1}{15}  \text { hours}$ . To convert this to minutes, we multiply by $60$
Thus, the solution to this problem is $\dfrac{1}{15}\times 60=\boxed{4}$


Answer: 28.0

Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$ , no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$ , where m and n are relatively prime positive integers. Find $m+n.$

Solution:
Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$ . Notice that there are $10!$ possible pairings.
Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad.
Note that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people.
Thus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$ ), check the left shoe it is paired with (say $L_i$ ), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$ , find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$ . We can imagine each right shoe "sending" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times.
Effectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$ ; thus, we need to count pairings where every cycle has length at least $5$ . This is only possible if there is a single cycle of length $10$ or two cycles of length $5$
The first case yields $9!$ working pairings. The second case yields $\frac{{10\choose 5}}{2}\cdot{4!}^2=\frac{10!}{2 \cdot {5!}^2} \cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$ , the probability is $\frac{1}{10}+\frac{1}{50} = \frac{3}{25}$ , for an answer of $\boxed{028}$


Answer: 581.0

Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs.

Solution: We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs.  There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent.  There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \cdot 4$ or $40$ have $4$ adjacent, and $10 \cdot {5\choose 2}$ or $100$ have $3$ adjacent.  With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \cdot {4\choose2}$ or $60$ have $4$ adjacent, $\frac{10 \cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \cdot \left({5\choose2} - 3\right)$ or $70$ have $1$ group of $3$ adjacent chairs.  All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\choose7}$ or $120$ ${10\choose8}$ or $45$ ${10\choose9}$ or $10$ , and ${10\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.$

Solution:
Starting with small cases, we see that four chairs give $4 + 1 = 5$ , five chairs give $5 + 5 + 1 = 11$ , and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$ , as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$
However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$ , and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = \boxed{581}$

Solution:
It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$ , where $T_n$ counts the ways to select a subset of chairs $\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row.
Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$ , and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$ . Now we know the complement is $274 + 81 + 88 = 443$ , and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \boxed{581}$


Answer: 480.0

Ten chairs are evenly spaced around a round table  and numbered clockwise from $1$ through $10$ . Five married couples are to sit in the chairs with men and women alternating, and no one is to sit either next to or across from his/her spouse. How many seating arrangements are possible?
$\mathrm{(A)}\ 240\qquad\mathrm{(B)}\ 360\qquad\mathrm{(C)}\ 480\qquad\mathrm{(D)}\ 540\qquad\mathrm{(E)}\ 720$

Solution: For the first man, there are $10$ possible seats. For each subsequent man, there are $4$ $3$ $2$ , or $1$ possible seats. After the men are seated, there are only two possible arrangements for the five women. The answer is $10\cdot 4\cdot 3\cdot 2\cdot 1\cdot 2 = \boxed{480}$

Solution: Label the seats ABCDEFGHIJ, where A is the top seat. The first man has $10$ possible seats. WLOG, assume he is in seat A in the diagram. Then, his wife can be in one of two seats, namely D or H. WLOG, assume she is in seat D. Now, in each structurally distinct solution we find, we know that there are $4! = 24$ ways to arrange the 4 other couples. Let there be x structurally distinct solutions under these conditions. We know the answer must be $10\cdot 2\cdot 24\cdot x = 480x$ possible seating arrangements, and x is a nonnegative integer. There is only one answer that is a multiple of $480$ . So, our answer is $\boxed{480}$


Answer: 190.0

Ten identical crates each of dimensions $3\mathrm{ft}\times 4\mathrm{ft}\times 6\mathrm{ft}$ .  The first crate is placed flat on the floor.  Each of the remaining nine crates is placed, in turn, flat on top of the previous crate, and the orientation of each crate is chosen at random.  Let $\frac {m}{n}$ be the probability that the stack of crates is exactly $41\mathrm{ft}$ tall, where $m$ and $n$ are relatively prime positive integers.  Find $m$

Solution:
Only the heights matter, and each crate is either 3, 4, or 6 feet tall with equal probability. We have the following:
\begin{align*}3a + 4b + 6c &= 41\\ a + b + c &= 10\end{align*}
Subtracting 3 times the second from the first gives $b + 3c = 11$ , or $(b,c) = (2,3),(5,2),(8,1),(11,0)$ . The last doesn't work, obviously. This gives the three solutions $(a,b,c) = (5,2,3),(3,5,2),(1,8,1)$ . In terms of choosing which goes where, the first two solutions are analogous.
For $(5,2,3),(3,5,2)$ , we see that there are $2\cdot\dfrac{10!}{5!2!3!} = 10\cdot9\cdot8\cdot7$ ways to stack the crates. For $(1,8,1)$ , there are $\dfrac{10!}{8!1!1!} = 90$ . Also, there are $3^{10}$ total ways to stack the crates to any height.
Thus, our probability is $\dfrac{10\cdot9\cdot8\cdot7 + 90}{3^{10}} = \dfrac{10\cdot8\cdot7 + 10}{3^{8}} = \dfrac{570}{3^8} = \dfrac{190}{3^{7}}$ . Our answer is the numerator, $\boxed{190}$

Solution: It would be helpful for this solution to be reformatted. To start with, let us observe the three numbers. Note that $3$ and $6$ are both divisible by $3$ , so the number of $4$ -crates must be congruent to $41\bmod{3}$ , which is also congruent to $2\bmod{3}$ . Our solutions for the number of $4$ -crates will repeat mod $3$ , so if $x$ is a solution, so is $x+3$ . By inspection, we have that $2$ is solution, and so are $5$ and $8$ . Each solution splits into its own case.We must solve the equation $41-4z=6x+3y$ , simultaneously with $x+y=10-z$ . Note that we already know the possible values of $z$ . Solving these (it's AIME $9$ , you should be able to do this and if anyone feels like they want to write a rundown of this please go ahead), we get the solution  sets $\{8,1,1\},\{5,2,3\},$ and $\{2,3,5\}$ . We can count the number of possible arrangements for each solution by taking $\dbinom{10}{z}$ and then multiplying by $\dbinom{10-z}{x}$ (the solution sets, for the sake of consistency, are in the form $z,x,y$ ). Summing the results for all the solutions gives us $5130$ . Finally, to calculate the probability we must determine our denominator. Since we have $3$ ways to arrange each block, our denominator is $3^{10}$ $\frac{5130}{3^{10}}=\frac{190}{3^7}$ . The answer is $m=\boxed{190}$

Solution:
Let's make two observations. We are trying to find the number of ways we can add $3\text{s}, 4\text{s}$ , and $6\text{s}$ to get $41$ , and the total number of (non-distinct) sums possible is $3^{10}$ .
Then we just use casework to easily and directly solve for the number of ways to get $41$ . To begin, the minimum sum is produced with $10$ threes, so WLOG we can solve for the number of ways to get $11$ with $0\text{s}, 1\text{s}$ , and $3\text{s}$
Case I: $0$ zeroes, $0$ threes, $11$ ones
Impossible, because there are only ten available spots.
Case II: $1$ zero, $1$ three, $8$ ones
This is just $\frac{10!}{8!}$ , so there are $90$ possibilities.
Case III: $3$ zeroes, $2$ threes, $5$ ones
This is just $\frac{10!}{3!2!5!}$ . This gives $2520$ possibilities.
Case IV: 5 zeroes, 3 threes, and 2 ones.
This is the same as case $3$ , so also $2520$ possibilities.
$90+2520+2520=5130$
$5130$ has three powers of $3$ , so $5130$ divided by $27$ is $\boxed{190}$

Solution:
Note we are placing 10 crates where each "height" is 3, 4, 6 and we want all the heights to sum to 41.
We can model this as the generating function \[\left(x^3+x^4+x^6\right)^{10}\] where we want the coefficient of $x^{41}.$ First off, factor this to get \[{x^3}^{10}\left(1+x+x^3\right)^{10}\] and then see that we want the coefficient of $x^{11}$ in $\left(1+x+x^3\right)^{10}.$ From multinomial theorem, this expansion is \[\sum_{a+b+c=10}\binom{10}{a,b,c}1^ax^bx^{3c}\] If we want the coefficient of $x^{11}$ then we need $b + 3c = 11.$ with $b + c \le 10$ (from the multinomial expansion).
This has the solutions $(b, c) = \{8, 1\}, \{5, 2\}, \{2, 3\}.$ Note that the denominator of the answer is just $3^{10}$ since there are 3 ways to orientate every crate and there are 10 creates. Thus, our answer is \[\frac{\binom{10}{8,1,1} + \binom{10}{5,2,3} + \binom{10}{2,3,5}}{3^{10}} = \frac{190}{3^7} \rightarrow \boxed{190}\]


Answer: 1.0

Ten people form a circle.  Each picks a number and tells it to the two neighbors adjacent to them in the circle.  Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person ( not the original number the person picked.) [asy] unitsize(2 cm);  for(int i = 1; i <= 10; ++i) {   label("``" + (string) i + "''", dir(90 - 360/10*(i - 1))); } [/asy] The number picked by the person who announced the average $6$ was
$\textbf{(A) } 1 \qquad  \textbf{(B) } 5 \qquad  \textbf{(C) } 6 \qquad  \textbf{(D) } 10 \qquad \textbf{(E) }\text{not uniquely determined from the given information}$

Solution:
For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$
Taking the indices modulo $10,$ we are given that $\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$
We have ten equations: five with odd-numbered indices and five with even-numbered indices. Note that these two sets of equations are independent. The set that involves $a_6$ is \begin{align*} a_2 + a_4 & = 6, &&(1) \\ a_4 + a_6 & = 10, &&(2) \\ a_6 + a_8 & = 14, &&(3) \\ a_8 + a_{10} & = 18, &&(4) \\ a_{10} + a_2 & = 2. &&(5) \end{align*} Summing these five equations, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50,$ from which \[a_2 + a_4 + a_6 + a_8 + a_{10} = 25. \hspace{15mm} (\bigstar)\] Subtracting $(1)+(4)$ from $(\bigstar),$ we obtain $a_6=\boxed{1}.$

Solution:
For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ announces the number $i.$
Let $x$ be the number picked by Person $6.$ We construct the following table: \[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \textbf{People} & \textbf{Average of \#s Picked} & \textbf{Sum of \#s Picked} & \multicolumn{1}{c}{\textbf{Conclusion}} \\ [0.5ex] \hline & & & \\ [-2ex] 6\text{ and }8 & 7 & 14 & \text{Person 8 picks } 14-x. \\ 8\text{ and }10 & 9 & 18 & \text{Person 10 picks } 4+x. \\ 10\text{ and }2 & 1 & 2 & \text{Person 2 picks } -2-x \\ 2\text{ and }4 & 3 & 6 & \text{Person 4 picks } 8+x \\ 4\text{ and }6 & 5 & 10 & \text{Person 6 picks } 2-x \\ \end{array}\] We have $x=2-x,$ from which $x=\boxed{1}.$


Answer: 968.0

Ten points are marked on a circle . How many distinct convex polygons of three or more sides can be drawn using some (or all) of the ten points as vertices

Solution: Any subset of the ten points with three or more members can be made into exactly one such polygon.  Thus, we need to count the number of such subsets.  There are $2^{10} = 1024$ total subsets of a ten-member set , but of these ${10 \choose 0} = 1$ have 0 members, ${10 \choose 1} = 10$ have 1 member and ${10 \choose 2} = 45$ have 2 members.  Thus the answer is $1024 - 1 - 10 - 45 = \boxed{968}$


Answer: 489.0

Ten points in the plane are given, with no three collinear .  Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely.  The probability that some three of the segments form a triangle whose vertices are among the ten given points is $m/n,$ where $m_{}$ and $n_{}$ are relatively prime positive integers .  Find $m+n.$

Solution:
Note that 4 points can NEVER form 2 triangles. Therefore, we just need to multiply the probability that the first three segments picked form a triangle by 4. We can pick any segment for the first choice, then only segments that share an endpoint with the first one, then the one segment that completes the triangle. Note that the fourth segment doesn't matter in this case.
Note that there are $(9 - 1) \times 2 = 16$ segments that share an endpoint with the first segment. The answer is then $4 \times \frac{16}{44} \times \frac{1}{43} = \frac{16}{11} \times \frac{1}{43} = \frac{16}{473} \implies m + n = \boxed{489}$ -whatRthose

Solution:
Instead of working with the four segments, let's focus on their endpoints. When we select these segments, we are working with $4, 5, 6, 7,$ or $8$ endpoints in total.
If we have $6, 7,$ or $8$ endpoints, it is easy to see that we cannot form a triangle by drawing four segments between them, because at least one point will be "left out".
However, if we have $5$ endpoints, we can use three segments form a triangle with three of the points and connect the remaining two points with the last segment. There are ${10\choose5}$ ways to select these $5$ points from the original $10,$ and ${5\choose3}$ ways to decide which three points are in the triangle.
Finally, if we have $4$ endpoints, we can also form a triangle with three of the points, then use the remaining segment to connect the last point to either of the previous three. We have ${10\choose4}$ ways to select the $4$ points and ${4\choose3}$ ways to choose three points for the triangle. Finally, we must connect the last point to one vertex of the triangle; we can do this in $3$ ways.
As in Solution 1, there are ${45\choose4}$ total ways to select four segments. So, our desired probability is
\[\dfrac{{10\choose5}{5\choose3}+{10\choose4}{4\choose3}\cdot 3}{{45\choose4}}=\dfrac{16}{473} \implies m + n = \boxed{489}.\]


Answer: 89.0

Ten women sit in $10$ seats in a line. All of the $10$ get up and then reseat themselves using all $10$ seats, each sitting in the seat she was in before or a seat next to the one she occupied before. In how many ways can the women be reseated?
$\textbf{(A)}\ 89\qquad \textbf{(B)}\ 90\qquad \textbf{(C)}\ 120\qquad \textbf{(D)}\ 2^{10}\qquad \textbf{(E)}\ 2^2 3^8$

Solution: Notice that either a woman stays in her own seat after the rearrangement, or two adjacent women swap places. Thus, our answer is counting the number of ways to arrange 1x1 and 2x1 blocks to form a 1x10 rectangle. This can be done via casework depending on the number of 2x1 blocks. The cases of 0, 1, 2, 3, 4, 5 2x1 blocks correspond to 10, 8, 6, 4, 2, 0 1x1 blocks, and so the sum of the cases is \[\binom{10}{0} + \binom{9}{1} + \binom{8}{2} + \binom{7}{3} + \binom{6}{4} + \binom{5}{5} = 1 + 9 + 28 + 35 + 15 + 1 = \boxed{89}.\]

Solution: Recall that the number of ways to arrange 1x1 and 2x1 blocks to form a 1xn rectangle results in Fibonacci numbers . Clearly, $\boxed{89}$ is the only fibonacci number, so no calculation is needed for this problem.

Solution:
Let $S_n$ be the number of possible seating arrangements with $n$ women. Consider $n \ge 3,$ and focus on the rightmost woman. If she returns back to her seat, then there are $S_{n-1}$ ways to seat the remaining $n-1$ women. If she sits in the second to last seat, then the woman who previously sat there must now sit at the rightmost seat. This gives us $S_{n-2}$ ways to seat the other $n-2$ women, so we obtain the recursion \[S_n = S_{n-1}+S_{n-2}.\]
Starting with $S_1=1$ and $S_2=2,$ we can calculate $S_{10}=\boxed{89}.$


Answer: 27.0

Terri produces a sequence of positive integers by following three rules.  She starts with a positive integer, then applies the appropriate rule to the result, and continues in this fashion.
Rule 1: If the integer is less than 10, multiply it by 9.
Rule 2: If the integer is even and greater than 9, divide it by 2.
Rule 3: If the integer is odd and greater than 9, subtract 5 from it.
A sample sequence: $23, 18, 9, 81, 76, \ldots .$
Find the $98^\text{th}$ term of the sequence that begins $98, 49, \ldots .$
$\text{(A)}\ 6 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 22 \qquad \text{(D)}\ 27 \qquad \text{(E)}\ 54$

Solution:
We could start by looking for a pattern.
$98, 49, 44, 22, 11, 6, 54, 27, 22, 11, 6, \ldots .$
From here, we see that we have a pattern of $22, 11, 6, 54, 27, \ldots .$ after $98, 49, 44$
Our problem is now really
Find the $95^\text{th}$ term of the sequence that goes $22, 11, 6, 54, 27, 22, 11, 6, 54, 27, 22, \ldots .$
There are 5 terms in each repetition of the pattern, and $95\equiv0\pmod{5}$ , so the answer is $\boxed{27}$


Answer: 682.0

Tetrahedron $ABCD$ has $AD=BC=28$ $AC=BD=44$ , and $AB=CD=52$ . For any point $X$ in space, suppose $f(X)=AX+BX+CX+DX$ . The least possible value of $f(X)$ can be expressed as $m\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution:
Let $M$ and $N$ be midpoints of $\overline{AB}$ and $\overline{CD}$ . The given conditions imply that $\triangle ABD\cong\triangle BAC$ and $\triangle CDA\cong\triangle DCB$ , and therefore $MC=MD$ and $NA=NB$ . It follows that $M$ and $N$ both lie on the common perpendicular bisector of $\overline{AB}$ and $\overline{CD}$ , and thus line $MN$ is that common perpendicular bisector. Points $B$ and $C$ are symmetric to $A$ and $D$ with respect to line $MN$ . If $X$ is a point in space and $X'$ is the point symmetric to $X$ with respect to line $MN$ , then $BX=AX'$ and $CX=DX'$ , so $f(X) = AX+AX'+DX+DX'$
Let $Q$ be the intersection of $\overline{XX'}$ and $\overline{MN}$ . Then $AX+AX'\geq 2AQ$ , from which it follows that $f(X) \geq 2(AQ+DQ) = f(Q)$ . It remains to minimize $f(Q)$ as $Q$ moves along $\overline{MN}$
Allow $D$ to rotate about $\overline{MN}$ to point $D'$ in the plane $AMN$ on the side of $\overline{MN}$ opposite $A$ . Because $\angle DNM$ is a right angle, $D'N=DN$ . It then follows that $f(Q) = 2(AQ+D'Q)\geq 2AD'$ , and equality occurs when $Q$ is the intersection of $\overline{AD'}$ and $\overline{MN}$ . Thus $\min f(Q) = 2AD'$ . Because $\overline{MD}$ is the median of $\triangle ADB$ , the Length of Median Formula shows that $4MD^2 = 2AD^2 + 2BD^2 - AB^2 = 2\cdot 28^2 + 2 \cdot 44^2 - 52^2$ and $MD^2 = 684$ . By the Pythagorean Theorem $MN^2 = MD^2 - ND^2 = 8$
Because $\angle AMN$ and $\angle D'NM$ are right angles, \[(AD')^2 = (AM+D'N)^2 + MN^2 = (2AM)^2 + MN^2 = 52^2 + 8 = 4\cdot 678.\] It follows that $\min f(Q) = 2AD' = 4\sqrt{678}$ . The requested sum is $4+678=\boxed{682}$


Answer: 88.0

The "Middle School Eight" basketball conference has $8$ teams. Every season, each team plays every other conference team twice (home and away), and each team also plays $4$ games against non-conference opponents. What is the total number of games in a season involving the "Middle School Eight" teams?
$\textbf{(A) }60\qquad\textbf{(B) }88\qquad\textbf{(C) }96\qquad\textbf{(D) }144\qquad \textbf{(E) }160$

Solution:
Within the conference, there are 8 teams, so there are $\dbinom{8}{2}=28$ pairings of teams, and each pair must play two games, for a total of $28\cdot 2=56$ games within the conference.
Each team also plays 4 games outside the conference, and there are 8 teams, so there are a total of $4\cdot 8 =32$ games outside the conference.
Therefore, the total number of games is $56+32 = \boxed{88}$ is our answer.


Answer: 9.0

The $16$ squares on a piece of paper are numbered as shown in the diagram.  While lying on a table, the paper is folded in half four times in the following sequence:
(1) fold the top half over the bottom half
(2) fold the bottom half over the top half
(3) fold the right half over the left half
(4) fold the left half over the right half.
Which numbered square is on top after step $4$
[asy] unitsize(18); for(int a=0; a<5; ++a)  {   draw((a,0)--(a,4));  } for(int b=0; b<5; ++b)  {   draw((0,b)--(4,b));  } label("$1$",(0.5,3.1),N); label("$2$",(1.5,3.1),N); label("$3$",(2.5,3.1),N); label("$4$",(3.5,3.1),N); label("$5$",(0.5,2.1),N); label("$6$",(1.5,2.1),N); label("$7$",(2.5,2.1),N); label("$8$",(3.5,2.1),N); label("$9$",(0.5,1.1),N); label("$10$",(1.5,1.1),N); label("$11$",(2.5,1.1),N); label("$12$",(3.5,1.1),N); label("$13$",(0.5,0.1),N); label("$14$",(1.5,0.1),N); label("$15$",(2.5,0.1),N); label("$16$",(3.5,0.1),N); [/asy]
$\text{(A)}\ 1 \qquad \text{(B)}\ 9 \qquad \text{(C)}\ 10 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution:
Suppose we undo each of the four folds, considering just the top square until we completely unfold the paper. $x$ will be marked in the square if the face that shows after all the folds is face up, $y$ if that face is facing down.
Step 0:  Step 1:  Step 2:  Step 3:  Step 4:  The marked square is in the same spot as the number $9\rightarrow \boxed{9}$


Answer: 16.0

The $2007 \text{ AMC }10$ will be scored by awarding $6$ points for each correct response, $0$ points for each incorrect response, and $1.5$ points for each problem left unanswered. After looking over the $25$ problems, Sarah has decided to attempt the first $22$ and leave only the last $3$ unanswered. How many of the first $22$ problems must she solve correctly in order to score at least $100$ points?
$\textbf{(A) } 13 \qquad\textbf{(B) } 14 \qquad\textbf{(C) } 15 \qquad\textbf{(D) } 16 \qquad\textbf{(E) } 17$

Solution: Sarah is leaving $3$ questions unanswered, guaranteeing her $3 \times 1.5 = 4.5$ points. She will either get $6$ points or $0$ points for the rest of the questions. Let $x$ be the number of questions Sarah answers correctly. \begin{align*} 6x+4.5 &\ge 100\\ 6x &\ge 95.5\\ x &\ge 15.92 \end{align*} The number of questions she answers correctly has to be a whole number, so round up to get $\boxed{16}$


Answer: 10.0

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Solution: Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\frac{50}{5}=\boxed{10}$

Solution:
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{10}$ as our answer.
~Baolan

Solution:
Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$ , or $\boxed{10}$ .
~Arctic_Bunny, edited by KINGLOGIC

Solution:
Let us consider the horizontal rows.  Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ .  Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ .
~cw357

Solution:
The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square  \square  2  \square  \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$


Answer: 10.0

Solution:
If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551


Answer: 10.0

Solution: Without loss of generality, consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the total value per row. The sum of the $25$ integers is $-10+-9+...+14=11+12+13+14=50$ , and the common sum is $\frac{50}{5}=\boxed{10}$

Solution:
Take the sum of the middle 5 values of the set (they will turn out to be the mean of each row). We get $0 + 1 + 2 + 3 + 4 = \boxed{10}$ as our answer.
~Baolan

Solution:
Taking the average of the first and last terms, $-10$ and $14$ , we have that the mean of the set is $2$ . There are 5 values in each row, column or diagonal, so the value of the common sum is $5\cdot2$ , or $\boxed{10}$ .
~Arctic_Bunny, edited by KINGLOGIC

Solution:
Let us consider the horizontal rows.  Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ .  Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ .
~cw357

Solution:
The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square  \square  2  \square  \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$


Answer: 10.0

Solution:
If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551


Answer: 3.0

The $5$ -digit number $\underline{2}$ $\underline{0}$ $\underline{1}$ $\underline{8}$ $\underline{U}$ is divisible by $9$ . What is the remainder when this number is divided by $8$
$\textbf{(A) }1\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution: We use the property that the digits of a number must sum to a multiple of $9$ if it are divisible by $9$ . This means $2+0+1+8+U$ must be divisible by $9$ . The only possible value for $U$ then must be $7$ . Since we are looking for the remainder when divided by $8$ , we can ignore the thousands. The remainder when $187$ is divided by $8$ is $\boxed{3}$


Answer: 263.0

The $52$ cards in a deck are numbered $1, 2, \cdots, 52$ . Alex, Blair, Corey, and Dylan each picks a card from the deck without replacement and with each card being equally likely to be picked, The two persons with lower numbered cards form a team, and the two persons with higher numbered cards form another team. Let $p(a)$ be the probability that Alex and Dylan are on the same team, given that Alex picks one of the cards $a$ and $a+9$ , and Dylan picks the other of these two cards. The minimum value of $p(a)$ for which $p(a)\ge\frac{1}{2}$ can be written as $\frac{m}{n}$ . where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Once the two cards are drawn, there are $\dbinom{50}{2} = 1225$ ways for the other two people to draw. Alex and Dylan are the team with higher numbers if Blair and Corey both draw below $a$ , which occurs in $\dbinom{a-1}{2}$ ways. Alex and Dylan are the team with lower numbers if Blair and Corey both draw above $a+9$ , which occurs in $\dbinom{43-a}{2}$ ways. Thus, \[p(a)=\frac{\dbinom{43-a}{2}+\dbinom{a-1}{2}}{1225}.\] Simplifying, we get $p(a)=\frac{(43-a)(42-a)+(a-1)(a-2)}{2\cdot1225}$ , so we need $(43-a)(42-a)+(a-1)(a-2)\ge (1225)$ . If $a=22+b$ , then \begin{align*}(43-a)(42-a)+(a-1)(a-2)&=(21-b)(20-b)+(21+b)(20+b)=2b^2+2(21)(20)\ge (1225) \\ b^2\ge \frac{385}{2} &= 192.5 >13^2 \end{align*} So $b> 13$ or $b< -13$ , and $a=22+b<9$ or $a>35$ , so $a=8$ or $a=36$ . Thus, $p(8) = \frac{616}{1225} = \frac{88}{175}$ , and the answer is $88+175 = \boxed{263}$


Answer: 19.0

The $5\times 5$ grid shown contains a collection of squares with sizes from $1\times 1$ to $5\times 5$ . How many of these squares contain the black center square?
2004 AMC 10A problem 16.png
$\mathrm{(A) \ } 12 \qquad \mathrm{(B) \ } 15 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ }  19\qquad \mathrm{(E) \ } 20$

Solution:
Since there are five types of squares: $1 \times 1, 2 \times 2, 3 \times 3, 4 \times 4,$ and $5 \times 5.$ We must find how many of each square contain the black shaded square in the center.
If we list them, we get that
Thus, the answer is $1+4+9+4+1=19\Rightarrow\boxed{19}$

Solution: We use complementary counting. There are only $2\times2$ and $1\times1$ squares that do not contain the black square. Counting, there are $12$ $2\times2$ squares, and $25-1 = 24$ $1\times1$ squares that do not contain the black square. That gives $12+24=36$ squares that don't contain it. There are a total of $25+16+9+4+1 = 55$ squares possible $(25$ $1\times1$ squares $16$ $2\times2$ squares $9$ $3\times3$ squares $4$ $4\times4$ squares and $1$ $5\times5$ square), therefore there are $55-36 = 19$ squares that contain the black square, which is $\boxed{19}$


Answer: 1.0

The $7$ -digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$ . Which of the following could be the value of $C$
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$

Solution: The number $\mod{3}$ is congruent to sum of a number's digits $\mod{3}$ is congruent to the number $\mod{3}$ $74A52B1 \pmod{3}$ must be congruent to 0, since it is divisible by 3. Therefore, $7+4+A+5+2+B+1 \pmod{3}$ is also congruent to 0. $7+4+5+2+1 \equiv 1 \pmod{3}$ , so $A+B\equiv 2 \pmod{3}$ . As we know, $326AB4C\equiv 0 \pmod{3}$ , so $3+2+6+A+B+4+C =15+A+B+C\equiv 0 \pmod{3}$ , and therefore $A+B+C\equiv 0 \pmod{3}$ . We can substitute 2 for $A+B$ , so $2+C\equiv 0 \pmod{3}$ , and therefore $C\equiv 1\pmod{3}$ . This means that C can be 1, 4, or 7, but the only one of those that is an answer choice is $\boxed{1}$

Solution: Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7 + 4 + 5 + 2 + 1 = 19$ . To be a multiple of $3$ $A + B$ has to be either $2$ or $5$ or $8$ ... and so on. We add up the numerical digits in the second number; $3 + 2 + 6 + 4 = 15$ . We then add two of the selected values, $5$ to $15$ , to get $20$ . We then see that C = $1, 4$ or $7, 10$ ... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$ , to get $23$ , which shows us that C = $1$ or $4$ or $7$ ... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1, 4,$ and $7$ . However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$ , but there is a $1$ , so $\boxed{1}$ is our answer.


Answer: 6.0

The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square.  What is $y$
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$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$

Solution:
Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.

As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{6}$

Solution:
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.

As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$

From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{6}$

Solution: Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$


Answer: 6.0

Solution:
Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$ . This means the square will have four sides of length 12. The only way to do this is shown below.

As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{6}$

Solution:
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.

As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$

From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{6}$

Solution: Because the two hexagons are congruent, we know that the perpendicular line to $A$ is half of $BC$ , or $4$ . Next, we plug the answer choices in to see which one works. Trying $A$ , we get the area of one hexagon is $72$ , as desired, so the answer is $\boxed{6}$


Answer: 841.0

The $\textit{taxicab distance}$ between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\] For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921  \qquad\textbf{(E)} \, 924$

Solution: Let us consider the number of points for a certain $x$ -coordinate. For any $x$ , the viable points are in the range $[-20 + |x|, 20 - |x|]$ . This means that our total sum is equal to \begin{align*} 1 + 3 + 5 + \cdots + 41 + 39 + 37 + \cdots + 1 &= (1 + 3 + 5 + \cdots + 39) + (1 + 3 + 5 + \cdots + 41) \\ & = 20^2 + 21^2 \\ & = 29^2 \\ &= \boxed{841} ~mathboy100

Solution:
Since the second point is the origin, this is equivalent to finding all points $(x, y)$ such that $|x| + |y| \leq 20$ . Due to the absolute values, the set of all such points will be symmetric about the origin meaning we can focus on the first quadrant and multiply by $4$
To avoid overcounts, ignore points on the axes. This means $x, y > 0$ . If $x = 1$ , there are $19$ solutions for $y$ $y = 1, 2, 3, \ldots, 19$ ). If $x = 2$ , there are $18$ solutions. This pattern repeats until $x = 19$ , at which point there is $1$ solution for $y$
So we get $19 + 18 + 17 + \cdots + 1 = \frac{19(20)}{2} = 190$ points in the first quadrant. Multiplying by $4$ gives $760$ . Now, the $x$ axis has $y = 0$ which gives $|x| \leq 20$ , meaning there are $41$ solutions. This is the same with the $y$ axis, but we overcounted the origin by $1$
Our final answer is $760 + 41 + 41 - 1 = \boxed{841}$

Solution: This solution refers to the Diagram section.  The problem can be visualized as depicted on the right split equally into four "triangular" parts excluding the origin. The "triangular" parts are identical the ones that would be used in a visual proof of the formula for triangular numbers. Becuase of this the number of points in each part is equal to $\frac{n(n+1)}{2}$ where $n$ is the length of a "leg" of the "triangle" which is $20$ for this problem. Substituting and computing, we get $210.$ Multiplying by $4$ and adding $1$ to account for all parts and the origin, we get $210\cdot4 + 1 = \boxed{841}.$

Solution:
This solution refers to the Diagram section.
As shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd.  Note that the red array consists of $21^2=441$ points, and the blue array consists of $20^2=400$ points.
Together, the answer is $441+400=\boxed{841}.$

Solution:
Let $P = (x, y)$ . Since the problem asks for taxicab distances from the origin, we want $|x| + |y| \le 20$ . The graph of all solutions to this equation on the $xy$ -plane is a square with vertices at $(0, \pm 20)$ and $(\pm 20, 0)$ (In order to prove this, one can divide the sections of this graph into casework on the four quadrants, and tie together the resulting branches.) We want the number of lattice points on the border of the square and inside the square.
Each side of the square goes through an equal number of lattice points, so if we focus on one side going from $(0,20)$ to $(20, 0)$ , we can see that it goes through $21$ points in total. In addition, each of the vertices gets counted twice, so the total number of border points is $21\cdot4 - 4 = 80$ . Also, the area of the square is $800$ , so when we plug this information inside Pick's theorem, we get $800 = i + \frac{80}{2} - 1 \implies i = 761$ . Then our answer is $761+80 = \boxed{841}.$

Solution:
Instead of considering all points with integer coordinates, first consider points with nonnegative coordinates only. Then, we want $x + y \le 20$ where $x$ and $y$ are nonnegative integers. We can introduce a third variable, say $z$ , such that $z = 20 - (x + y)$ . Note that counting the number of ways to have $x + y + z = 20$ is the same as counting the number of ways to have $x + y \le 20$ . Therefore, by stars and bars, there are $\dbinom{20 + 3 - 1}{3 - 1} = 231$ solutions with nonnegative integer coordinates.
Then, we can copy our solutions over to the other four quadrants. First, so as not to overcount, we remove all points on the axes. There are $20 + 20 + 1 = 41$ such points with nonnegative integer coordinates. We multiply the $190$ remaining points by $4$ to get $760$ points that are not on the axes. Then, we can add back the $41$ nonnegative points on the axes, as well as the $40$ other points on the negative axes to get $760 + 41 + 40 = \boxed{841}.$


Answer: 60.0

The $y$ -intercepts, $P$ and $Q$ , of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$

Solution: Note that if the $y$ -intercepts have a sum of $0$ , the distance from the origin to each of the intercepts must be the same. Call this distance $a$ . Since the $\angle PAQ = 90^\circ$ , the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\sqrt{6^2+8^2} = 10$ , this means $a=10$ , and the length of the hypotenuse is $2a = 20$ . Since the $x$ -coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{60}$

Solution:
We can let the two lines be \[y=mx+b\] \[y=-\frac{1}{m}x-b\] This is because the lines are perpendicular, hence the $m$ and $-\frac{1}{m}$ , and the sum of the y-intercepts is equal to 0, hence the $b, -b$
Since both lines contain the point $(6,8)$ , we can plug this into the two equations to obtain \[8=6m+b\] and \[8=-6\frac{1}{m}-b\]
Adding the two equations gives \[16=6m+\frac{-6}{m}\] Multiplying by $m$ gives \[16m=6m^2-6\] \[\implies 6m^2-16m-6=0\] \[\implies 3m^2-8m-3=0\] Factoring gives \[(3m+1)(m-3)=0\]
Plugging $m=3$ into one of our original equations, we obtain \[8=6(3)+b\] \[\implies b=8-6(3)=-10\]
Since $\bigtriangleup APQ$ has hypotenuse $2|b|=20$ and the altitude to the hypotenuse is equal to the the x-coordinate of point $A$ , or 6, the area of $\bigtriangleup APQ$ is equal to \[\frac{20\cdot6}{2}=\boxed{60}\]

Solution:
Since their sum is $0$ , let the y intercepts be P $(0,a)$ and Q $(0,-a)$ . The slope of $AP$ is $\frac{8-a}{6}$ . The slope of AQ is $\frac{8+a}{6}$ . Since multiplying the slopes of perpendicular lines yields a product of $-1$ , we have $\frac{64-a^2}{36}=-1$ , which results in $a^2=100$ . We can use either the positive or negative solution because if we choose $10$ , then the other y-intercept is $-10$ ; but if we choose $-10$ , then the other y-intercept is $10$ . For simplicity, we choose that $a=10$ in this solution.
Now we have a triangle APQ with points A $(6,8)$ , P $(0,10)$ , and Q $(0,-10)$ . By the Pythagorean theorem, we have that $AP=\sqrt{6^2+2^2}=2\sqrt{10}$ , and that $AQ=\sqrt{6^2+18^2}=6\sqrt{10}$ $PQ$ is obviously $10-(-10)=20$ since they have the same $x$ coordinate. Now using Heron's formula, we have $\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(4\sqrt{10}+10)(4\sqrt{10}-10)(10+2\sqrt{10})(10-2\sqrt{10})}=\sqrt{60^2}=60 \implies \boxed{60}$

Solution: Using point-slope form, the first line has the equation \[y-8=m\left(x-6\right) \longrightarrow y=mx-6m+8\] The second line has the equation \[y-8=-\frac{1}{m}\left(x-6\right) \longrightarrow y=-\frac{x}{m}+\frac{6}{m}+8\] At the y-intercept, the value of the x-coordinate is $0$ , hence: the first equation is $y=-6m+8$ and the second is $y=\frac{6}{m}+8$ . Since the y-intercepts sum to $0$ , they are opposites, so: \[-6m+8=-\left(\frac{6}{m}+8\right)=-\frac{6}{m}-8\] \[6m-\frac{6}{m}=16\] Multiply both sides by m: $6m^{2}-6=16m \longrightarrow 3m^{2}-8m-3=0$ . The solution to this quadratic, using the quadratic formula, is: $\frac{8\pm\sqrt{64-4\left(3\right)\left(-3\right)}}{6}=\frac{8\pm\sqrt{100}}{6}=\frac{8\pm10}{6}$ This yields $m=-\frac{1}{3}$ and $m=3$ . Plugging $m=3$ into the second equation, we get $y=\frac{6}{3}+8=10$ . Plugging $m=-\frac{1}{3}$ into the first equation, we get $y=-10$ So the base is $20$ and the height is $6$ , the area is $60 \Longrightarrow \boxed{60}$


Answer: 130.0

The 64 whole numbers from 1 through 64 are written, one per square, on a checkerboard (an 8 by 8 array of 64 squares).  The first 8 numbers are written in order across the first row, the next 8 across the second row, and so on.  After all 64 numbers are written, the sum of the numbers in the four corners will be
$\text{(A)}\ 130 \qquad \text{(B)}\ 131 \qquad \text{(C)}\ 132 \qquad \text{(D)}\ 133 \qquad \text{(E)}\ 134$

Solution: Obviously $1$ is in the top left corner, $8$ is in the top right corner, and $64$ is in the bottom right corner. To find the bottom left corner, subtract $7$ from $64$ which is $57$ . Adding the results gives $1+8+57+64=130$ which is answer $\boxed{130}$


Answer: 790.0

The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
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Solution: The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\dfrac{80}{3}$ mm.  According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}$ mm.  Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side.  Therefore, the minimum of the lace in millimeters is \[6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.\]


Answer: 150.0

The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling?

Solution:
Let $r$ represent the rate Tom swims in miles per minute. Then we have
$\frac{1/2}{r} + \frac{8}{5r} + \frac{30}{10r} = 255$
Solving for $r$ , we find $r = 1/50$ , so the time Tom spends biking is $\frac{30}{(10)(1/50)} = \boxed{150}$ minutes.


Answer: 4.0

The Amaco Middle School bookstore sells pencils costing a whole number of cents. Some seventh graders each bought a pencil, paying a total of $1.43$ dollars. Some of the $30$ sixth graders each bought a pencil, and they paid a total of $1.95$ dollars. How many more sixth graders than seventh graders bought a pencil?
$\textbf{(A)}\  1  \qquad \textbf{(B)}\   2  \qquad \textbf{(C)}\   3  \qquad \textbf{(D)}\   4  \qquad \textbf{(E)}\   5$

Solution: Because the pencil costs a whole number of cents, the cost must be a factor of both $143$ and $195$ . They can be factored into $11\cdot13$ and $3\cdot5\cdot13$ . The common factor cannot be $1$ or there would have to be more than $30$ sixth graders, so the pencil costs $13$ cents. The difference in costs that the sixth and seventh graders paid is $195-143=52$ cents, which is equal to $52/13 = \boxed{4}$ sixth graders.


Answer: 346.0

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings.  At meetings, committee members sit at a round table with chairs numbered from $1$ to $15$ in clockwise order.  Committee rules state that a Martian must occupy chair $1$ and an Earthling must occupy chair $15$ ,  Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling.  The number of possible seating arrangements for the committee is $N \cdot (5!)^3$ .  Find $N$

Solution:
Since the 5 members of each planet committee are distinct we get that the number of arrangement of sittings is in the form $N*(5!)^3$ because for each $M, V, E$ sequence we have $5!$ arrangements within the Ms, Vs, and Es.
Pretend the table only seats $3$ "people", with $1$ "person" from each planet. Counting clockwise, only the arrangement M, V, E satisfies the given constraints. Therefore, in the actual problem, the members must sit in cycles of M, V, E, but not necessarily with one M, one V, and one E in each cycle(for example, MMVVVE, MVVVEEE, MMMVVVEE all count as cycles). These cycles of MVE must start at seat $1$ , since an M is at seat $1$ . We simply count the number of arrangements through casework.
1. The entire arrangement is one cycle- There is only one way to arrange this, MMMMMVVVVVEEEEE
2. Two cycles - There are 3 Ms, Vs and Es left to distribute among the existing MVEMVE. Using stars and bars, we get $\binom{4}{1}=4$ ways for the members of each planet. Therefore, there are $4^3=64$ ways in total.
3. Three cycles - 2 Ms, Vs, Es left, so $\binom{4}{2}=6$ , making there $6^3=216$ ways total.
4. Four cycles - 1 M, V, E left, each M can go to any of the four MVE cycles and likewise for V and E, $4^3=64$ ways total
5. Five cycles - MVEMVEMVEMVEMVE is the only possibility, so there is just $1$ way.
Combining all these cases, we get $1+1+64+64+216= \boxed{346}$

Solution: The arrangements must follow the pattern MVEMVE....... where each MVE consists of some Martians followed by some Venusians followed by some Earthlings, for $1, 2, 3, 4,$ or $5$ MVE's. If there are $k$ MVE's, then by stars and bars, there are ${4 \choose k-1}$ choices for the Martians in each block, and the same goes for the Venusians and the Earthlings. Thus, we have $N = 1^3+4^3+6^3+4^3+1^3 = \boxed{346}$ - aops5234


Answer: 428.0

The Binomial Expansion is valid for exponents that are not integers. That is, for all real numbers $x,y$ and $r$ with $|x|>|y|$
\[(x+y)^r=x^r+rx^{r-1}y+\dfrac{r(r-1)}{2}x^{r-2}y^2+\dfrac{r(r-1)(r-2)}{3!}x^{r-3}y^3 \cdots\]
What are the first three digits to the right of the decimal point in the decimal representation of $(10^{2002}+1)^{\frac{10}{7}}$

Solution:
$1^n$ will always be 1, so we can ignore those terms, and using the definition ( $2002 / 7 = 286$ ):
\[(10^{2002} + 1)^{\frac {10}7} = 10^{2860}+\dfrac{10}{7}10^{858}+\dfrac{15}{49}10^{-1144}+\cdots\]
Since the exponent of the $10$ goes down extremely fast, it suffices to consider the first few terms. Also, the $10^{2860}$ term will not affect the digits after the decimal, so we need to find the first three digits after the decimal in
\[\dfrac{10}{7}10^{858}\]
(The remainder after this term is positive by the Remainder Estimation Theorem ). Since the repeating decimal of $\dfrac{10}{7}$ repeats every 6 digits, we can cut out a lot of 6's from $858$ to reduce the problem to finding the first three digits after the decimal of
$\dfrac{10}{7}$
That is the same as $1+\dfrac{3}{7}$ , and the first three digits after $\dfrac{3}{7}$ are $\boxed{428}$


Answer: 12.0

The Blue Bird High School chess team consists of two boys and three girls. A photographer wants to take a picture of the team to appear in the local newspaper. She decides to have them sit in a row with a boy at each end and the three girls in the middle. How many such arrangements are possible?
$\textbf{(A) }2\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad \textbf{(E) }12$

Solution: There are $2! = 2$ ways to order the boys on the ends, and there are $3!=6$ ways to order the girls in the middle. We get the answer to be $2 \cdot 6 = \boxed{12}$


Answer: 6.0

The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$\text{(A)}\ 2 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 5 \qquad \text{(E)}\ 6$

Solution:
Let $x$ be the number of the children and the mom. The father, who is $48$ , plus the sum of the ages of the kids and mom divided by the number of kids and mom plus $1$ (for the dad) = $20$ . This is because the average age of the entire family is $20.$ This statement, written as an equation, is: \[\frac{48+16x}{x+1}=20\] \[48+16x=20x+20\] \[4x=28\] \[x=7\]
$7$ people - $1$ mom = $6$ children.
Therefore, the answer is $\boxed{6}$

Solution:
Let $m$ be the Mom's age.
Let the number of children be $x$ and their average age be $y$ . Their age totaled up is simply $xy$
We have the following two equations:
$\frac{m+48+xy}{2+x}=20$ , where $m+48+xy$ is the family's total age and $2+x$ is the total number of people in the family.
$\frac{m+48+xy}{2+x}=20$
$m+48+xy=40+20x$
The next equation is $\frac{m+xy}{1+x}=16$ , where $m+xy$ is the total age of the Mom and the children, and $1+x$ is the number of children along with the Mom.
$\frac{m+xy}{1+x}=16$
$m+xy=16+16x$
We know the value for $m+xy$ , so we substitute the value back in the first equation.
$m+48+xy=40+20x$
$(16+16x)+48=40+20x$
$x=6$
Earlier, we set $x$ to be the number of children. Therefore, there are $\boxed{6}$ children.


Answer: 6.0

The Fibonacci sequence $1,1,2,3,5,8,13,21,\ldots$ starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
$\textbf{(A)} \ 0  \qquad \textbf{(B)} \ 4  \qquad \textbf{(C)} \ 6  \qquad \textbf{(D)} \ 7  \qquad \textbf{(E)} \ 9$

Solution:
Note that any digits other than the units digit will not affect the answer.  So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{6}$


Answer: 6.0

Solution:
Note that any digits other than the units digit will not affect the answer.  So to make computation quicker, we can just look at the Fibonacci sequence in $\bmod{10}$
$1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6,....$
The last digit to appear in the units position of a number in the Fibonacci sequence is $6 \Longrightarrow \boxed{6}$