Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Given a square side of length $s$ . On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$

Solution: The area of the square is $s^2$ .  The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\sqrt{2}$
The area of the triangle is $\frac{1}{2}bh$ .  The base $b$ of the triangle is the diagonal of the square, which is $b = s\sqrt{2}$ .  If the area of the triangle is equal to the area of the square, we have:
$s^2 = \frac{1}{2}bh$
$s^2 = \frac{1}{2}s\sqrt{2}\cdot h$
$s = \frac{\sqrt{2}}{2}h$
$h = \frac{2}{\sqrt{2}}s$
$h = s\sqrt{2}$
This is option $\boxed{2}$

Answer: A 2
Problem: Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$ , find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$

Solution: Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$ , we have:
$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$
Note that if $a^b = a^c$ , then $b=c$ .  Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$ .  Plugging the first equation into the second equation gives:
$2y = (3y + 3) - 9$
$2y = 3y - 6$
$0 = y - 6$
$y = 6$
Plugging that back in to $x = 3y + 3$ gives $x = 3(6) + 3$ , or $x = 21$ .  Thus, $x+y = 21 + 6$ , or $x+y=27$ , which is option $\boxed{27}$

Answer: D 27
Problem: A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep. 
The remaining $49$ sheep were sold at the same price per head as the other $700$ . 
Based on the cost, the percent gain on the entire transaction is:
$\textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$

Solution: Let us say each sheep cost $x$ dollars.  The farmer paid $749x$ for the sheep.  He sold $700$ of them for $749x$ , so each sheep sold for $\frac{749}{700} = 1.07x$
Since every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$ , there is an increase of $\frac{1.07x - 1x}{1x} = 0.07$ , or $7\%$ , which is option $\boxed{7}$

Answer: C 7
Problem: Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ . 
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$

Solution: Note that for all polynomials $f(x)$ $f(x + 6) \equiv f(x) \pmod 6$
Proof:
If $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$ .  In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\pmod 6$ .  This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , which is $f(x)$ , so $f(x+6) \equiv f(x) \pmod 6$
So, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$ .  The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets.  The remainders are $2, 0, 0, 2, 0, 0$
This means for $s=1, 2, 4, 5$ $f(s)$ is divisible by $6$ .  Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$ , which is $5$ values of $s$ that work.  Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$ , which is $4$ more values of $s$ that work.  The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$ .  This is a total of $17$ values of $s$ , for an answer of $\boxed{17}$

Answer: E 17
Problem: Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ . 
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$

Solution: $f(x)$ $x^2$ $3x$ $2$ is $0$ congruent modulo 6 that implies $x+1$ or/and $x+2$ is $0$ congruent modulo $6$ .The numbers are of the form $6k+5$ and $6k+4$ for some integer $k$ and due to restrictions of number of elements in the set $S$ we get the inequality $k<4$ .Then we calculate the possible combinations to get $17$ as answer i.e. option $\boxed{17}$

Answer: E 17
Problem: The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$

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

Answer: B 1
Problem: Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$ . The product of the two numbers is:
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$

Solution: Set the two numbers as $x$ and $y$ . Therefore, $x+y=7(x-y),  xy=24(x-y)$ , and $24(x+y)=7xy$ . Simplifying the first equation gives $y=\frac{3}{4}x$ . Substituting for $y$ in the second equation gives $\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$ . Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$ . Substituting into $y=\frac{3}{4}x$ gives $y=6$ and $xy=\boxed{48}$

Answer: D 48
Problem: The sum of $n$ terms of an arithmetic progression is $153$ , and the common difference is $2$ . 
If the first term is an integer, and $n>1$ , then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$

Solution: Let the progression start at $a$ , have common difference $2$ , and end at $a + 2(n-1)$
The average term is $\frac{a + (a + 2(n-1))}{2}$ , or $a + n - 1$ .  Since the number of terms is $n$ , and the sum of the terms is $153$ , we have:
$n(a+n-1) = 153$
Since $n$ is a positive integer, it must be a factor of $153$ .  This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities.  We are given $n>1$ , leaving the other five factors.
We now must check if $a$ is an integer.  We have $a = \frac{153}{n} + 1 - n$ .  If $n$ is a factor of $153$ , then $\frac{153}{n}$ will be an integer.  Adding $1-n$ wil keep it an integer.
Thus, there are $5$ possible values for $n$ , which is answer $\boxed{5}$

Answer: D 5
Problem: The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$ , in inches, is:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$

Solution: By the Median Formula $PM = \frac12\sqrt{2PQ^2+2PR^2-QR^2}$
Plugging in the numbers given in the problem, we get \[\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2}\]
Solving, \[7=\sqrt{2(16)+2(49)-QR^2}\] \[49=32+98-QR^2\] \[QR^2=81\] \[QR=9=\boxed{9}\]

Answer: D 9
Problem: let $n=x-y^{x-y}$ . Find $n$ when $x=2$ and $y=-2$
$\textbf{(A)}\ -14 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 256$

Solution: Substitute the variables to determine the value of $n$ \[n = 2 - (-2)^{2-(-2)}\] \[n = 2 - (-2)^4\] \[n = 2 - 16\] \[n = -14\] The answer is $\boxed{14}$

Answer: A 14
Problem: In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:
$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$

Solution: By the Binomial Theorem , each term of the expansion is $\binom{7}{n}(a)^{7-n}(\frac{-1}{\sqrt{a}})^n$
We want the exponent of $a$ to be $-\frac{1}{2}$ , so \[(7-n)-\frac{1}{2}n=-\frac{1}{2}\] \[-\frac{3}{2}n = -\frac{15}{2}\] \[n = 5\]
If $n=5$ , then the corresponding term is \[\binom{7}{5}(a)^{2}(\frac{-1}{\sqrt{a}})^5\] \[-21a^{-\frac{1}{2}}\]
The answer is $\boxed{21}$

Answer: C 21
Problem: The arithmetic mean of a set of $50$ numbers is $38$ . If two numbers of the set, namely $45$ and $55$ , are discarded, 
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$

Solution: If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ .  Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ .  Therefore, the new mean is $37.5$ , which is answer choice $\boxed{37.5}$

Answer: B 37.5
Problem: Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite. 
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$

Solution: [asy]  import graph; size(7.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=10.2,ymin=-6.2,ymax=5.2;  pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0);   /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs);  Label laxis; laxis.p=fontsize(10);  xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);  dot((-3,-2)); label("P",(-3,-2),NW); dot((1,-5)); label("Q",(1,-5),NW); dot((9,1)); label("R",(9,1),NW);  [/asy] Graph the three points on the coordinate grid.  Noting that the opposite sides of a parallelogram are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$ .  The sum of the x-coordinates and y-coordinates is $9$ , so the answer is $\boxed{9}$

Answer: E 9
Problem: If $2^a+2^b=3^c+3^d$ , the number of integers $a,b,c,d$ which can possibly be negative, is, at most:
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$

Solution: Assume $c,d \ge 0$ , and WLOG, assume $a<0$ and $a \le b$ .  This also takes into account when $b$ is negative.  That means \[\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\] Multiply both sides by $2^{-a}$ to get \[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\] Note that both sides are integers.  If $a \ne b$ , then the right side is even while the left side is odd, so equality can not happen.  If $a = b$ , then $2^{-a} (3^c + 3^d) = 2$ , and since $a<0$ $a = -1$ and $3^c + 3^d = 1$ .  No nonnegative value of $c$ and $d$ works, so equality can not happen.  Thus, $a$ and $b$ can not be negative when $c,d \ge 0$
Assume $a,b \ge 0$ , and WLOG, assume $c < 0$ and $c \le d$ .  This also takes into account when $d$ is negative.  That means \[2^a + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $3^{-c}$ to get \[3^{-c} (2^a + 2^b) = 1 + 3^{d-c}\] That makes both sides integers.  The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen.  Thus, $c$ and $d$ can not be negative when $a,b \ge 0$
Assume $a,c < 0$ , and WLOG, let $a \le b$ and $c \le d$ .  This also takes into account when $b$ or $d$ is negative.  That means \[\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $2^{-a} \cdot 3^{-c}$ to get \[3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})\] That makes both sides integers.  The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen.  Thus, $a$ and $c$ can not be negative.
Putting all the information together, none of $a,b,c,d$ can be negative, so the answer is $\boxed{0}$

Answer: E 0
Problem: Three numbers $a,b,c$ , none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results 
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$

Solution: Let $d$ be the common difference of the arithmetic sequence , so $a = b-d$ and $c = b+d$
Since increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence \[\frac{b}{b-d+1} = \frac{b+d}{b}\] \[\frac{b}{b-d} = \frac{b+d+2}{b}\] Cross-multiply in both equations to get a system of equations \[b^2 = b^2 - d^2 + b + d\] \[b^2 = b^2 - d^2 + 2b - 2d\] Rearranging terms results in \[d^2 = b+d\] \[d^2 = 2b-2d\] Substitute and solve for $d$ \[b+d = 2b-2d\] \[d = \frac{b}{3}\] Finally, substitute $d$ and solve for $b$ .  Since $b \ne 0$ , dividing by $b$ is allowed. \[(\frac{b}{3})^2 = b + \frac{b}{3}\] \[\frac{b^2}{9} = \frac{4b}{3}\] \[\frac{b}{9} = \frac{4}{3}\] \[b = 12\] The answer is $\boxed{12}$

Answer: C 12
Problem: In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red. 
Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 200 \qquad \textbf{(D)}\ 180 \qquad \textbf{(E)}\ 175$

Solution: The desired percentage of red balls is more than $90$ percent, so write an inequality
\[\frac{49+7x}{50+8x} \ge 0.9\]
Since $x >0$ , the sign does not need to be swapped after multiplying both sides by $50+8x$
\[49+7x \ge 45+7.2x\] \[4 \ge 0.2x\] \[20 \ge x\]
Thus, up to $20$ batches of balls can be used, so a total of $20 \cdot 8 + 50 = 210$ balls can be counted while satisfying the requirements.  The answer is $\boxed{210}$

Answer: B 210
Problem: Two men at points $R$ and $S$ $76$ miles apart, set out at the same time to walk towards each other. 
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant 
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the second hour,
and so on, in arithmetic progression. If the men meet $x$ miles nearer $R$ than $S$ in an integral number of hours, then $x$ is:
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 2$

Solution: First, find the number of hours it takes for the two to meet together.  After $h$ hours, the person at $R$ walks $4.5h$ miles.  In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\frac{h(6.5+0.5(h-1))}{2}$ miles.
In order for both to meet, the sum of both of the distances walked must total $76$ miles, so \[4.5h + \frac{h(6+0.5h)}{2} = 76\] \[4.5h + 3h + 0.25h^2 = 76\] \[0.25h^2 + 7.5h - 76 = 0\] \[h^2 + 30h - 304 = 0\] \[(h + 38)(h - 8) = 0\] Since $h$ must be positive, $h = 8$ .  Because it takes $8$ hours to meet, the person from $R$ traveled $36$ miles while the person from $S$ traveled $40$ miles.  Thus, they are $4$ miles closer to $R$ than $S$ , so the answer is $\boxed{4}$

Answer: D 4
Problem: A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$

Solution: Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.
After $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.
After $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cents, and $C$ has $4c$ cents.
After $C$ gave his money away, $A$ has $4a-4b-4c$ cents, $B$ has $-2a+6b-2c$ cents, and $C$ has $-a-b+7c$ cents.
Since all of them have $16$ cents in the end, we can write a system of equations \[4a-4b-4c=16\] \[-2a+6b-2c=16\] \[-a-b+7c=16\] Note that adding the three equation yields $a+b+c=48$ , so $4a+4b+4c=192$ .  Therefore, $8a=208$ , so $a = 26$ .  Solving for $a$ can also be done traditionally.
Thus, $A$ started out with $26$ cents, which is answer choice $\boxed{26}$

Answer: B 26
Problem: A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$

Solution: We know that people not giving away money on the previous turn now have twice what they previously had. Using the fact that the sum of their money is $48$ cents, we can work backward \[16,16,16\] \[8,8,32\] \[4,28,16\] \[26,14,8\] Thus at the beginning $A$ has $26\boxed{26}$ cents.

Answer: B 26
Problem: Consider equations of the form $x^2 + bx + c = 0$ . How many such equations have real roots and have coefficients $b$ and $c$ selected 
from the set of integers $\{1,2,3, 4, 5,6\}$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 19 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 16$

Solution: The discriminant of the quadratic is $b^2 - 4c$ .  Since the quadratic has real roots, \[b^2 - 4c \ge 0\] \[b^2 \ge 4c\] If $b = 6$ , then $c$ can be from $1$ to $6$ .  If $b = 5$ , then $c$ can also be from $1$ to $6$ .  If $b=4$ , then $c$ can be from $1$ to $4$ .  If $b=3$ , then $c$ can be $1$ or $2$ .  If $b=2$ , then $c$ can only be $1$ .  If $b = 1$ , no values of $c$ in the set would work.
Thus, there are a total of $19$ equations that work.  The answer is $\boxed{19}$

Answer: B 19
Problem: Point $F$ is taken in side $AD$ of square $ABCD$ . At $C$ a perpendicular is drawn to $CF$ , meeting $AB$ extended at $E$ . 
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$

Solution: Because $ABCD$ is a square $DC = CB = 16$ $DC \perp DA$ , and $CB \perp BA$ .  Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$ $\angle DCF = \angle BCE$ .  Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$
From the congruency, $CF = CE$ .  Using the area formula for a triangle, $CE = 20$ .  Finally, by the Pythagorean Theorem $BE = \sqrt{20^2 - 16^2} = 12$ , which is answer choice $\boxed{12}$

Answer: A 12
Problem: Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$ , and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

Solution: Statement $1$ states that $p$ is true and $q$ is false.  Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false.  This means that $(p \rightarrow q) \rightarrow X$ , where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true.  So statement $1$ implies the truth of the given statement.
Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.
Statement $2$ states that $p$ and $q$ are both false.  This in turn means that $p \rightarrow q$ is true.  Since $r$ is also true from statement $2$ , this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$ .  Thus statement $2$ implies the truth of the given statement.
Statement $4$ states that $p$ is false and $q$ is true.  In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false.  And, since $r$ is also true from statement $4$ , this means $(p \rightarrow q) \rightarrow r$ is true.  Thus, statement $4$ implies the truth of the given statement.
All four statements imply the truth of the given statement, so the answer is $\boxed{4}$

Answer: E 4
Problem: Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

Solution: The first line divides the plane into two regions.  The second line intersects one line, creating two regions.  The third line intersects two lines, creating three regions.  Similarly, the fourth line intersects three lines and creates four regions, the fifth line intersects four lines and creates five regions, and the sixth line intersects five lines and creates six regions.
Totaling the regions created results in $2 + 2 + 3 + 4 + 5 + 6 = 22$ regions, which is answer choice $\boxed{22}$

Answer: C 22
Problem: Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

Solution: 1963 AHSME Problem 27.png
With careful drawing, one can draw all six lines and count the regions.  There are $22$ regions in total, which is answer choice $\boxed{22}$

Answer: C 22
Problem: Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$

Solution: We can use the fact that the number of regions that $n$ lines divide a plane is given by the equation $L_n = \frac{n^2 + n +2}{2}$ , and in this problems, $n=6$ , from which the answer is $\boxed{22}$

Answer:  22
Problem: A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s = 160 t - 16t^2$ . The highest elevation is:
$\textbf{(A)}\ 800 \qquad \textbf{(B)}\ 640\qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 320 \qquad \textbf{(E)}\ 160$

Solution: The highest elevation a particle can reach is the vertex of the quadratic.  The x-value that can get the maximum is $\frac{-160}{-2 \cdot 16} = 5$ , so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\boxed{400}$

Answer: C 400
Problem: The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$

Solution: Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$ .  Solving the inequality results in $y \le 254 \frac{1}{3}$ .  From the two conditions, $y$ can be an odd number from $1$ to $253$ , so there are $127$ solutions where $x$ and $y$ are integers.  The answer is $\boxed{127}$

Answer: D 127