Answer: 216.0

In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$ , for some fixed $b\ge2$ . A Martian student writes down \[3\log(\sqrt{x}\log x)=56\] \[\log_{\log x}(x)=54\] and finds that this system of equations has a single real number solution $x>1$ . Find $b$

Solution:
Using change of base on the second equation to base b, \[\frac{\log x}{\log \log_{b}{x}}=54\] \[\log x = 54 \cdot \log\log_{b}{x}\] \[x = (\log_{b}{x})^{54}\] Note by dolphin7 - you could also just rewrite the second equation in exponent form.
Substituting this into the $\sqrt x$ of the first equation, \[3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56\] \[3\log_{b}{(\log_{b}{x})^{28}} = 56\] \[\log_{b}{(\log_{b}{x})^{84}} = 56\]
We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times: \[\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}\] \[\log_{b}{x} = 36\] \[(\log_{b}{x})^{54} = 36^{54}\] \[x = 6^{108}\]
However, since we found that $\log_{b}{x} = 36$ $x$ is also equal to $b^{36}$ . Equating these, \[b^{36} = 6^{108}\] \[b = 6^3 = \boxed{216}\]

Solution: We start by simplifying the first equation to \[3\log_{b}{(\sqrt{x}\log x)}=\log_{b}{(x^{\frac{3}{2}}\log^3x)}=56\] \[x^\frac{3}{2}\cdot \log_b^3x=b^{56}\] Next, we simplify the second equation to \[\log_{\log(x)}(x)=\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] \[\log_bx=54\log_b(\log_b(x))=\log_b(\log_b^{54}(x))\] \[x=\log_b^{54}x\] Substituting this into the first equation gives \[\log_b^{54\cdot \frac{3}{2}}(x)\cdot \log_b^3x=\log_b^{84}x=b^{56}\] \[x=b^{b^{\frac{56}{84}}}=b^{b^{\frac{2}{3}}}\] Plugging this into $x=\log_b^{54}x$ gives \[b^{b^{\frac{2}{3}}}=\log_b^{54}(b^{b^\frac{2}{3}})=b^{\frac{2}{3}\cdot 54}=b^{36}\] \[b^{\frac{2}{3}}=36\] \[b=36^{\frac{3}{2}}=6^3=\boxed{216}\] -ktong

Solution: Apply change of base to \[\log_{\log x}(x)=54\] to yield: \[\frac{\log_b(x)}{\log_b(\log_b(x))}=54\] which can be rearranged as: \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] Apply log properties to \[3\log(\sqrt{x}\log x)=56\] to yield: \[3(\frac{1}{2}\log_b(x)+\log_b(\log_b(x)))=56\Rightarrow\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}\] Substituting \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] into the equation $\frac{1}{2}\log_b(x)+\log_b(\log_b(x))=\frac{56}{3}$ yields: \[\frac{1}{2}\log_b(x)+\frac{\log_b(x)}{54}=\frac{28\log_b(x)}{54}=\frac{56}{3}\] So \[\log_b(x)=36.\] Substituting this back in to \[\frac{\log_b(x)}{54}=\log_b(\log_b(x))\] yields \[\frac{36}{54}=\log_b(36).\] So, \[b^{\frac{2}{3}}=36\Rightarrow \boxed{216}\]

Solution: 1st equation: \[\log (\sqrt{x}\log x)=\frac{56}{3}\] \[\log(\sqrt x)+\log(\log x)=\frac{1}{2}\log x+\log(\log x)=\frac{56}{3}\] 2nd equation: \[x=(\log x)^{54}\] So now substitute $\log x=a$ and $x=b^a$ \[b^a=a^{54}\] \[b=a^{\frac{54}{a}}\] We also have that \[\frac{1}{2}a+\log_{a^\frac{54}{a}} a=\frac{56}{3}\] \[\frac{1}{2}a+\frac{1}{54}a=\frac{56}{3}\] This means that $\frac{14}{27}a=\frac{56}{3}$ , so \[a=36\] \[b=36^{\frac{54}{36}}=36^\frac{3}{2}=\boxed{216}\]

Solution:
This system of equations looks complicated to work with, so we let $a=\log_bx$ to make it easier for us to read.
Now, the first equation becomes $3\log(\sqrt x \cdot a) = 56 \implies \log(\sqrt{x}\cdot a)=\frac{56}3$
The second equation, $\log_{\log(x)}(x)=54$ gives us $\underline{a^{54} = x}$
Let's plug this back into the first equation to see what we get: $\log_b(\sqrt{a^{54}}\cdot a)=\frac{56}3$ , and simplifying, $\log_b(a^{27}\cdot a^1)=\log_b(a^{28})=\frac{56}{3}$ , so $b^{\frac{56}3}=a^{28}\implies \underline{b^{\frac 23}=a}$
Combining this new finding with what we had above $a^{54} = (b^{\frac 23})^{54} = x\implies \mathbf{b^{36} =x}$
Now that we've expressed one variable in terms of the other, we can plug this into either equation, say equation 1. Then we get $\log_b(\sqrt{b^{36}}\cdot\log_b(b^{36})=\frac{56}3\implies$ $\log_b(b^{18}\cdot 36)=\frac{56}3\implies b^{\frac{56}3}=b^{18}\cdot 36$
Finally, that gives us that $\frac{b^{\frac{56}3}}{b^{18}}=36\implies b^{\frac{56}{3}-18}=b^{\frac{56}{3}-\frac{54}{3}}=b^{\frac 23}=36\implies b=36^{\frac 32}=6^3$ . Thus, $b=\boxed{216}$

Solution:
Using change of base on the second equation, we have
\[\frac{\log_{b} x}{\log_{b} \log_{b} x} = 54\]
Using log rules on the first equation, we have
\[\frac{3}{2} \log_{b} x + 3 \log_{b} \log_{b} x = 56\]
We notice that $\log_{b} x$ and $\log_{b} \log_{b} x$ are in both equations. Thus, we set $m = \log_{b} x$ and $n = \log_{b} \log_{b} x$ and we have
\[\frac{3}{2} m + 3n = 56\] \[\frac{m}{n} = 54\]
Solving this yields $m = 36$ $n = \frac{2}{3}$
Now, $n = \log_{b} \log_{b} x = \log_{b} m = \log_{b} 36$ , so we have $\log_{b} 36 = \frac{2}{3}$ . Solving this yields $b = \boxed{216}$

Solution:
The second equation implies that \[\log_{\log_b x} x=54\implies (\log_b x)^{54}=x \implies \log_b x=x^{\frac{1}{54}}\] The first equation implies that \[3\log_b(\sqrt{x} \log_b x)=56 \implies b^{\frac{56}{3}}=\sqrt{x} \log_b x\] Substituting the first result into the second gives us \[b^{\frac{56}{3}}=x^{\frac{1}{2}}\cdot x^{\frac{1}{54}} \implies b=x^{\frac{1}{36}}.\] Because $b^{36}=x,$ $\log_b x=36$ by the definition of a logarithm.
Substituting this into the second equation, \[\log_{36} x=54 \implies x=36^{54}.\] Finally, \[b=(36^{54})^{\frac{1}{36}}=36^{54\cdot\frac{1}{36}}=6^{2*\frac{3}{2}}=6^3=\boxed{216}.\]


Answer: 2.0

In a certain card game, a player is dealt a hand of $10$ cards from a deck of $52$ distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as $158A00A4AA0$ . What is the digit $A$
$\textbf{(A) } 2 \qquad\textbf{(B) } 3 \qquad\textbf{(C) } 4 \qquad\textbf{(D) } 6 \qquad\textbf{(E) } 7$

Solution:
$158A00A4AA0 \equiv 1+5+8+A+0+0+A+4+A+A+0 \equiv 4A \pmod{9}$
We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
We need to get rid of the multiples of $3$ , which will subsequently get rid of the multiples of $9$ (if we didn't, the zeroes would mess with the equation since you can't divide by 0)
$9\cdot5=45$ $8\cdot6=48$ $\frac{51}{3}$ leaves us with 17.
$\frac{52\cdot\cancel{51}^{17}\cdot50\cdot49\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}\cdot44\cdot43}{10\cdot\cancel{9}\cdot\cancel{8}\cdot7\cdot\cancel{6}\cdot\cancel{5}\cdot4\cdot\cancel{3}\cdot2\cdot1}$
Converting these into $\pmod{9}$ , we have
$\binom{52}{10}\equiv \frac{(-2)\cdot(-1)\cdot(-4)\cdot4\cdot2\cdot1\cdot(-1)\cdot(-2)}{1\cdot(-2)\cdot4\cdot2\cdot1} \equiv (-1)\cdot(-4)\cdot(-1)\cdot(-2) \equiv 8 \pmod{9}$
$4A\equiv8\pmod{9} \implies A=\boxed{2}$ ~quacker88

Solution:
We're looking for the amount of ways we can get $10$ cards from a deck of $52$ , which is represented by $\binom{52}{10}$
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
And after simplifying, we get $26\cdot17\cdot7\cdot47\cdot46\cdot5\cdot11\cdot43$ .
Now, if we examine the number $158A00A4AA0$ , we can notice that it is equal to some number $n$ times 10.
Therefore, we can divide 10 from the aforementioned expression and find the units digit, which will be $A$
Now, after dividing ten, we will have $26\cdot17\cdot7\cdot47\cdot23\cdot11\cdot43$ .
We can then use modulo 10 and find that the unit digit of the expression is $\boxed{2}$ ~lucaswujc

Solution:
$\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43$
Since this number is divisible by $4$ but not $8$ , the last $2$ digits must be divisible by $4$ but the last $3$ digits cannot be divisible by $8$ . This narrows the options down to $2$ and $6$
Also, the number cannot be divisible by $3$ . Adding up the digits, we get $18+4A$ . If $A=6$ , then the expression equals $42$ , a multiple of $3$ . This would mean that the entire number would be divisible by $3$ , which is not what we want. Therefore, the only option is $\boxed{2}$ -PCChess

Solution: It is not hard to check that $13$ divides the number, \[\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43.\] As $10^3\equiv-1\pmod{13}$ , using $\pmod{13}$ we have $13|\overline{AA0}-\overline{0A4}+\overline{8A0}-\overline{15}=110A+781$ . Thus $6A+1\equiv0\pmod{13}$ , implying $A\equiv2\pmod{13}$ so the answer is $\boxed{2}$

Solution: As mentioned above, \[\binom{52}{10}=\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = {10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43} = 158A00A4AA0.\] We can divide both sides of $10 \cdot 17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA0$ by 10 to obtain \[17 \cdot 13 \cdot 7 \cdot 47 \cdot 46 \cdot 11 \cdot 43 = 158A00A4AA,\] which means $A$ is simply the units digit of the left-hand side. This value is \[7 \cdot 3 \cdot 7 \cdot 7 \cdot 6 \cdot 1 \cdot 3 \equiv \boxed{2}.\] i_equal_tan_90 , revised by emerald_block

Solution:
We note that: \[\frac{(52)(51)(50)(49)(48)(47)(46)(45)(44)(43)}{(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)} = (13)(17)(7)(47)(46)(5)(22)(43).\] Let $K=(13)(17)(7)(47)(46)(5)(22)(43)$ . This will help us find the last two digits modulo $4$ and modulo $25$ .
It is obvious that $K \equiv 0 \pmod{4}$ . Also (although this not so obvious), \[K \equiv (13)(17)(7)(47)(46)(5)(22)(43)\] \[\equiv (13)(-8)(7)(-3)(-4)(5)(-3)(-7)\] \[\equiv (13)(-96)(21)(35)\] \[\equiv (13)(4)(-4)(10)\] \[\equiv (13)(-16)(10)\] \[\equiv (13)(9)(10)\] \[\equiv (117)(10)\] \[\equiv (-8)(10)\] \[\equiv 20 \pmod{25}.\] Therefore, $K \equiv 20 \mod 100$ . Thus $K=20$ , implying that $\boxed{2}$

Solution:
As in Solution 2, we see that
\[\binom{52}{10}=26\cdot17\cdot5\cdot7\cdot47\cdot46\cdot11\cdot43,\]
which contains no factors of $3.$ Therefore, the sum of the digits must not be a multiple of $3.$ This sum is
\[1+5+8+A+0+0+A+4+A+A+0=18+4A.\]
It follows that $4A$ cannot be a multiple of $3,$ ruling out choices $(B)$ and $(D).$ Therefore, our possibilities are $A=2,4,$ and $7.$ Now, notice that $\binom{52}{10}$ is divisible by $7.$ Therefore, we can plug each possible value of $A$ into $158A00A4AA0$ and test for divisibility by $7.$ Conveniently, we see that the first value, $A=2,$ works. Thus, the answer is $\boxed{2}.$ (To make our argument more rigorous, we can also test divisibility by $7$ for $A=4$ and $7$ to show that these values do not work.)

Solution:
The total number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Using divisibility rules, we have that A is not a multiple of $3$ . Then, divide this equation by 10. This implies that the new number $158A00A4AA0$ is divisible by $2$ but not $4$ . This means that $A$ is either $2$ or $6$ . However, $6$ is a multiple of $3$ , meaning $A$ has to be $\boxed{2}$

Solution:
As stated in previous solutions, the number of ways to choose $10$ from $52$ is $\binom{52}{10}=\frac{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46\cdot45\cdot44\cdot43}{10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}$
Canceling out common factors $(\dfrac{52}{2} = 26 \text{, } \dfrac{51}{3} = 17 \text{, } 5 \cdot 10 = 50 \text{, } \dfrac{49}{7} = 7 \text{, } 48 = 6 \cdot 8 \text{, } \dfrac{45}{9} = 5 \text{, } \dfrac{44}{11} = 4)$ , you get this - $\frac{\cancel{52}^{26}\cdot\cancel{51}^{17}\cdot\cancel{50}\cdot\cancel{49}^{7}\cdot\cancel{48}\cdot47\cdot46\cdot\cancel{45}^5\cdot\cancel{44}^{4}\cdot43}{\cancel{10}\cdot\cancel{9}\cdot\cancel{8}\cdot\cancel{7}\cdot\cancel{6}\cdot\cancel{5}\cdot\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$
When you multiply the remaining numbers, you get the product as $15820024220$ . From this product, we can then determine that $A$ is equal to $\boxed{2}$

Solution: Compute $\frac{52!}{10!42!} = 15820024220.$ Therefore our answer is $\boxed{2}.$


Answer: 174.0

In a certain circle , the chord of a $d$ -degree arc is $22$ centimeters long, and the chord of a $2d$ -degree arc is $20$ centimeters longer than the chord of a $3d$ -degree arc, where $d < 120.$ The length of the chord of a $3d$ -degree arc is $- m + \sqrt {n}$ centimeters, where $m$ and $n$ are positive integers.  Find $m + n.$

Solution:
Note that a cyclic quadrilateral in the form of an isosceles trapezoid can be formed from three chords of three $d$ -degree arcs and one chord of one $3d$ -degree arc. The diagonals of this trapezoid turn out to be two chords of two $2d$ -degree arcs. Let $AB$ $AC$ , and $BD$ be the chords of the $d$ -degree arcs, and let $CD$ be the chord of the $3d$ -degree arc. Also let $x$ be equal to the chord length of the $3d$ -degree arc. Hence, the length of the chords, $AD$ and $BC$ , of the $2d$ -degree arcs can be represented as $x + 20$ , as given in the problem.
Using Ptolemy's theorem,
\[AB(CD) + AC(BD) = AD(BC)\] \[22x + 22(22)  = (x + 20)^2\] \[22x + 484 = x^2 + 40x + 400\] \[0 = x^2 + 18x - 84\]
We can then apply the quadratic formula to find the positive root to this equation since polygons obviously cannot have sides of negative length. \[x = \frac{-18 + \sqrt{18^2 + 4(84)}}{2}\] \[x = \frac{-18 + \sqrt{660}}{2}\]
$x$ simplifies to $\frac{-18 + 2\sqrt{165}}{2},$ which equals $-9 + \sqrt{165}.$ Thus, the answer is $9 + 165 = \boxed{174}$

Solution: Let $z=\frac{d}{2},$ and $R$ be the circumradius. From the given information, \[2R\sin z=22\] \[2R(\sin 2z-\sin 3z)=20\] Dividing the latter by the former, \[\frac{2\sin z\cos z-(3\cos^2z\sin z-\sin^3 z)}{\sin z}=2\cos z-(3\cos^2z-\sin^2z)=1+2\cos z-4\cos^2z=\frac{10}{11}\] \[4\cos^2z-2\cos z-\frac{1}{11}=0 (1)\] We want to find \[\frac{22\sin (3z)}{\sin z}=22(3-4\sin^2z)=22(4\cos^2z-1).\] From $(1),$ this is equivalent to $44\cos z-20.$ Using the quadratic formula, we find that the desired length is equal to $\sqrt{165}-9,$ so our answer is $\boxed{174}$

Solution: Let $z=\frac{d}{2}$ $R$ be the circumradius, and $a$ be the length of 3d degree chord. Using the extended sine law, we obtain: \[22=2R\sin(z)\] \[20+a=2R\sin(2z)\] \[a=2R\sin(3z)\] Dividing the second from the first we get $\cos(z)=\frac{20+a}{44}$ By the triple angle formula we can manipulate the third equation as follows: $a=2R\times \sin(3z)=\frac{22}{\sin(z)} \times (3\sin(z)-4\sin^3(z)) = 22(3-4\sin^2(z))=22(4\cos^2(z)-1)=\frac{(20+a)^2}{22}-22$ Solving the quadratic equation gives the answer to be $\boxed{174}$


Answer: 13.0

In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$ th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?
(A) 10 (B) 13 (C) 27 (D) 120 (E) 126

Solution: The scores of all ten runners must sum to $55$ . So a winning score is anything between $1+2+3+4+5=15$ and $\lfloor\tfrac{55}{2}\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$ $1+2+3+x+10$ and $1+2+x+9+10$ , so the answer is $\boxed{13}$


Answer: 17.0

In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$
$\mathrm{(A)}\ 12\qquad \mathrm{(B)}\ 17\qquad \mathrm{(C)}\ 20\qquad \mathrm{(D)}\ 25\qquad \mathrm{(E)}\ 35$

Solution: Let $p$ be the price at the beginning of January.  The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$

Solution:
Without loss of generality, we can assume the price at the beginning of January was $$100$
When it rose by $20\%$ , it became $$120$ , when it fell by $20\%$ , it became $$96$ , and when it rose by $25\%$ , it became $$120$ again.
In order for the price at the end of April to be the same as it was at the beginning of January ( $$100$ ), the price must decrease by $$20$
20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{17}$ . ~azc1027


Answer: 17.0

Solution: Let $p$ be the price at the beginning of January.  The price at the end of March was $(1.2)(0.8)(1.25)p = 1.2p.$ Because the price at the end of April was $p$ , the price decreased by $0.2p$ during April, and the percent decrease was \[x = 100 \cdot \frac{0.2p}{1.2p} = \frac {100}{6} \approx 16.7.\] So to the nearest integer $x$ is $\boxed{17}$

Solution:
Without loss of generality, we can assume the price at the beginning of January was $$100$
When it rose by $20\%$ , it became $$120$ , when it fell by $20\%$ , it became $$96$ , and when it rose by $25\%$ , it became $$120$ again.
In order for the price at the end of April to be the same as it was at the beginning of January ( $$100$ ), the price must decrease by $$20$
20 is $\frac{1}{6}th$ of 120, and $\frac{1}{6} \approx 0.167 \approx 17\%$ So to the nearest integer, $x = 17$ and the answer is $\boxed{17}$ . ~azc1027


Answer: 378.0

In a circle of radius $42$ , two chords of length $78$ intersect at a point whose distance from the center is $18$ .  The two chords divide the interior of the circle into four regions.  Two of these regions are bordered by segments of unequal lengths, and the area of either of them can be expressed uniquely in the form $m\pi-n\sqrt{d},$ where $m, n,$ and $d_{}$ are positive integers and $d_{}$ is not divisible by the square of any prime number.  Find $m+n+d.$

Solution:
Let the center of the circle be $O$ , and the two chords be $\overline{AB}, \overline{CD}$ and intersecting at $E$ , such that $AE = CE < BE = DE$ . Let $F$ be the midpoint of $\overline{AB}$ . Then $\overline{OF} \perp \overline{AB}$
By the Pythagorean Theorem $OF = \sqrt{OB^2 - BF^2} = \sqrt{42^2 - 39^2} = 9\sqrt{3}$ , and $EF = \sqrt{OE^2 - OF^2} = 9$ . Then $OEF$ is a $30-60-90$ right triangle , so $\angle OEB = \angle OED = 60^{\circ}$ . Thus $\angle BEC = 60^{\circ}$ , and by the Law of Cosines
It follows that $\triangle BCO$ is an equilateral triangle , so $\angle BOC = 60^{\circ}$ . The desired area can be broken up into two regions, $\triangle BCE$ and the region bounded by $\overline{BC}$ and minor arc $\stackrel{\frown}{BC}$ . The former can be found by Heron's formula to be $[BCE] = \sqrt{60(60-48)(60-42)(60-30)} = 360\sqrt{3}$ . The latter is the difference between the area of sector $BOC$ and the equilateral $\triangle BOC$ , or $\frac{1}{6}\pi (42)^2 - \frac{42^2 \sqrt{3}}{4} = 294\pi - 441\sqrt{3}$
Thus, the desired area is $360\sqrt{3} + 294\pi - 441\sqrt{3} = 294\pi - 81\sqrt{3}$ , and $m+n+d = \boxed{378}$


Answer: 49.0

In a circle parallel chords of lengths 2, 3, and 4 determine central angles of $\alpha$ $\beta$ , and $\alpha + \beta$ radians , respectively, where $\alpha + \beta < \pi$ . If $\cos \alpha$ , which is a positive rational number , is expressed as a fraction in lowest terms, what is the sum of its numerator and denominator?

Solution:
All chords of a given length in a given circle subtend the same arc and therefore the same central angle.  Thus, by the given, we can re-arrange our chords into a triangle with the circle as its circumcircle
This triangle has semiperimeter $\frac{2 + 3 + 4}{2}$ so by Heron's formula it has area $K = \sqrt{\frac92 \cdot \frac52 \cdot \frac32 \cdot \frac12} = \frac{3}{4}\sqrt{15}$ .  The area of a given triangle with sides of length $a, b, c$ and circumradius of length $R$ is also given by the formula $K = \frac{abc}{4R}$ , so $\frac6R = \frac{3}{4}\sqrt{15}$ and $R = \frac8{\sqrt{15}}$
Now, consider the triangle formed by two radii and the chord of length 2.  This isosceles triangle has vertex angle $\alpha$ , so by the Law of Cosines
\[2^2 = R^2 + R^2 - 2R^2\cos \alpha \Longrightarrow \cos \alpha = \frac{2R^2 - 4}{2R^2} = \frac{17}{32}\] and the answer is $17 + 32 = \boxed{049}$

Solution: It’s easy to see in triangle which lengths 2, 3, and 4, that the angle opposite the side 2 is $\frac{\alpha}{2}$ , and using the Law of Cosines , we get: \[2^2 = 3^2 + 4^2 - 2\cdot3\cdot4\cos\frac{\alpha}{2}\] Which, rearranges to: \[21 = 24\cos\frac{\alpha}{2}\] And, that gets us: \[\cos\frac{\alpha}{2} = 7/8\] Using $\cos 2\theta = 2\cos^2 \theta - 1$ , we get that: \[\cos\alpha = 17/32\] Which gives an answer of $\boxed{049}$

Solution:
Using the first diagram above, \[\sin \frac{\alpha}{2} = \frac{1}{r}\] \[\sin \frac{\beta}{2} = \frac{1.5}{r}\] \[\sin(\frac{\alpha}{2}+\frac{\beta}{2})=\frac{2}{r}\] by the Pythagorean trig identities, \[\cos\frac{\alpha}{2}=\sqrt{1-\frac{1}{r^2}}\] \[\cos\frac{\beta}{2}=\sqrt{1-\frac{2.25}{r^2}}\] so by the composite sine identity \[\frac{2}{r}=\frac{1}{r}\sqrt{1-\frac{2.25}{r^2}}+\frac{1.5}{r}\sqrt{1-\frac{1}{r^2}}\] multiply both sides by $2r$ , then subtract $\sqrt{4-\frac{9}{r^2}}$ from both sides
squaring both sides, we get \[16 - 8\sqrt{4-\frac{9}{r^2}} + 4 - \frac{9}{r^2}=9 - \frac{9}{r^2}\] \[\Longrightarrow 16+4=9+8\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{11}{8}=\sqrt{4-\frac{9}{r^2}}\Longrightarrow\frac{121}{64}=4-\frac{9}{r^2}\] \[\Longrightarrow\frac{(256-121)r^2}{64}=9\Longrightarrow r^2= \frac{64}{15}\] plugging this back in, \[\cos^2(\frac{\alpha}{2})=1-\frac{15}{64}=\frac{49}{64}\] so \[\cos(\alpha)=2(\frac{49}{64})-1=\frac{34}{64}=\frac{17}{32}\] and the answer is $17+32=\boxed{049}$


Answer: 5.0

In a circle with center $O$ $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$ . Then the length of $BC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^^Circle(O,1)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$O$", O, dir(305)); label("$5$", B--O, dir(O--B)*dir(90)); label("$60^\circ$", dir(185), dir(185)); label("$60^\circ$", B+0.05*dir(-25), dir(-25));[/asy]
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{3} \qquad \mathrm{(C) \  } 5-\frac{\sqrt{3}}{2} \qquad \mathrm{(D) \  } 5 \qquad \mathrm{(E) \  }\text{none of the above}$

Solution: Since $\angle CAD$ is an angle inscribed in a $60{^\circ}$ arc, we obtain $\angle CAD =\frac{60^{\circ}}{2} = 30^{\circ}$ , so $\triangle ABO$ is a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle. This gives $AO = BO\sqrt{3} = 5\sqrt{3}$ and $AB = 2BO = 10$ , and now since $AD$ is a diameter, $AD = 2AO = 10\sqrt{3}$ . The fact that $AD$ is a diameter also means that $\angle ACD = 90^{\circ}$ , so $\triangle ACD$ is again a $30^{\circ}$ $60^{\circ}$ $90^{\circ}$ right triangle, yielding $CD = \frac{1}{2}AD = 5\sqrt{3}$ and $AC = \frac{\sqrt{3}}{2}AD = 15$ , which finally gives $BC = AC-AB = 15-10 = \boxed{5}$


Answer: 3.4

In a collection of red, blue, and green marbles, there are $25\%$ more red marbles than blue marbles, and there are $60\%$ more green marbles than red marbles. Suppose that there are $r$ red marbles. What is the total number of marbles in the collection?
$\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r$

Solution:
Let the number of red marbles be 100. You have $b = 80,$ and $g = 160.$
The answer is $\frac{100+80+160}{100} = \boxed{3.4}$


Answer: 341.0

In a drawer Sandy has $5$ pairs of socks, each pair a different color.  On Monday Sandy selects two individual socks at random from the $10$ socks in the drawer.  On Tuesday Sandy selects $2$ of the remaining $8$ socks at random and on Wednesday two of the remaining $6$ socks at random.  The probability that Wednesday is the first day Sandy selects matching socks is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers,  Find $m+n$

Solution:
Let the fifth sock be arbitrary; the probability that the sixth sock matches in color is $\dfrac{1}{9}$
Assuming this, then let the first sock be arbitrary; the probability that the second sock does not match is $\dfrac{6}{7}.$
The only "hard" part is the third and fourth sock. But that is simple casework. If the third sock's color matches the color of one of the first two socks (which occurs with probability $\dfrac{2}{6} = \dfrac{1}{3}$ ), then the fourth sock can be arbitrary. Otherwise (with probability $\dfrac{2}{3}$ ), the fourth sock can be chosen with probability $\dfrac{4}{5}$ (5 socks left, 1 sock that can possibly match the third sock's color). The desired probability is thus \[\frac{1}{9} \cdot \frac{6}{7} \cdot \left(\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{4}{5}\right) = \frac{26}{315}.\] The sum is therefore $26+315=\boxed{341}.$

Solution: The key is to count backwards. First, choose the pair which you pick on Wednesday in $5$ ways. Then there are four pairs of socks for you to pick a pair of on Tuesday, and you don't want to pick a pair. Since there are $4$ pairs, the number of ways to do this is $\dbinom{8}{2}-4$ . Then, there are two pairs and two nonmatching socks for you to pick from on Monday, a total of $6$ socks. Since you don't want to pick a pair, the number of ways to do this is $\dbinom{6}{2}-2$ . Thus the answer is \[\dfrac{\left(5\right)\left(\dbinom{8}{2}-4\right)\left(\dbinom{6}{2}-2\right)}{\dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}}=\dfrac{26}{315}.\] $26 + 315 = \boxed{341}$

Solution:
For the first sock, note that to pick two different socks, can complementary count to get the total, $\binom{10}{2}$ minus the number of pairs (5) to get \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\] The next steps aren't quite as simple, though.
WLOG suppose the socks are (a, a, b, b, c, c, d, d, e, e) and that we chose ab on the first day.
This leaves abccddee as the remaining socks. We can think of our next choice as how many "pairs" we "break".
Note that right now, in abccddee there are 3 pairs, namely {cc, dd, ee}. Our next choice could either leave all of these pairs (if we choose ab), leave 2 of these pairs, (choose say ac to keep dd and ee as pairs), or leave only 1 as a pair, say choosing de to get abccde. \[\bold{\text{Case 1: Break zero pairs}}.\] In this case, we want to keep 3 pairs remaining, so our only choice is to choose AB, which would make the remaining ccddee. The probability that we choose AB is $\frac{1}{\binom{8}{2}}$ and the probability that we get a pair after this is $\frac{3}{\binom{6}{2}}$ since we just choose CC, DD, or EE. This case has a probability $\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}.$ \[\bold{\text{Case 2: Break one pair}}.\] In this case, we want to choose A and one of {C,C,D,D,E,E} or B and one of {C,C,D,D,E,E}. This yields $\frac{6 + 6}{\binom{8}{2}}$ probability. After this, we end up with 2 pairs, so the total probability of choosing a pair on Wednesday in this case is $\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}.$ \[\bold{\text{case 3: Break 2 pairs}}.\] In this case we'd choose two distinct characters from {C,C,D,D,E,E} which would be $\frac{\binom{6}{2} - 3}{\binom{6}{2}}.$ After this, only one pair remains so the total probability for this case is
$\frac{\binom{6}{2} - 3}{\binom{6}{2}}    \frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}.$ Adding all the cases and multiplying by $\frac{\binom{10}{2} - 5}{\binom{10}{2}}$ yields \[\frac{\binom{10}{2} - 5}{\binom{10}{2}}\left(\frac{1}{\binom{8}{2}}\frac{3}{\binom{6}{2}}+\frac{6 + 6}{\binom{8}{2}}\frac{2}{\binom{6}{2}}+ \frac{\binom{6}{2} - 3}{\binom{6}{2}}\frac{6}{\binom{8}{2}}\frac{1}{\binom{6}{2}}\right) = \frac{26}{315} \rightarrow 26 + 315 = \boxed{341}\]


Answer: 49.0

In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$

Solution:
We can use complementary counting : finding the probability that at least one team wins all games or at least one team loses all games.
No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.
Now we use PIE
The probability that one team wins all games is $5\cdot \left(\frac{1}{2}\right)^4=\frac{5}{16}$
Similarity, the probability that one team loses all games is $\frac{5}{16}$
The probability that one team wins all games and another team loses all games is $\left(5\cdot \left(\frac{1}{2}\right)^4\right)\left(4\cdot \left(\frac{1}{2}\right)^3\right)=\frac{5}{32}$
$\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}$
Since this is the opposite of the probability we want, we subtract that from 1 to get $\frac{17}{32}$
$17+32=\boxed{049}$

Solution:
There are $\dbinom{5}{2} = 10$ games in total, and every game can either end in a win or a loss. Therefore, there are $2^{10} = 1024$ possible outcomes.
Now, computing this probability directly seems a little hard, so let's compute the complement -- the probability that there is an undefeated team, a winless team, or both.
The number of ways that we can have an undefeated team is $5 \cdot 2^6,$ as there are five ways to choose the team itself and $2^6$ outcomes for the games that are not concerning the undefeated team. Reversing all the wings to losses yields a winless team, so this situation is symmetrical to the first one. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6$ ways for an undefeated or winless team.
In the process, however, we have accidentally overcounted the scenario of both an undefeated and winless team. There are $5 \cdot 4 \cdot 2^3$ ways for this scenario to happen, because there are five ways to choose the undefeated team, four ways for the winless, and three games that don't concern either of the teams. Therefore, there are $5 \cdot 2^6 + 5 \cdot 2^6 - 5 \cdot 4 \cdot 2^3 = 480$ ways to have an undefeated and/or winless team, so there are $1024 - 480 = 544$ to not have any.
Our probability is thus $\dfrac{544}{1024},$ which simplifies to $\dfrac{17}{32},$ for our answer of $17 + 32 = \boxed{049}.$


Answer: 792.0

In a game of Chomp , two players alternately take bites from a 5-by-7 grid of unit squares . To take a bite, a player chooses one of the remaining squares , then removes ("eats") all squares in the quadrant defined by the left edge (extended upward) and the lower edge (extended rightward) of the chosen square. For example, the bite determined by the shaded square in the diagram would remove the shaded square and the four squares marked by $\times.$ (The squares with two or more dotted edges have been removed form the original board in previous moves.)
AIME 1992 Problem 12.png
The object of the game is to make one's opponent take the last bite. The diagram shows one of the many subsets of the set of 35 unit squares that can occur during the game of Chomp. How many different subsets are there in all? Include the full board and empty board in your count.

Solution:
By drawing possible examples of the subset, one can easily see that making one subset is the same as dividing the game board into two parts.
One can also see that it is the same as finding the shortest route from the upper left hand corner to the lower right hand corner; Such a route would require 5 lengths that go down, and 7 that go across, with the shape on the right "carved" out by the path a possible subset.
Therefore, the total number of such paths is $\binom{12}{5}=\boxed{792}$

Solution: If any square is eaten, the squares to the right of it must also be eaten. Thus, if $a_i$ is the number of squares remaining on row $i$ , there is exactly one way the row can be configured (the leftmost $a_i$ squares are uneaten and the ones to the right are eaten.) Additionally, the squares above an eaten square must be eaten, so $\{a_i\}$ is nondecreasing. We can thus write $0 \leq a_1 \leq a_2 \leq \dots \leq a_5 \leq 7$ ; the number of sequences $\{a_i\}$ satisfying this inequality may be found by Stars and Bars to be $\binom{7+6-1}{6-1} = \boxed{792}$


Answer: 28.0

In a geometric sequence of real numbers, the sum of the first $2$ terms is $7$ , and the sum of the first $6$ terms is $91$ . The sum of the first $4$ terms is
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 35\qquad\textbf{(D)}\ 49\qquad\textbf{(E)}\ 84$

Solution: Denote the sum of the first $2$ terms as $x$ . Since we know that the sum of the first $6$ terms is $91$ which is $7 \cdot 13$ , we have $x$ $xy$ $xy^2$ $13x$ because it is a geometric series. We can quickly see that $y$ $3$ , and therefore, the sum of the first $4$ terms is $4x = 4 \cdot 7 = \boxed{28}$


Answer: 0.0

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?
$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$

Solution:
Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$ , we have to make sure that its $y$ -coordinate is $\pm20$ , because that's the only way the area of the triangle can be $100$
Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$ . We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$ . The perimeter of this minimal triangle is $2\sqrt{425} + 10$ , which is larger than $50$ . Since the minimum perimeter is greater than $50$ , there is no triangle that satisfies the condition, giving us $\boxed{0}$

Solution: Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $CD$ , we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ , which is clearly impossible, again giving the answer as $\boxed{0}$


Answer: 0.0

Solution:
Notice that whatever point we pick for $C$ $AB$ will be the base of the triangle. Without loss of generality, let points $A$ and $B$ be $(0,0)$ and $(10,0)$ , since for any other combination of points, we can just rotate the plane to make them $(0,0)$ and $(10,0)$ under a new coordinate system. When we pick point $C$ , we have to make sure that its $y$ -coordinate is $\pm20$ , because that's the only way the area of the triangle can be $100$
Now when the perimeter is minimized, by symmetry, we put $C$ in the middle, at $(5, 20)$ . We can easily see that $AC$ and $BC$ will both be $\sqrt{20^2+5^2} = \sqrt{425}$ . The perimeter of this minimal triangle is $2\sqrt{425} + 10$ , which is larger than $50$ . Since the minimum perimeter is greater than $50$ , there is no triangle that satisfies the condition, giving us $\boxed{0}$

Solution: Without loss of generality, let $AB$ be a horizontal segment of length $10$ . Now realize that $C$ has to lie on one of the lines parallel to $AB$ and vertically $20$ units away from it. But $10+20+20$ is already 50, and this doesn't form a triangle. Otherwise, without loss of generality, $AC<20$ . Dropping altitude $CD$ , we have a right triangle $ACD$ with hypotenuse $AC<20$ and leg $CD=20$ , which is clearly impossible, again giving the answer as $\boxed{0}$


Answer: 7.0

In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$

Solution: Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{7}\]


Answer: 16.0

In a group of nine people each person shakes hands with exactly two of the other people from the group. Let $N$ be the number of ways this handshaking can occur. Consider two handshaking arrangements different if and only if at least two people who shake hands under one arrangement do not shake hands under the other arrangement. Find the remainder when $N$ is divided by $1000$

Solution:
Given that each person shakes hands with two people, we can view all of these through graph theory as 'rings'.  This will split it into four cases:  Three rings of three, one ring of three and one ring of six, one ring of four and one ring of five, and one ring of nine. (All other cases that sum to nine won't work, since they have at least one 'ring' of two or fewer points, which doesn't satisfy the handshaking conditions of the problem.)
Case 1: To create our groups of three, there are $\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}$ .  In general, the number of ways we can arrange people within the rings to count properly is $\dfrac{(n-1)!}{2}$ , since there are $(n-1)!$ ways to arrange items in the circle, and then we don't want to want to consider reflections as separate entities.  Thus, each of the three cases has $\dfrac{(3-1)!}{2}=1$ arrangements.  Therefore, for this case, there are $\left(\dfrac{\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}{3!}\right)(1)^3=280$
Case 2: For three and six, there are $\dbinom{9}{6}=84$ sets for the rings.  For organization within the ring, as before, there is only one way to arrange the ring of three.  For six, there is $\dfrac{(6-1)!}{2}=60$ .  This means there are $(84)(1)(60)=5040$ arrangements.
Case 3: For four and five, there are $\dbinom{9}{5}=126$ sets for the rings.  Within the five, there are $\dfrac{4!}{2}=12$ , and within the four there are $\dfrac{3!}{2}=3$ arrangements.  This means the total is $(126)(12)(3)=4536$
Case 4: For the nine case, there is $\dbinom{9}{9}=1$ arrangement for the ring.  Within it, there are $\dfrac{8!}{2}=20160$ arrangements.
Summing the cases, we have $280+5040+4536+20160=30016 \to \boxed{016}$

Solution:
Let $f_N$ be the number of ways a group of $N$ people can shake hands with exactly two of the other people from the group, where $N \ge 3$
We can easily find that $f_3=1$
Continuing on, we will label the people as $A$ $B$ $C$ ,... corresponding with $N$
$f_4$ : There are $\dbinom{3}{2}=3$ possible ways for person $A$ to shake hands with two others (WLOG assume $A$ shakes hands with $B$ and $C$ ), and there is only one possible outcome thereon ( $B$ and $C$ both shakes hands with $D$ ).
Therefore, we can conclude that $f_4=3$
$f_5$ : There are $\dbinom{4}{2}=6$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ). Then, $B$ and $C$ must also shake hands with the remaining two people ( $2$ ways), and those last $2$ people must shake hands with each other ( $1$ way).
Therefore, $f_5=\dbinom{4}{2}\ast(2)\ast(1)=12$
$f_6$ : There are $\dbinom{5}{2}=10$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ). However, $B$ and $C$ shake hands with each other, then there are $f_3$ ways for the rest of the people to shake hands. If $B$ and $C$ shake hands with two others out of the three remaining people (3 $\ast$ 2 ways), those $2$ people must shake hands with the last person ( $1$ way).
Therefore, $f_5=\dbinom{5}{2}\ast(f_3)+3\ast(2)\ast(1)=70$
Now we have enough information to find $f_9$
$f_9$ : There are $\dbinom{8}{2}=28$ possible ways for person $A$ to shake hands with two others (WLOG assume they are $B$ and $C$ ).
Case 1: $B$ and $C$ shake hands with each other
There are $f_6$ ways for $D$ $E$ $F$ $G$ $H$ , and $I$ to shake hands
Case 2: $B$ and $C$ shake hands with one of the others out of the $6$ remaining people ( $6$ ways)
There are $f_5$ ways for $E$ $F$ $G$ $H$ , and $I$ to shake hands
Case 3: If $B$ and $C$ shake hands with two different people (there are 6 $\ast$ 5 ways to choose the people but WLOG assume they are $D$ and $E$ ). Calculations of Case 3 are in its subcases.
Subcase 3.1: $D$ and $E$ shake hands with each other
There are $f_4$ ways for $F$ $G$ $H$ , and $I$ to shake hands
Subcase 3.2: $D$ and $E$ each shake hands with one other person (there are $4$ ways to choose the person, WLOG let that be $F$ )
There are $f_3$ ways for $G$ $H$ , and $I$ to shake hands
Subcase 3.3: $D$ and $E$ each shake hands with two different people (there are 4 $\ast$ 3 ways, WLOG let that be $F$ and $G$ )
There are $2\ast1$ ways for $F$ and $G$ to then shake hands with $H$ and $I$ , and $H$ and $I$ will shake hands with each other.
We have
\begin{align*} 28(f_6+6\ast(f_5)+6\ast(5)\ast(f_4)+4\ast(f_3)+4\ast(3)\ast(2)\ast(1) &= 28(70+6\ast(12)+6\ast(5)\ast(3+4\ast(1)+4\ast(3)\ast(2)\ast(1) \\ &= 30016 \end{align*}
which gives us the answer of $\boxed{016}$


Answer: 154.0

In a high school with $500$ students, $40\%$ of the seniors play a musical instrument, while $30\%$ of the non-seniors do not play a musical instrument. In all, $46.8\%$ of the students do not play a musical instrument. How many non-seniors play a musical instrument?
$\textbf{(A) } 66 \qquad\textbf{(B) } 154 \qquad\textbf{(C) } 186 \qquad\textbf{(D) } 220 \qquad\textbf{(E) } 266$

Solution:
$60\%$ of seniors do not play a musical instrument. If we denote $x$ as the number of seniors, then \[\frac{3}{5}x + \frac{3}{10}\cdot(500-x) = \frac{468}{1000}\cdot500\]
\[\frac{3}{5}x + 150 - \frac{3}{10}x = 234\] \[\frac{3}{10}x = 84\] \[x = 84\cdot\frac{10}{3} = 280\]
Thus there are $500-x = 220$ non-seniors. Since 70% of the non-seniors play a musical instrument, $220 \cdot \frac{7}{10} = \boxed{154}$

Solution:
Let $x$ be the number of seniors, and $y$ be the number of non-seniors. Then \[\frac{3}{5}x + \frac{3}{10}y = \frac{468}{1000}\cdot500 = 234\]
Multiplying both sides by $10$ gives us \[6x + 3y = 2340\]
Also, $x + y = 500$ because there are 500 students in total.
Solving these system of equations give us $x = 280$ $y = 220$
Since $70\%$ of the non-seniors play a musical instrument, the answer is simply $70\%$ of $220$ , which gives us $\boxed{154}$

Solution: We can clearly deduce that $70\%$ of the non-seniors do play an instrument, but, since the total percentage of instrument players is $46.8\%$ , the non-senior population is quite low. By intuition, we can therefore see that the answer is around $\text{B}$ or $\text{C}$ . Testing both of these gives us the answer $\boxed{154}$

Solution:
We know that $40\%$ of the seniors play a musical instrument, and $30\%$ of the non-seniors do not. In addition, we know that the number of people who do not play a musical instrument is \[46.8\% \cdot 500 = 46.8 \cdot 5 = \frac{468}{2} = 234\] We can also conclude that $60\%$ of the seniors do not play an instrument, $70\%$ of the non seniors do play an instrument, and $500-234 = 266$ people do play an instrument.
We can now set up the following equations, where $s$ is the number of seniors and $n$ is the number of non-seniors: \[0.3n + 0.6s = 234\] \[0.7n + 0.4s = 266\] By elimination, we get $1.5n$ to be equal to $330$ . This means that $n = \frac{330}{1.5} = 220$ .
The answer is $70$ percent of $220$ . This is equal to \[0.7*220 = 7*22 = 154\] Therefore, the answer is $\boxed{154}$


Answer: 9.0

In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?
$\textbf{(A)}\hspace{.05in}6\qquad\textbf{(B)}\hspace{.05in}8\qquad\textbf{(C)}\hspace{.05in}9\qquad\textbf{(D)}\hspace{.05in}10\qquad\textbf{(E)}\hspace{.05in}12$

Solution:
$6$ are blue and green - $b+g=6$
$8$ are red and blue - $r+b=8$
$4$ are red and green- $r+g=4$
We can do trial and error. Let's make blue $5$ . That makes green $1$ and red $3$ because $6-5=1$ and $8-5=3$ . To check this, let's plug $1$ and $3$ into $r+g=4$ , which works. Now count the number of marbles - $5+3+1=9$ . So the answer is $\boxed{9}.$

Solution:
We already knew the facts: $6$ are blue and green, meaning $b+g=6$ $8$ are red and blue, meaning $r+b=8$ $4$ are red and green, meaning $r+g=4$ . Then we need to add these three equations: $b+g+r+b+r+g=2(r+g+b)=6+8+4=18$ . It gives us all of the marbles are $r+g+b = 18/2 = 9$ . So the answer is $\boxed{9}$ .
~LarryFlora

Solution:
We may draw three Venn diagrams to represent these three cases, respectively.
Screen Shot 2021-08-29 at 9.14.51 AM.png
Let the amount of all the marbles be $x$ , meaning $R+G+B = x$
The Venn diagrams give us the equation: $x = (x-6)+(x-8)+(x-4)$ .
So $x = 3x-18$ $x = 18/2 =9$ .
Thus, the answer is $\boxed{9}$ .
~LarryFlora

Solution: Since we know all but $8$ marbles in the jar are green, the jar must have at least $9$ marbles. Then we can just start from $C$ and keep going. If there are $9$ marbles total, there are $3$ red marbles $(9-6)$ $1$ green marble $(9-8)$ , and $5$ blue marbles $(9-4)$ . Since we assumed there were $9$ marbles and $3+1+5=9$ , the answer is $\boxed{9}$

Solution:
Let $x$ be the number of total marbles. There are $x – 6$ red marbles, $x – 8$ green marbles, and $x – 4$ blue marbles.
We can create an equation: $(x – 6)+(x – 8)+(x – 4)=x$ Solving, we get $x=9$ , which means the total number of marbles is $\boxed{9}$ .
-J.L.L
(Feel free to edit)

Solution: Let $x$ be the number of total marbles, $r$ be the number of red marbles, $g$ be the number of green marbles, and $b$ be the number of blue marbles. Then we have $x - r = 6$ $x - g = 8$ $x - b = 4$ , and $r + g + b = x$ . Adding the first three equations together, we get $3x - r - g - b = 18$ or $3x - (r + g + b) = 18$ . Substituting in the fourth equation, we have $3x - x = 18$ $\implies$ $\boxed{9}$


Answer: 200.0

In a magic square , the sum of the three entries in any row, column, or diagonal is the same value. The figure shows four of the entries of a magic square. Find $x$

Solution:
Let's make a table.
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\]
\begin{eqnarray*} x+19+96=x+1+c\Rightarrow c=19+96-1=114,\\ 114+96+a=x+1+114\Rightarrow a=x-95 \end{eqnarray*}
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&d&e\\\hline \end{array}\]
\begin{eqnarray*}19+x-95+d=x+d-76=115+x\Rightarrow  d=191,\\ 114+191+e=x+115\Rightarrow e=x-190 \end{eqnarray*} \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table in progress}}\\\hline x&19&96\\\hline 1&x-95&b\\\hline 114&191&x-190\\\hline \end{array}\]
\[3x-285=x+115\Rightarrow 2x=400\Rightarrow x=\boxed{200}\]

Solution:
Use the table from above. Obviously $c = 114$ . Hence $a+e = 115$ . Similarly, $1+a = 96 + e \Rightarrow a = 95+e$
Substitute that into the first to get $2e = 20 \Rightarrow e=10$ , so $a=105$ , and so the value of $x$ is just $115+x = 210 + 105 \Rightarrow x = \boxed{200}$

Solution:
\[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&d&e\\\hline \end{array}\] The formula \[e=\frac{1+19}{2}\] can be used. Therefore, $e=10$ . Similarly, \[96=\frac{1+d}{2}\] So $d=191$
Now we have this table: \[\begin{array}{|c|c|c|} \multicolumn{3}{c}{\text{Table}}\\\hline x&19&96\\\hline 1&a&b\\\hline c&191&10\\\hline \end{array}\] By property of magic squares, observe that \[x+a+10=19+a+191\] The $a$ 's cancel! We now have \[x+10=19+191\] Thus $x=\boxed{200}.$


Answer: 39.0

Solution:
Let the number in the top circle be $a$ and then $b$ $c$ $d$ $e$ , and $f$ , going in clockwise order. Then, we have \[S=a+b+c\] \[S=c+d+e\] \[S=e+f+a\]
Adding these equations together, we get
\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}
where the last step comes from the fact that since $a$ $b$ $c$ $d$ $e$ , and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$
The left hand side is divisible by $3$ and $75$ is divisible by $3$ , so $a+c+e$ must be divisible by $3$ .  The largest possible value of $a+c+e$ is then $15+14+13=42$ , and the corresponding value of $S$ is $\frac{75+42}{3}=39$ , which is choice $\boxed{39}$

Solution:
To make the sum the greatest, put the three largest numbers $(13,14$ and $15)$ in the corners. Then, balance the sides by putting the least integer $(10)$ between the greatest sum $(14$ and $15)$ . Then put the next least integer $(11)$ between the next greatest sum ( $13 +15$ ). Fill in the last integer $(12)$ and you can see that the sum of any three numbers on a side is (for example) $14 +10 + 15 = 39$ $\boxed{39}$ .
-by goldenn


Answer: 3.0

In a magical swamp there are two species of talking amphibians: toads, whose statements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements.
Brian: "Mike and I are different species."
Chris: "LeRoy is a frog."
LeRoy: "Chris is a frog."
Mike: "Of the four of us, at least two are toads."
How many of these amphibians are frogs?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution:
Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
Hence we must have one toad and $\boxed{3}$ frogs.

Solution:
Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$ , or $n=0, 1$ ). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.
Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{3}$


Answer: 3.0

Solution:
Start with Brian. If he is a toad, he tells the truth, hence Mike is a frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog.
As Mike is a frog, his statement is false, hence there is at most one toad.
As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toad.
Hence we must have one toad and $\boxed{3}$ frogs.

Solution:
Notice that one of Chris and LeRoy must be a frog: if Chris is a frog, then he lies about LeRoy being a frog. Hence LeRoy is a toad. Alternatively, if Chris is a toad, then he tells the truth about LeRoy being a frog.
Assume Brian is a toad. Then Mike is a frog, and he lies about at least two being toads. This means that none or one of the amphibians is a toad (the opposite of the statement $n\geq2$ is $n<2$ , or $n=0, 1$ ). However, this is absurd because we assumed Brian is a toad, and we know one of Chris and LeRoy is a toad. So our assumption leads to a contradiction.
Hence Brian must be a frog, and he and Mike are the same species. Mike is also a frog. One of Chris and LeRoy is a frog. There are $3$ frogs in total $\Longrightarrow \boxed{3}$


Answer: 7.0

In a mathematics contest with ten problems, a student gains 5 points for a correct answer and loses 2 points for an incorrect answer. If Olivia answered every problem and her score was 29, how many correct answers did she have?
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$

Solution: We can try to guess and check to find the answer. If she got five right, her score would be $(5*5)-(5*2)=15$ . If she got six right her score would be $(6*5)-(2*4)=22$ . That's close, but it's still not right! If she got 7 right, her score would be $(7*5)-(2*3)=29$ . Thus, our answer is $\boxed{7}$ . ~avamarora

Solution:
We can start with the full score, 50, and subtract not only 2 points for each incorrect answer but also the 5 points we gave her credit for. This expression is equivalent to her score, 29.
Let $x$ be the number of questions she answers correctly. Then, we will represent the number incorrect by $10-x$
\begin{align*} 50-7(10-x)&=29\\ 50-70+7x&=29\\ 7x&=49\\ x&=\boxed{7}

Solution: Suppose she got $x$ questions right. Then she got $10 - x$ questions wrong. Since she gains 5 points for a correct answer and loses 2 for an incorrect one, we can solve $5x - 2(10 - x) = 29$ to get that $x = \boxed{7}$


Answer: 25.0

In a new school $40$ percent of the students are freshmen, $30$ percent are sophomores, $20$ percent are juniors, and $10$ percent are seniors. All freshmen are required to take Latin, and $80$ percent of the sophomores, $50$ percent of the juniors, and $20$ percent of the seniors elect to take Latin. The probability that a randomly chosen Latin student is a sophomore is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We see that $40\% \cdot 100\% + 30\% \cdot 80\% + 20\% \cdot 50\% + 10\% \cdot 20\% = 76\%$ of students are learning Latin. In addition, $30\% \cdot 80\% = 24\%$ of students are sophomores learning Latin. Thus, our desired probability is $\dfrac{24}{76}=\dfrac{6}{19}$ and our answer is $6+19=\boxed{025}$

Solution: Assume that there are $100$ students in the school. There are $40$ freshmen taking Latin, $24$ sophomores taking Latin, $10$ juniors taking Latin, and $2$ seniors taking Latin. We get the probability to be the number of sophomores taking Latin over the total number of students taking Latin, or $\dfrac{24}{76}$ . Simplifying, we get $\dfrac{6}{19}$ . Adding, we get $\boxed{025}$


Answer: 358.0

In a parlor game, the magician asks one of the participants to think of a three digit number $(abc)$ where $a$ $b$ , and $c$ represent digits in base $10$ in the order indicated. The magician then asks this person to form the numbers $(acb)$ $(bca)$ $(bac)$ $(cab)$ , and $(cba)$ , to add these five numbers, and to reveal their sum, $N$ . If told the value of $N$ , the magician can identify the original number, $(abc)$ . Play the role of the magician and determine $(abc)$ if $N= 3194$

Solution:
Let $m$ be the number $100a+10b+c$ . Observe that $3194+m=222(a+b+c)$ so
\[m\equiv -3194\equiv -86\equiv 136\pmod{222}\]
This reduces $m$ to one of $136, 358, 580, 802$ . But also $a+b+c=\frac{3194+m}{222}>\frac{3194}{222}>14$ so $a+b+c\geq 15$ .
Recall that $a, b, c$ refer to the digits the three digit number $(abc)$ , so of the four options, only $m = \boxed{358}$ satisfies this inequality.

Solution:
As in Solution 1, $3194 + m \equiv 222(a+b+c) \pmod{222}$ , and so as above we get $m \equiv 136 \pmod{222}$ .
We can also take this equation modulo $9$ ; note that $m \equiv a+b+c \pmod{9}$ , so
\[3194 + m \equiv 222m \implies 5m \equiv 8 \implies m \equiv 7 \pmod{9}.\]
Therefore $m$ is $7$ mod $9$ and $136$ mod $222$ . There is a shared factor in $3$ in both, but the Chinese Remainder Theorem still tells us the value of $m$ mod $666$ , namely $m \equiv 358$ mod $666$ . We see that there are no other 3-digit integers that are $358$ mod $666$ , so $m = \boxed{358}$

Solution: Let $n=abc$ then \[N=222(a+b+c)-n\] \[N=222(a+b+c)-100a-10b-c=3194\] Since $0<100a+10b+c<1000$ , we get the inequality \[N<222(a+b+c)<N+1000\] \[3194<222(a+b+c)<4194\] \[14<a+b+c<19\] Checking each of the multiples of $222$ from $15\cdot222$ to $18\cdot222$ by subtracting $N$ from each $222(a+b+c)$ , we quickly find $n=\boxed{358}$

Solution:
The sum of the five numbers is $222(a+b+c)-100a-10b-c=122a+212b+221c=122(a+b+c)+9(10b+11c)=3194$ We can see that $3194 \equiv 8$ （mod $9$ ） and $122 \equiv 5$ （mod $9$ ） so we need to make sure that $a+b+c \equiv 7$ （mod $9$ ） by some testing. So we let $a+b+c=9k+7$
Then, we know that $1\leq a+b+c \leq 27$ so only $7,16,25$ lie in the interval
When we test $a+b+c=25, 10b+11c=16$ , impossible
When we test $a+b+c=16, 10b+11c=138, b=5,c=8,a=3$
When we test $a+b+c=7, 10b+11c=260$ , well, it's impossible
The answer is $\boxed{358}$ then


Answer: 1979.0

In a party with $1982$ people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else?

Solution:
We induct on $n$ to prove that in a party with $n$ people, there must be at least $(n-3)$ people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people $A, B, C$ , who do not know each other.)
Base case: $n = 4$ is obvious.
Inductive step: Suppose in a party with $k$ people (with $k \ge 4$ ), at least $(k-3)$ people know everyone else. Consider a party with $(k+1)$ people. Take $k$ of the people (leaving another person, $A$ , out) and apply the inductive step to conclude that at least $(k-3)$ people know everyone else in the $k$ -person group, $G$
Now suppose that everyone in the group $G$ knows each other. Then take $3$ of these people and $A$ to deduce that $A$ knows a person $B \in G$ , which means $B$ knows everyone else. Then apply the inductive step on the remaining $k$ people (excluding $B$ ) to find $(k-3)$ people out of them that know everyone else (including $B$ , of course). Then these $(k-3)$ people and $B$ , which enumerate $(k-2)$ people, know everyone else.
Suppose that there exist two people $B, C \in G$ who do not know each other. Because $k-3 \ge 1$ , there exist at least one person in $G$ , person $D$ , who knows everyone else in $G$ . Now, take $A, B, C, D$ and observe that because $B, C$ do not know each other, either $A$ or $D$ knows everyone else of $A, B, C, D$ (by the problem condition), so in particular $A$ and $D$ know each other. Then apply the inductive step on the remaining $k$ people (excluding $D$ ) to find $(k-3)$ people out of them that know everyone else (including $D$ , of course). Then these $(k-3)$ people and $D$ , which enumerate $(k-2)$ people, know everyone else.
This completes the inductive step and thus the proof of this stronger result, which easily implies that at least $1982 - 3 = \boxed{1979}$ people know everyone else.


Answer: 65.0

In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$ . Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$
$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$

Solution:
Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:
The diagram below shows one possible configuration of the four circles:  Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{65}.$


Answer: 18.0

In a recent basketball game, Shenille attempted only three-point shots and two-point shots.  She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots.  Shenille attempted $30$ shots.  How many points did she score?
$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$

Solution:
Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$ .  We know that $x+y=30$ , and we need to evaluate $3(0.2x) + 2(0.3y)$ , as we know that the three-point shots are worth $3$ points and that she made $20$ % of them and that the two-point shots are worth $2$ and that she made $30$ % of them.
Simplifying, we see that this is equal to $0.6x + 0.6y = 0.6(x+y)$ .  Plugging in $x+y=30$ , we get $0.6(30) = \boxed{18}$

Solution: The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{18}$ points. If we assume Shenille only attempts two-pointers, then she makes $0.3 \cdot 30 = 9$ shots, which are worth $9 \cdot 2 = 18$ points.


Answer: 18.0

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?
$\textbf{(A)}\ 12\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 36$

Solution:
(similar to Solution 1, however a slightly more obvious way)
Say that
x = # of 2-pt shots
y = # of 3-pt shots
Because the total number of shots is $30$ $x + y = 30$
However, Shenille was only successful on $20\%$ of the 3-pt shots, and $30\%$ of the 2-pt shots, so
$0.2x + 0.3y$ = #number of successful shots
For each successful shot, there is an associated number of points with it.
Therefore, $0.2(x)(3) + (0.3)(y)(2)$ = her score
this evaluates to $0.6 (x + y)$ = her score
$x + y$ is already determined to be 30, so her score is $0.6 (30) = \boxed{18}$


Answer: 149.0

In a rectangular array of points, with 5 rows and $N$ columns, the points are numbered consecutively from left to right beginning with the top row.  Thus the top row is numbered 1 through $N,$ the second row is numbered $N + 1$ through $2N,$ and so forth.  Five points, $P_1, P_2, P_3, P_4,$ and $P_5,$ are selected so that each $P_i$ is in row $i.$ Let $x_i$ be the number associated with $P_i.$ Now renumber the array consecutively from top to bottom, beginning with the first column.  Let $y_i$ be the number associated with $P_i$ after the renumbering.  It is found that $x_1 = y_2,$ $x_2 = y_1,$ $x_3 = y_4,$ $x_4 = y_5,$ and $x_5 = y_3.$ Find the smallest possible value of $N.$

Solution:
Let each point $P_i$ be in column $c_i$ . The numberings for $P_i$ can now be defined as follows. \begin{align*}x_i &= (i - 1)N + c_i\\ y_i &= (c_i - 1)5 + i \end{align*}
We can now convert the five given equalities. \begin{align}x_1&=y_2 & \Longrightarrow & & c_1 &= 5 c_2-3\\ x_2&=y_1 & \Longrightarrow & & N+c_2 &= 5 c_1-4\\ x_3&=y_4 & \Longrightarrow & & 2 N+c_3 &= 5 c_4-1\\ x_4&=y_5 & \Longrightarrow & & 3 N+c_4 &= 5 c_5\\ x_5&=y_3 & \Longrightarrow & & 4 N+c_5 &= 5 c_3-2 \end{align} Equations $(1)$ and $(2)$ combine to form \[N = 24c_2 - 19\] Similarly equations $(3)$ $(4)$ , and $(5)$ combine to form \[117N +51 = 124c_3\] Take this equation modulo 31 \[24N+20\equiv 0 \pmod{31}\] And substitute for N \[24 \cdot 24 c_2 - 24 \cdot 19 +20\equiv 0 \pmod{31}\] \[18 c_2 \equiv 2 \pmod{31}\]
Thus the smallest $c_2$ might be is $7$ and by substitution $N = 24 \cdot 7 - 19 = 149$
The column values can also easily be found by substitution \begin{align*}c_1&=32\\ c_2&=7\\ c_3&=141\\ c_4&=88\\ c_5&=107 \end{align*} As these are all positive and less than $N$ $\boxed{149}$ is the solution.

Solution:
If we express all the $c_i$ in terms of $N$ , we have \[24c_1=5N+23\] \[24c_2=N+19\] \[124c_3=117N+51\] \[124c_4=73N+35\] \[124c_5=89N+7\]
It turns out that there exists such an array satisfying the problem conditions if and only if \[N\equiv 149 \pmod{744}\]
In addition, the first two equation can be written $n = 5mod24$ , and chasing variables in the last three equation gives us $89n + 7 = 124e$ . With these two equations you may skip a lot of rewriting and testing. $\boxed{149}$ still appears as our answer.


Answer: 28.0

In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution:
Embed the tetrahedron in 4-space to make calculations easier.
Its vertices are $(1,0,0,0)$ $(0,1,0,0)$ $(0,0,1,0)$ $(0,0,0,1)$
To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$ $(\frac{1}{3}, \frac{1}{3},0, \frac{1}{3})$ $(\frac{1}{3},0, \frac{1}{3}, \frac{1}{3})$ $(0,\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$
The side length of the large tetrahedron is $\sqrt{2}$ by the distance formula.
The side length of the smaller tetrahedron is $\frac{\sqrt{2}}{3}$ by the distance formula.
Their ratio is $1:3$ , so the ratio of their volumes is $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$
$m+n = 1 + 27 = \boxed{28}$

Solution: Let the large tetrahedron be $ABCD$ , and the small tetrahedron be $WXYZ$ , with $W$ on $ABC$ $X$ on $BCD$ $Y$ on $ACD$ , and $Z$ on $ABD$ . Clearly, the two regular tetrahedrons are similar, so if we can find the ratio of the sides, we can find the ratio of the volumes. Let $AB=1$ , for our convenience. Dropping an altitude from $W$ to $BC$ , and calling the foot $M$ , we have $WM=XM=\frac{\sqrt3}{6}$ . Since $\cos\angle{WMX}=\cos\angle{AMX}=MX/AM=1/3$ . By Law of Cosines, we have $WX=\sqrt{1/12+1/12-2(1/12)(1/3)}=1/3$ . Hence, the ratio of the volumes is $\left(\frac{1}{3}\right)^3=1/27$ $m+n=1+27=\boxed{028}$


Answer: 45.0

In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$

Solution: WLOG, by scaling, that the hypotenuse has length 1. Let $\theta$ be an angle opposite from some leg. Then the two legs have length $\sin\theta$ and $\cos\theta$ respectively, so we have $2\sin\theta\cos\theta = 1^2$ . From trigonometry, we know that this equation is true when $\theta = 45^{\circ}$ , so our answer is $\boxed{45}$ and we are done.

Solution: Look at the options. Note that if $\textbf{(A)}$ is the correct answer, one of the acute angles of the triangle will measure to $15$ degrees. This implies that the other acute angle of the triangle would measure to be $75$ degrees, which would imply that $\textbf{(E)}$ is another correct answer. However, there is only one correct answer per question, so $\textbf{(A)}$ can't be a correct answer. Using a similar argument, neither $\textbf{(B)}$ $\textbf{(D)}$ , nor $\textbf{(E)}$ can be a correct answer. Since $\textbf{(C)}$ is the only answer choice left and there must be one correct answer, the answer must be $\boxed{45}$


Answer: 3.0

In a room, $2/5$ of the people are wearing gloves, and $3/4$ of the people are wearing hats. What is the minimum number of people in the room wearing both a hat and a glove?
$\textbf{(A)}\ 3 \qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 20$

Solution:
Let $x$ be the number of people wearing both a hat and a glove. Since the number of people wearing a hat or a glove must be whole numbers, the number of people in the room must be a multiple of (4,5), but we do lcm of 4,5 = 20. Since we are trying to find the minimum $x$ , we must use the smallest possible value for the number of people in the room. Similarly, we can assume that there are no people present who are wearing neither of the two items since this would unnecessarily increase the number of people in the room. Thus, we can say that there are $20$ people in the room, all of which are wearing at least a hat or a glove.
It follows that there are $\frac{2}{5}\cdot 20 = 8$ people wearing gloves and $\frac{3}{4}\cdot 20 = 15$ people wearing hats. Then by applying the Principle of Inclusion Exclusion (PIE), the total number of people in the room wearing either a hat or a glove or both is $8+15-x = 23-x$ . Since we know that this equals $20$ , it follows that $23-x = 20$ , which implies that $x=3$ . Thus, $\boxed{3}$ is the correct answer.


Answer: 5.0

In a round-robin tournament with 6 teams, each team plays one game against each other team, and each game results in one team winning and one team losing. At the end of the tournament, the teams are ranked by the number of games won. What is the maximum number of teams that could be tied for the most wins at the end of the tournament?
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

Solution:
The total number of games (and wins) in the tournament is $\frac{6 \times 5}{2}= 15$ . A six-way tie is impossible as this would imply each team has 2.5 wins, so the maximum number of tied teams is five. Here's a chart of 15 games where five teams each have 3 wins:
The "X's" are for when it is where a team is set against itself, which cannot happen. The chart says that Team 6 has lost all of its matches, which means that each of the other teams won against it. Then, alternating Wins and Losses were tried. It shows that it is possible for 5 teams to tie for the same amount of wins, which in this case is 3 wins.
Thus, the answer is $\boxed{5}$


Answer: 560.0

In a sequence of coin tosses, one can keep a record of instances in which a tail is immediately followed by a head, a head is immediately followed by a head, and etc. We denote these by TH HH , and etc. For example, in the sequence TTTHHTHTTTHHTTH of 15 coin tosses we observe that there are two HH , three HT , four TH , and five TT subsequences. How many different sequences of 15 coin tosses will contain exactly two HH , three HT , four TH , and five TT subsequences?

Solution:
Let's consider each of the sequences of two coin tosses as an operation instead; this operation takes a string and adds the next coin toss on (eg, THHTH HT THHTHT ). We examine what happens to the last coin toss. Adding HH or TT is simply an identity for the last coin toss, so we will ignore them for now. However, adding HT or TH switches the last coin. switches to three times, but switches to four times; hence it follows that our string will have a structure of THTHTHTH
Now we have to count all of the different ways we can add the identities back in. There are 5 TT subsequences, which means that we have to add 5 into the strings, as long as the new s are adjacent to existing s. There are already 4 s in the sequence, and since order doesn’t matter between different tail flips this just becomes the ball-and-urn argument. We want to add 5 balls into 4 urns, which is the same as 3 dividers; hence this gives ${{5+3}\choose3} = 56$ combinations. We do the same with 2 s to get ${{2+3}\choose3} = 10$ combinations; thus there are $56 \cdot 10 = \boxed{560}$ possible sequences.


Answer: 5.0

In a sequence of positive integers, each term after the second is the product of the previous two terms. The sixth term is $4000$ . What is the first term?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 10$

Solution: In this solution, we will use trial and error to solve. $4000$ can be expressed as $200 \times 20$ . We divide $200$ by $20$ and get $10$ , divide $20$ by $10$ and get $2$ , and divide $10$ by $2$ to get $\boxed{5}$ . No one said that they have to be in ascending order!

Solution: Consider the first term is $a$ and the second term is $b$ . Then, the following term will be $ab$ $ab^2$ $a^2b^3$ and $a^3b^5$ . Notice that $4000=2^5\times 5^3$ , then we obtain $a=\boxed{5}$ and $b=2$


Answer: 560.0

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:
1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.
If the rules are followed, in how many different orders can the eight targets be broken?

Solution:
Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.
From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$
Since the letter arrangements of $LLLMMRRR$ and the shooting orders have one-to-one correspondence, we count the letter arrangements: \[\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.\]


Answer: 8.0

In a sign pyramid a cell gets a "+" if the two cells below it have the same sign, and it gets a "-" if the two cells below it have different signs. The diagram below illustrates a sign pyramid with four levels. How many possible ways are there to fill the four cells in the bottom row to produce a "+" at the top of the pyramid?
[asy] unitsize(2cm); path box = (-0.5,-0.2)--(-0.5,0.2)--(0.5,0.2)--(0.5,-0.2)--cycle; draw(box); label("$+$",(0,0)); draw(shift(1,0)*box); label("$-$",(1,0)); draw(shift(2,0)*box); label("$+$",(2,0)); draw(shift(3,0)*box); label("$-$",(3,0)); draw(shift(0.5,0.4)*box); label("$-$",(0.5,0.4)); draw(shift(1.5,0.4)*box); label("$-$",(1.5,0.4)); draw(shift(2.5,0.4)*box); label("$-$",(2.5,0.4)); draw(shift(1,0.8)*box); label("$+$",(1,0.8)); draw(shift(2,0.8)*box); label("$+$",(2,0.8)); draw(shift(1.5,1.2)*box); label("$+$",(1.5,1.2)); [/asy]
$\textbf{(A) } 2 \qquad \textbf{(B) } 4 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16$

Solution:
You could just make out all of the patterns that make the top positive. In this case, you would have the following patterns:
+−−+, −++−, −−−−, ++++, −+−+, +−+−, ++−−, −−++. There are 8 patterns and so the answer is $\boxed{8}$

Solution:
The top box is fixed by the problem.
Choose the left 3 bottom-row boxes freely. There are $2^3=8$ ways.
Then the left 2 boxes on the row above are determined.
Then the left 1 box on the row above that is determined
Then the right 1 box on that row is determined.
Then the right 1 box on the row below is determined.
Then the right 1 box on the bottom row is determined, completing the diagram.
So the answer is $\boxed{8}$

Solution:
Let the plus sign represent 1 and the negative sign represent -1.
The four numbers on the bottom are $a$ $b$ $c$ , and $d$ , which are either 1 or -1.

Which means $ab^3c^3d$ = 1. Since $b$ and $c$ are either 1 or -1, $b^3 = b$ and $c^3 = c$ . This shows that $abcd$ = 1.
Therefore either $a$ $b$ $c$ , and $d$ are all positive or negative, or 2 are positive and 2 are negative.
There are 2 ways where $a$ $b$ $c$ , and $d$ are 1 (1, 1, 1, 1) and (-1, -1, -1, -1)
There are 6 ways where 2 variables are positive and 2 are negative: (1, 1, -1, -1), (1, -1, 1, -1), (-1, 1, 1, -1), (-1, -1, 1, 1), (-1, 1, -1, 1), and (-1, -1, 1, 1).
So the answer is $\boxed{8}$

Solution: The pyramid is built on the basic 3 blocks pattern: one above and two below. The basic pattern have four possible symbols and half of them have a $+$ on the above, half of them have a $-$ above. So, For the lowest layer with $4$ blocks, there are $2^4=16$ possible combination and half of them will lead a $+$ (or $-$ ) on the top. The answer is $16/2=\boxed{8}$


Answer: 36.0

In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?
$\textbf{(A) }15\qquad\textbf{(B) }36\qquad\textbf{(C) }45\qquad\textbf{(D) }48\qquad\textbf{(E) }66$

Solution: We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$ , the sum of the first $n-1$ triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$ , so the answer is $\boxed{36}$

Solution: First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$ . Thus, the total number of games must be divisible by $12$ . Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is $\boxed{36}$

Solution:
Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ .
We note that the answer is some number $x$ choose $2$ . This means the answer is in the form $x(x-1)/2$ . Since answer choice D gives $48 = x(x-1)/2$ , and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{36}$ is the only possible choice.

Solution: If there are $n$ players, the total number of games played must be $\binom{n}{2}$ , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$ , so the number of games played must also be divisible by $12$ . Finally, we notice that only $\boxed{36}$ satisfies both of these conditions.


Answer: 36.0

Solution: We know that the total amount of games must be the sum of games won by left and right handed players. Then, we can write $g = l + r$ , and since $l = 1.4r$ $g = 2.4r$ . Given that $r$ and $g$ are both integers, $g/2.4$ also must be an integer. From here we can see that $g$ must be divisible by 12, leaving only answers B and D. Now we know the formula for how many games are played in this tournament is $n(n-1)/2$ , the sum of the first $n-1$ triangular numbers. Now, setting 36 and 48 equal to the equation will show that two consecutive numbers must have a product of 72 or 96. Clearly, $72=8*9$ , so the answer is $\boxed{36}$

Solution: First, we know that every player played every other player, so there's a total of $\dbinom{n}{2}$ games since each pair of players forms a bijection to a game. Therefore, that rules out D. Also, if we assume the right-handed players won a total of $x$ games, the left-handed players must have won a total of $\dfrac{7}{5}x$ games, meaning that the total number of games played was $\dfrac{12}{5}x$ . Thus, the total number of games must be divisible by $12$ . Therefore leaving only answer choices B and D. Since answer choice D doesn't satisfy the first condition, the only answer that satisfies both conditions is $\boxed{36}$

Solution:
Let $r$ be the amount of games the right-handed won. Since the left-handed won $1.4r$ games, the total number of games played can be expressed as $(2.4)r$ , or $12/5r$ , meaning that the answer is divisible by 12. This brings us down to two answer choices, $B$ and $D$ .
We note that the answer is some number $x$ choose $2$ . This means the answer is in the form $x(x-1)/2$ . Since answer choice D gives $48 = x(x-1)/2$ , and $96 = x(x-1)$ has no integer solutions, we know that $\boxed{36}$ is the only possible choice.

Solution: If there are $n$ players, the total number of games played must be $\binom{n}{2}$ , so it has to be a triangular number. The ratio of games won by left-handed to right-handed players is $7:5$ , so the number of games played must also be divisible by $12$ . Finally, we notice that only $\boxed{36}$ satisfies both of these conditions.


Answer: 25.0

In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded $1$ point, the loser got $0$ points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points earned by each player were earned against the ten players with the least number of points. (In particular, each of the ten lowest scoring players earned half of her/his points against the other nine of the ten). What was the total number of players in the tournament?

Solution: Let us suppose for convenience that there were $n + 10$ players overall.  Among the $n$ players not in the weakest 10 there were $n \choose 2$ games played and thus $n \choose 2$ points earned.  By the givens, this means that these $n$ players also earned $n \choose 2$ points against our weakest 10.  Now, the 10 weakest players playing amongst themselves played ${10 \choose 2} = 45$ games and so earned 45 points playing each other.  Then they also earned 45 points playing against the stronger $n$ players.  Since every point earned falls into one of these categories, It follows that the total number of points earned was $2{n \choose 2} + 90 = n^2 - n + 90$ .  However, there was one point earned per game, and there were a total of ${n + 10 \choose 2} = \frac{(n + 10)(n + 9)}{2}$ games played and thus $\frac{(n + 10)(n + 9)}{2}$ points earned.  So we have $n^2 -n + 90 = \frac{(n + 10)(n + 9)}{2}$ so $2n^2 - 2n + 180 = n^2 + 19n + 90$ and $n^2 -21n + 90 = 0$ and $n = 6$ or $n = 15$ .  Now, note that the top $n$ players got $n(n - 1)$ points in total (by our previous calculation) for an average of $n - 1$ , while the bottom 10 got 90 points total, for an average of 9.  Thus we must have $n > 10$ , so $n = 15$ and the answer is $15 + 10 = \boxed{25}$

Solution: Suppose that there are $n$ players participating in the tournament. We break this up into a group of the weakest ten, and the other $n-10$ people. Note that the $10$ players who played each other generated a total of $\dbinom{10}{2} = 45$ points playing each other. Thus, they earned $45$ playing the $n-10$ other people. Thus, the $n-10$ people earned a total of $10(n-10)-45 = 10n-145$ points playing vs. this group of 10 people, and also earned a total of $10n-145$ playing against themselves. Since each match gives a total of one point, we must have that $\dbinom{n-10}{2}=10n-145$ . Expanding and simplifying gives us $n^2-41n+400=0$ . Thus, $n=16$ or $n=25$ . Note however that if $n=16$ , then the strongest $16$ people get a total of $16*10-145=15$ playing against the weakest $10$ who gained $45$ points vs them, which is a contradiction since it must be larger. Thus, $n=\boxed{25}$

Solution: Note that the total number of points accumulated must sum to ${p \choose 2} = \frac{p(p-1)}{2}$ . Say the number of people is $n$ . Consider the number of points gained when the 10 lowest scoring people play each other. The problem tells us that each of these 10 people must earn exactly half of the total number of points they will earn during the whole game. This implies that this group of 10 people must accumulate half their total combined points after they (the 10 people) all play each other, meaning they must earn the other half of their points by playing the $n-10$ stronger players. The problem also tells us that the $n-10$ people who aren't part of the losers group will earn half of their points by playing the $10$ losers.  Since the $n-10$ group and $10$ losers will earn half their points by playing each other, the sum of the number of points that they gain playing each other must then be half of the total amount of points earned by everyone in the game. Therefore, $\frac{p(p-1)}{4} = 10(p-10)$ . This equation is the same as above, and by the same logic, the answer is $n=\boxed{25}$


Answer: 24.0

In a tournament there are six teams that play each other twice. A team earns $3$ points for a win, $1$ point for a draw, and $0$ points for a loss. After all the games have been played it turns out that the top three teams earned the same number of total points. What is the greatest possible number of total points for each of the top three teams?
$\textbf{(A) }22\qquad\textbf{(B) }23\qquad\textbf{(C) }24\qquad\textbf{(D) }26\qquad\textbf{(E) }30$

Solution:
This isn't finished
to another. This gives equality, as each team wins once and loses once as well. For a win, we have $3$ points, so a team gets $3\times2=6$ points if they each win a game and lose a game. This case brings a total of $18+6=24$ points.
Therefore, we use Case 2  since it brings the greater amount of points, or $\boxed{24}$

Solution: We can name the top three teams as $A$ $B$ , and $C$ . We can see that (respective scores of) $A=B=C$ because these teams have the same points. If we look at the matches that involve the top three teams, we see that there are some duplicates: $AB$ $BC$ , and $AC$ come twice. In order to even out the scores and get the maximum score, we can say that in match $AB$ $A$ and $B$ each win once out of the two games that they play. We can say the same thing for $AC$ and $BC$ . This tells us that each team $A$ $B$ , and $C$ win and lose twice. This gives each team a total of $3 + 3 + 0 + 0 = 6$ points. Now, we need to include the other three teams. We can label these teams as $D$ $E$ , and $F$ . We can write down every match that $A, B,$ or $C$ plays in that we haven't counted yet: $AD$ $AD$ $AE$ $AE$ $AF$ $AF$ $BD$ $BD$ $BE$ $BE$ $BF$ $BF$ $CD$ $CD$ $CE$ $CE$ $CF$ , and $CF$ . We can say $A$ $B$ , and $C$ win each of these in order to obtain the maximum score that $A$ $B$ , and $C$ can have. If $A$ $B$ , and $C$ win all six of their matches, $A$ $B$ , and $C$ will have a score of $18$ $18 + 6$ results in a maximum score of $\boxed{24}$

Solution: To start, we calculate how many games each team plays. Each team can play against $5$ people twice, so there are $10$ games that each team plays. So the answer is $10\cdot 3$ which is $30!$ But wait... if we want $3$ teams to have the same amount of points, there can't possibly be a player who wins all their games. Let the top three teams be $A,B$ , and $C.$ $A$ plays $B$ and $C$ twice so in order to maximize the games being played, we can split it $50-50$ between the $4$ games $A$ plays against $B$ or $C$ . We find that we just subtract $2$ games or $6$ points. Therefore the answer is $30-6$ $24$ or $\boxed{24}$


Answer: 306.0

In a town of $351$ adults, every adult owns a car, motorcycle, or both. If $331$ adults own cars and $45$ adults own motorcycles, how many of the car owners do not own a motorcycle?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 306 \qquad \textbf{(E)}\ 351$

Solution: By PIE , the number of adults who own both cars and motorcycles is $331+45-351=25.$ Out of the $331$ car owners, $25$ of them own motorcycles and $331-25=\boxed{306}$ of them don't.

Solution: There are $351$ total adults, and $45$ own a motorcycle. The number of adults that don't own a motorcycle is $351 - 45 = 306$ . Since everyone owns a car or motorcycle and one who doesn't own a motorcycle owns a car, the answer is $\boxed{306}$

Solution: Clearly, we can eliminate answer choice $E$ , for at least one adult owns motorcycles. It is also fairly obvious that the answer must be in the 300 range, giving us $\boxed{306}$


Answer: 43.0

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is $15$ . What is the greatest possible perimeter of the triangle?
$\textbf{(A) } 43\qquad \textbf{(B) } 44\qquad \textbf{(C) } 45\qquad \textbf{(D) } 46\qquad \textbf{(E) } 47$

Solution:
Let $x$ be the length of the first side.
The lengths of the sides are: $x$ $3x$ , and $15$
By the Triangle Inequality
$3x < x + 15$
$2x < 15$
$x < \frac{15}{2}$
The greatest integer satisfying this inequality is $7$
So the greatest possible perimeter is $7 + 3\cdot7 + 15 =\boxed{43}$


Answer: 43.0

In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle?
$\text {(A) } 43 \qquad \text {(B) } 44 \qquad \text {(C) } 45 \qquad \text {(D) } 46 \qquad \text {(E) } 47$

Solution: If the second size has length x, then the first side has length 3x, and we have the third side which has length 15. By the triangle inequality, we have: \[\\ x+15>3x \Rightarrow 2x<15 \Rightarrow x<7.5\] Now, since we want the greatest perimeter, we want the greatest integer x, and if $x<7.5$ then $x=7$ . Then, the first side has length $3*7=21$ , the second side has length $7$ , the third side has length $15$ , and so the perimeter is $21+7+15=43 \Rightarrow \boxed{43}$


Answer: 60.0

In a triangle with sides of lengths $a$ $b$ , and $c$ $(a+b+c)(a+b-c) = 3ab$ . The measure of the angle opposite the side length $c$ is
$\textbf{(A)}\ 15^\circ\qquad\textbf{(B)}\ 30^\circ\qquad\textbf{(C)}\ 45^\circ\qquad\textbf{(D)}\ 60^\circ\qquad\textbf{(E)}\ 150^\circ$

Solution: \[(a+b+c)(a+b-c)=3ab\] \[a^2+2ab+b^2-c^2=3ab\] \[a^2+b^2-c^2=ab\] \[c^2=a^2+b^2-ab\] This looks a lot like Law of Cosines, which is $c^2=a^2+b^2-2ab\cos{c}$ \[c^2=a^2+b^2-ab=a^2+b^2-2ab\cos{c}\] \[ab=2ab\cos{c}\] \[\frac{1}{2}=\cos{c}\] $\cos{c}$ is $\frac{1}{2}$ , so the angle opposite side $c$ is $\boxed{60}$


Answer: 574.0

In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$ , respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$ . Suppose $XP=10$ $PQ=25$ , and $QY=15$ . The value of $AB\cdot AC$ can be written in the form $m\sqrt n$ where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$

Solution:
First we have $a\cos A=PQ=25$ , and $(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400$ by PoP. Similarly, $(a\cos A)(b\cos B)=15(10+25)=525,$ and dividing these each by $a\cos A$ gives $b\cos B=21,c\cos C=16$
It is known that the sides of the orthic triangle are $a\cos A,b\cos B,c\cos C$ , and its angles are $\pi-2A$ $\pi-2B$ , and $\pi-2C$ . We thus have the three sides of the orthic triangle now.
Letting $D$ be the foot of the altitude from $A$ , we have, in $\triangle DPQ$ \[\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.\] \[\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},\] similarly, we get \[\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.\] To finish, \[bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.\] The requested sum is $\boxed{574}$ .
- crazyeyemoody907

Solution:
Let $BC=a$ $AC=b$ , and $AB=c$ . Let $\cos\angle A=k$ . Then $AP=bk$ and $AQ=ck$
By Power of a Point theorem, \begin{align}     AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\     AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 \end{align} Thus $bck = (bk)^2+400=(ck)^2+525 = u$ . Then $bk=\sqrt{u-400}$ $ck=\sqrt{u-525}$ , and \[k=\sqrt{\frac{(u-400)(u-525)}{u^2}}\] Use the Law of Cosines in $\triangle APQ$ to get $25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3$ , which rearranges to \[775=bck - k^2\cdot bck  = u-\frac{(u-400)(u-525)}{u}\] Upon simplification, this reduces to a linear equation in $u$ , with solution $u=1400$ . Then \[AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}\] So the final answer is $560 + 14 = \boxed{574}$

Solution: Let $AP=p$ $PB=q$ $AQ=r$ , and $QC=s$ .  By Power of a Point, \begin{align}  AP\cdot PB=XP\cdot YP \quad &\Longrightarrow \quad pq=400\\  AQ\cdot QC=YQ\cdot XQ \quad &\Longrightarrow \quad rs=525  \end{align} Points $P$ and $Q$ lie on the circle, $\omega$ , with diameter $BC$ , and pow $(A,\omega) = AP\cdot AB = AQ\cdot AC$ , so \[p(p+q)=r(r+s)\quad \Longrightarrow \quad p^2-r^2=125\] Use Law of Cosines in $\triangle APQ$ to get $25^2=p^2+r^2-2pr\cos A$ ; since $\cos A = \frac r{p+q}$ , this simplifies as \[500 \ =\  2r^2-\frac{2pr^2}{p+q} \ =\  2r^2-\frac{2p^2r^2}{p^2+400} \ =\ \frac{800r^2}{r^2+525}\] We get $r=5\sqrt{35}$ and thus \[r=5\sqrt{35}, \quad p = \sqrt{r^2+125} = 10\sqrt{10}, \quad q = \frac{400}{p} =4\sqrt{10}, \quad s= \frac{525}{r} = 3\sqrt{35}.\] Therefore $AB\cdot AC = (p+q)\cdot(r+s) = 560\sqrt{14}$ . So the answer is $560 + 14 = \boxed{574}$

Solution:
This solution is directly based of @CantonMathGuy's solution.
We start off with a key claim.
Claim. $XB \parallel AC$ and $YC \parallel AB$
Proof.
Let $E$ and $F$ denote the reflections of the orthocenter over points $P$ and $Q$ , respectively. Since $EF \parallel XY$ and \[EF = 2 PQ = XP + PQ + QY = XY,\] we have that $E X Y F$ is a rectangle. Then, since $\angle XYF = 90^\circ$ we obtain $\angle XBF = 90^\circ$ (which directly follows from $XBYF$ being cyclic); hence $\angle XBQ = \angle AQB$ , or $XB \parallel AQ \Rightarrow XB \parallel AC$
Similarly, we can obtain $YC \parallel AB$ $\ \blacksquare$
A direct result of this claim is that $\triangle BPX \sim \triangle APQ \sim \triangle CYQ$
Thus, we can set $AP = 5k$ and $BP = 2k$ , then applying Power of a Point on $P$ we get $10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}$ . Also, we can set $AQ = 5l$ and $CQ = 3l$ and once again applying Power of a Point (but this time to $Q$ ) we get
$\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}$
Hence,
$\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}$
and the answer is $560 + 14 = \boxed{574}$ . ~rocketsri


Answer: 2.0

In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?
$\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$

Solution: Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\frac{h+d}s$ . In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$ . This is less than $h$ , therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{2}$


Answer: 43.0

In an All-Area track meet, $216$ sprinters enter a $100-$ meter dash competition. The track has $6$ lanes, so only $6$ sprinters can compete at a time. At the end of each race, the five non-winners are eliminated, and the winner will compete again in a later race. How many races are needed to determine the champion sprinter?
$\textbf{(A)}\mbox{ }36\qquad\textbf{(B)}\mbox{ }42\qquad\textbf{(C)}\mbox{ }43\qquad\textbf{(D)}\mbox{ }60\qquad\textbf{(E)}\mbox{ }72$

Solution:
From any $n-$ th race, only $\frac{1}{6}$ will continue on. Since we wish to find the total number of races, a column representing the races over time is ideal.
Starting with the first race: \[\frac{216}{6}=36\] \[\frac{36}{6}=6\] \[\frac{6}{6}=1\] Adding all of the numbers in the second column yields $\boxed{43}$

Solution: Every race eliminates $5$ players. The winner is decided when there is only $1$ runner left. You can construct the equation: $216$ $5x$ $1$ . Thus, $215$ players have to be eliminated. Therefore, we need $\frac{215}{5}$ games to decide the winner, or $\boxed{43}$

Solution: Since $216$ is a power of $6,$ the answer will be in the form of $6x+1.$ We can see that this is odd and the only option is $\boxed{43}$ ~sanaops9


Answer: 8.0

In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors and $10\%$ of the seniors are on the debate team. How many juniors are in the program?
$\textbf{(A)} ~5 \qquad\textbf{(B)} ~6 \qquad\textbf{(C)} ~8 \qquad\textbf{(D)} ~11 \qquad\textbf{(E)} ~20$

Solution:
Say there are $j$ juniors and $s$ seniors in the program. Converting percentages to fractions, $\frac{j}{4}$ and $\frac{s}{10}$ are on the debate team, and since an equal number of juniors and seniors are on the debate team, $\frac{j}{4} = \frac{s}{10}.$
Cross-multiplying and simplifying we get $5j=2s.$ Additionally, since there are $28$ students in the program, $j+s = 28.$ It is now a matter of solving the system of equations \[5j=2s\] \[j+s=28,\] and the solution is $j = 8, s = 20.$ Since we want the number of juniors, the answer is \[\boxed{8}.\]

Solution: We immediately see that $E$ is the only possible amount of seniors, as $10\%$ can only correspond with an answer choice ending with $0$ . Thus the number of seniors is $20$ and the number of juniors is $28-20=8\rightarrow \boxed{8}$ . ~samrocksnature

Solution: Since there are an equal number of juniors and seniors on the debate team, suppose there are $x$ juniors and $x$ seniors. This number represents $25\% =\frac{1}{4}$ of the juniors and $10\%= \frac{1}{10}$ of the seniors, which tells us that there are $4x$ juniors and $10x$ seniors. There are $28$ juniors and seniors in the program altogether, so we get \[10x+4x=28,\] \[14x=28,\] \[x=2.\] Which means there are $4x=8$ juniors on the debate team, $\boxed{8}$

Solution: The amount of juniors must be a multiple of $4$ , since exactly $\frac{1}{4}$ of the students are on the debate team. Thus, we can immediately see that $\boxed{8}$