Navvye/Olympiad-Problems-CSV huggingface dataset:
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: We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get $x_0$ $380$ and $y_0$ $1$ . then the general solution of the given diophanitine equation will be $x$ $x_0$ $3t$ and $y$ $y_0$ $2t$ . Since we need only positive integer solutions So we solve $380$ $3t$ $>$ $0$ and $1$ $2t$ $>$ $0$ to get $t$ $>$ $0$ (applying Greatest integer function) also we can clearly see that $t_{(min)}$ $=$ $0$ so,t $<$ $GIF$ $383$ $3$ ). That implies $t$ ranges from $0$ to $127$ . Hence,the correct answer is $127$ $\boxed{127}$

Answer: D 127
Problem: The dimensions of a rectangle $R$ are $a$ and $b$ $a < b$ . It is required to obtain a rectangle with dimensions $x$ and $y$ $x < a, y < a$ , 
so that its perimeter is one-third that of $R$ , and its area is one-third that of $R$ . The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$

Solution: Using the perimeter and area formulas, \[2(x+y) = \frac{2}{3}(a+b)\] \[x+y = \frac{a+b}{3}\] \[xy = \frac{ab}{3}\] Dividing the second equation by the last equation results in \[\frac1y + \frac1x = \frac1b + \frac1a\] Since $x,y < a$ $\tfrac1a < \tfrac1x, \tfrac1y$ .  Since $a < b$ $\tfrac1b < \tfrac1a$ .  That means \[\tfrac1x + \tfrac1y > \tfrac1a + \tfrac1a > \tfrac1a + \tfrac1b\] This is a contradiction, so there are $\boxed{0}$ rectangles that satisfy the conditions.

Answer: A 0
Problem: In $\triangle ABC$ , side $a = \sqrt{3}$ , side $b = \sqrt{3}$ , and side $c > 3$ . Let $x$ be the largest number such that the magnitude, 
in degrees, of the angle opposite side $c$ exceeds $x$ . Then $x$ equals:
$\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf{(D)}\ 90^{\circ} \qquad \textbf{(E)}\ 60^{\circ}$

Solution: Using the Law of Cosines \[\sqrt{3 + 3 - 2\cdot 3 \cdot \cos{x^\circ}}>3\]
Both sides are positive, so squaring both sides will not affect the inequality.
\[6 - 6 \cos{x^\circ}>9\] \[\cos{x^\circ} < -\frac{1}{2}\]
Note that $\cos{120^\circ} = -\frac{1}{2}$ .  As $x$ gets closer to $180^{\circ}$ $\cos{x}$ decreases towards $-1$ .  Thus, $x > 120$ , so the answer is $\boxed{120}$

Answer: B 120
Problem: The lengths of the sides of a triangle are integers, and its area is also an integer. 
One side is $21$ and the perimeter is $48$ . The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$

Solution: Let $b$ and $c$ be the other two sides of the triangle.  The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.
By Heron's Formula , the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$ .  Plug in the answer choices for $b$ and write the prime factorization of the product to make sure it is a perfect square.
Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$ , so $8$ does not work.  However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$ , so $10$ works.  The third side is $17$ , and the sides satisfy the Triangle Inequality , so the answer is $\boxed{10}$

Answer: B 10
Problem: Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$ $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$ . 
If $EF = 32$ and $GF = 24$ , then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$

Solution: Let $BE = x$ and $BC = y$ .  Since $AF \parallel BC$ , by AA Similarity, $\triangle AFE \sim \triangle CBE$ .  That means $\frac{AF}{CB} = \frac{FE}{BE}$ .  Substituting in values results in \[\frac{AF}{y} = \frac{32}{x}\] Thus, $AF = \frac{32y}{x}$ , so $FD = \frac{32y - xy}{x}$
In addition, $DC \parallel AB$ , so by AA Similarity, $\triangle FDG = \triangle FAB$ .  That means \[\frac{\frac{32y-xy}{x}}{\frac{32y}{x}} = \frac{24}{x+32}\] Cross multiply to get \[\frac{y(32-x)}{x} (x+32) = \frac{32y}{x} \cdot 24\] Since $x \ne 0$ and $y \ne 0$ \[(32-x)(32+x) = 32 \cdot 24\] \[32^2 - x^2 = 32 \cdot 24\] \[32 \cdot 8 = x^2\] Thus, $x = 16$ , which is answer choice $\boxed{16}$

Answer: E 16
Problem: The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$ , in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$

Solution: Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\implies y=2$
Substituting gives us $(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0$
Therefore, our answer is $\boxed{0}$

Answer: B 0
Problem: When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$

Solution: Obviously, we can factor out an $x$ first to get $x(x^8-1)$ .
Next, we repeatedly factor differences of squares: \[x(x^4+1)(x^4-1)\] \[x(x^4+1)(x^2+1)(x^2-1)\] \[x(x^4+1)(x^2+1)(x+1)(x-1)\] None of these 5 factors can be factored further, so the answer is $\boxed{5}$

Answer: B 5
Problem: When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$

Solution: This is equivalent to $\frac{(a-1)^6}{a^6}.$ Its expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$ .
By the Binomial Theorem, the sum of the last three coefficients is $\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10}$

Answer: C 10
Problem: $R$ varies directly as $S$ and inversely as $T$ . When $R = \frac{4}{3}$ and $T = \frac {9}{14}$ $S = \frac37$ . Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$

Solution: \[R=c\cdot\frac{S}T\]
for some constant $c$
You know that
\[\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,\]
so
\[c=\frac{4/3}{2/3}=2\,.\]
When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have
\[\sqrt{48}=\frac{2S}{\sqrt{75}}\,,\]
so
\[S=\frac12\sqrt{48\cdot75}=30\,.\] $\boxed{30}$

Answer: B 30
Problem: Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$ , as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$

Solution: The infinite sum of a geometric series with first term $a$ and common ratio $r$ $-1<r<1$ ) is $\frac{a}{1-r}$ .
Now, in this geometric series, $a=4$ , and $r=-\frac23$ . Plugging these into the formula, we get $\frac4{1-(-\frac23)}$ , which simplifies to $\frac{12}5$ , or $\boxed{2.4}$

Answer: B 2.4
Problem: If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$ $(0, 5)$ , and $(2, - 3)$ , the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$

Solution: Substituting in the $(x, y)$ pairs gives the following system of equations: \[a-b+c=12\] \[c=5\] \[4a+2b+c=-3\] We know $c=5$ , so plugging this in reduces the system to two variables: \[a-b=7\] \[4a+2b=-8\] Dividing the second equation by 2 gives $2a+b=-4$ , which can be added to the first equation to get $3a=3$ , or $a=1$ . So the solution set is $(1, -6, 5)$ , and the sum is $\boxed{0}$

Answer: C 0
Problem: The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$

Solution: If the angles are in an arithmetic progression, they can be expressed as $a$ $a+n$ $a+2n$ $a+3n$ , and $a+4n$ for some real numbers $a$ and $n$ .
Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$ .
Adding our expressions for the five angles together, we get $5a+10n=540$ .
We now divide by 5 to get $a+2n=108$ . It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108}$ .
(In fact, for any arithmetic progression with an odd number of terms,
the middle term is equal to the average of all the terms.)

Answer: A 108
Problem: It is given that one root of $2x^2 + rx + s = 0$ , with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$ . The value of $s$ is:
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$

Solution: If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.
This means the other root of the given quadratic is $\overline{3+2i}=3-2i$ . 
Now Vieta's formulas say that $s/2$ is equal to the product of the two roots, so $s = 2(3+2i)(3-2i) = \boxed{26}$

Answer: E 26
Problem: Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$ , one additional hour; and $R$ $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

Solution: Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.
We also know that all three working together take $x$ hours. 
Now the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \[x=\frac1{\frac1{x+6}+\frac1{x+1}+\frac1{2x}}\] \[\frac{x}{x+6}+\frac{x}{x+1}+\frac{x}{2x}=1\] \[\frac{x}{x+6}+\frac{x}{x+1}=\frac12\] \[2x(x+1)+2x(x+6)=(x+1)(x+6)\] \[2x^2+2x+2x^2+12x=x^2+7x+6\] \[3x^2+7x-6=0\] \[(3x-2)(x+3)=0\] \[x\in\{\frac23, -3\}\] Obviously, the number of hours is positive, so the answer is $\boxed{23}$

Answer: A 23
Problem: Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$ . The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$

Solution: Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$ . The length of $OE$ is $EF-OF=-x+8$ . By Pythagorean Theorem, $OA=OD=\sqrt{x^2+4^2}=\sqrt{x^2+16}$ . Because $OA$ $OD$ , and $OE$ are radii of the same circle, $-x+8=OE=OA=AD=\sqrt{x^2+16}$ . So, $\sqrt{x^2+16}=-x+8$ . Squaring both sides, we obtain $x^2+16=x^2-16x+64$ . Subtracting $x^2+16$ from both sides and adding $16x$ , our equation becomes $16x=80$ , so our answer is $x=\boxed{5}$

Answer: C 5
Problem: The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$ . How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$

Solution: The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$ .  Cross multiplying both sides, we have $2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0$ . Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24$ . Because $k$ and $r$ are both more than $2$ , we know that $6-k < r+4$ . Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$ $2$ $3$ , or $4$ and $r+4$ to $24$ $12$ $8$ , or $6$ respectively to get the following pairs for $(k, r)$ $(5, 20)$ $(4, 8)$ $(3, 4)$ , and $(2, 2)$ . However, we have to take out the solution with $k = 2$ , because $k$ and $r$ are both more than $2$ , leaving us with $\boxed{3}$ as the correct answer.

Answer: C 3
Problem: A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$ . Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$ . The number of minutes that he has been away is:
$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45$

Solution: Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$ . This is equivalent to \[180-\frac{11n}2=\pm110\] \[-\frac{11n}2\in\{-70,-290\}\] \[\frac{11n}2\in\{70,290\}\] \[11n\in\{140,580\}\] \[n\in\{\frac{140}{11},\frac{580}{11}\}\] The difference between the two values of $n$ is $\frac{440}{11}=\boxed{40}$

Answer: B 40
Problem: If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$

Solution: The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$ .  (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$ , for a total of $\boxed{2}$ solutions.

Answer: C 2
Problem: The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$

Solution: Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count 
We first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$ .
We then factor the right side getting $99$ $=$ $(b-a)(b+a)$ . 
Since we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$ .
We factor $99$ into $3^2 \cdot 11$ $=$ $(b-a)(b+a)$ There are a few cases to look at: $1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$ .
Adding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$ , which means $a$ $=$ $1$ .
But looking at the restriction that the second population + $100$ $=$ third population... $10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\neq$ a perfect square.
$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$ .
Adding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$ , which means $a$ $=$ $15$ .
Looking at the same restriction, we get $18^2$ $1$ $100$ $=$ $324$ $101$ $=$ $425$ , which is NOT a perfect square.
Finally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$ $2b$ $=$ $100$ or $b$ $=$ $50$ , which means $a$ $=$ $49$ .
Looking at the same restriction, we get $50^2$ $1$ $100$ $=$ $2500$ $101$ $=$ $2601$ $=$ $51^2$ . Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$ . Or $a^2$ is a multiple of $\boxed{7}$

Answer: B 7
Problem: The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$

Solution: Let $P$ $=$ original population. Translating the word problem into a system of equations, we got: \begin{align} P &= x^2 \\ P + 100 &= y^2 + 1 \\ P + 200 &= z^2 \end{align} for some positive integers $x$ $y$ and $z$ .
Now, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$ ), we got: \begin{align*} 100 &= z^2 - y^2 - 1 \\ 101 &= z^2 - y^2 \\ 101 &= (z - y)(z + y) \end{align*} Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$ .
Plugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$ , a multiple of $7$ .
Therefore, the answer is $\boxed{7}$ . -nullptr07

Answer: B 7
Problem: An automobile travels $a/6$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in $3$ minutes?
$\textbf{(A)}\ \frac{a}{1080r}\qquad \textbf{(B)}\ \frac{30r}{a}\qquad \textbf{(C)}\ \frac{30a}{r}\qquad \textbf{(D)}\ \frac{10r}{a}\qquad \textbf{(E)}\ \frac{10a}{r}$

Solution: Use dimensional analysis. \[\frac{a/6 \text{ feet}}{r \text{ seconds}} \cdot \frac{1 \text{ yard}}{3 \text{ feet}} \cdot \frac{60 \text{ seconds}}{1 \text{ minute}} \cdot 3 \text{ minutes}\] \[\frac{10a}{r} \text{ yards}\] The answer is $\boxed{10}$

Answer: E 10
Problem: Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$ , 
and $E$ is the midpoint of $AD$ . The length of $BE$ , in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$

Solution: [asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label("$B$",(0,0),SW); dot((50,0)); label("$C$",(50,0),SE); dot((25,43.301)); label("$A$",(25,43.301),N); dot((25,0)); label("$D$",(25,0),S); dot((25,21.651)); label("$E$",(25,21.651),E); draw((25,21.651)--(0,0));  label("$12$",(10,25)); label("$6$",(12.5,-5)); label("$6$",(37.5,-5)); label("$12$",(40,25)); draw((25,3)--(28,3)--(28,0)); label("$3\sqrt{3}$",(30.5,11)); [/asy] Note that $\triangle ABC$ is an equilateral triangle .  From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$ .  That means $DE = 3\sqrt{3}$ .  Using the Pythagorean Theorem again, $BE = \sqrt{63}$ , which is answer choice $\boxed{63}$

Answer: D 63
Problem: Two tangents are drawn to a circle from an exterior point $A$ ; they touch the circle at points $B$ and $C$ respectively. 
A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$ , and touches the circle at $Q$ . If $AB=20$ , then the perimeter of $\triangle APR$ is
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 39\frac{7}{8} \qquad \textbf{(E)}\ \text{not determined by the given information}$

Solution: Since $Q$ can be anywhere on the circle between $A$ and $B$ , it can basically be "on top" of $A$ . Then $R$ will be at the same point as $A$ , so $APR$ form a degenerate triable with side length $AB=20$ . So its perimeter will be $40$ . Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in length, $PR=PQ+QR$ will "grow" to compensate, so the perimeter will stay constant with a value of $\boxed{40}$

Answer: C 40
Problem: In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$ . 
In the Land of Mathesis, however, numbers are written in the base $r$ . 
Jones purchases an automobile there for $440$ monetary units (abbreviated m.u). 
He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The base $r$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 12$

Solution: If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then \[440_r + 340_r = 1000_r\] This equation can be rewritten as \[4r^2 + 4r + 3r^2 + 4r = r^3\] Bring all of the terms to one side to get \[r^3 - 7r^2 - 8r = 0\] Factor to get \[r(r-8)(r+1)=0\] Since base numbers are always positive, base $r$ is $8$ , which is answer choice $\boxed{8}$

Answer: D 8
Problem: The yearly changes in the population census of a town for four consecutive years are, 
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease. 
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$

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

Answer: A 12
Problem: Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$ , 
and $Q$ divides $AB$ in the ratio $3:4$ . If $PQ=2$ , then the length of $AB$ is:
$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$

Solution: [asy] draw((0,0)--(70,0)); dot((0,0)); dot((28,0)); dot((30,0)); dot((70,0)); label("2",(29,0),N); label("x",(14,0),N); label("y",(50,0),N); label("A",(0,0),S); label("P",(28,0),SW); label("Q",(30,0),SE); label("B",(70,0),S); [/asy]
Draw diagram as shown, where $P$ and $Q$ are on the same side.  Let $AP = x$ and $QB = y$
Since $P$ divides $AB$ in the ratio $2:3$ $\frac{x}{y+2} = \frac{2}{3}$ .  Since $Q$ divides $AB$ in the ratio $3:4$ $\frac{x+2}{y} = \frac{3}{4}$ .  Cross multiply to get a system of equations \[3x = 2y+4\] \[4x+8=3y\] Solve the system to get $x = 28$ and $y = 40$ .  Thus, $AB = 40 + 28 + 2 = 70$ , so the answer is $\boxed{70}$

Answer: B 70
Problem: For a given arithmetic series the sum of the first $50$ terms is $200$ , and the sum of the next $50$ terms is $2700$ . 
The first term in the series is:
$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$

Solution: Let the first term of the arithmetic sequence be $a$ and the common difference be $d$
The $50^{\text{th}}$ term of the sequence is $a+49d$ , so the sum of the first $50$ terms is $\frac{50(a + a + 49d)}{2}$
The $51^{\text{th}}$ term of the sequence is $a+50d$ and the $100^{\text{th}}$ term of the sequence is $a+99d$ , so the sum of the next $50$ terms is $\frac{50(a+50d+a+99d)}{2}$
Substituting in values results in this system of equations \[2a+49d=8\] \[2a+149d=108\] Solving for $a$ yields $a = \frac{-41}{2}$ .  The first term is $-20.5$ , which is answer choice $\boxed{20.5}$

Answer: C 20.5
Problem: If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$

Solution: $7^1$ has a unit digit of $7$ $7^2$ has a unit digit of $9$ $7^3$ has a unit digit of $3$ $7^4$ has a unit digit of $1$ $7^5$ has a unit digit of $7$
Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$ .  It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit.  Since $753$ leaves a remainder of $1$ after being divided by $4$ , the units digit of $2137^{753}$ is $7$ , which is answer choice $\boxed{7}$

Answer: D 7