Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A regular polygon of $n$ sides is inscribed in a circle of radius $R$ . The area of the polygon is $3R^2$ . Then $n$ equals:
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$

Solution: Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$ .  Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.
That means each of these angles between the two lines from the center to the two vertices of a side equals $\frac{360}{n}$ degrees.  Thus, the area of the polygon is \[n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2\] Dividing both sides by $R^2$ yields \[\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3\] Multiply both sides by $\frac{2}{n}$ to get \[\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}\] At this point, use trial-and-error for each of the answer choices.  When checking $n = 12$ , the equation results in $\sin(30^{\circ}) = \frac{1}{2}$ , which is correct.  Thus, the answer is $\boxed{12}$

Answer: C 12
Problem: The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$ . Find $a_4$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$

Solution: This problem can be approached similarly to other base number problems.
Since $120 < 695 < 720$ , divide $695$ by $120$ .  The quotient is $5$ and the remainder is $95$ , so rewrite the number as \[695 = 5 \cdot 120 + 95\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$ , so the number can be rewritten as \[695 = 5 \cdot 120 + 3 \cdot 24 + 23\] Repeat the steps to get \[695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1\] The answer is $\boxed{3}$ .  One can also stop at the second step by noting $23 < 24$

Answer: D 3
Problem: If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$

Solution: Substitute $2$ for $x$ . We are given that this equation is true. Solving for $h$ gives $h=-9$ . The answer is $\boxed{9}$

Answer: E 9
Problem: Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$ % and two successive discounts of $36$ % and $4$ %, 
expressed in dollars, is:
$\textbf{(A)}0\qquad \textbf{(B)}144\qquad \textbf{(C)}256\qquad \textbf{(D)}400\qquad \textbf{(E)}416$

Solution: Taking the discount of $40$ % means you're only paying $60$ % of the bill.  That results in $10,000\cdot0.6=\textdollar{6,000}$
Likewise, taking two discounts of $36$ % and $4$ % means taking $64$ % of the original amount and then $96$ % of the result.  That results in $10,000\cdot0.64\cdot0.96=\textdollar{6,144}$
Taking the difference results in $6,144-6,000=\textdollar{144}$ , or answer choice $\boxed{144}$

Answer: B 144
Problem: The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$

Solution: By the Binomial Theorem , each term of the expansion is $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n$
We want the exponent of $x$ to be $7$ , so \[2(8-n)-n=7\] \[16-3n=7\] \[n=3\]
If $n=3$ , then the corresponding term is \[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]
The answer is $\boxed{14}$

Answer: D 14
Problem: Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ . 
The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

Solution: First, factor the difference of squares \[(m+n)(m-n)\] Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$ , where $a$ and $b$ can be any integer. \[(2a+2b+2)(2a-2b)\] Factor the resulting expression. \[4(a+b+1)(a-b)\] If $a$ and $b$ are both even, then $a-b$ is even.  If $a$ and $b$ are both odd, then $a-b$ is even as well.  If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even.  Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$
This can be confirmed by setting $m=3$ and $n=1$ , making $m^2-n^2=9-1=8$ .  Since $8$ is not a multiple of $3$ and is less than $16$ , we can confirm that the answer is $\boxed{8}$

Answer: D 8
Problem: Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the 
exterior angle at the same vertex. Then
$\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\ \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \\ \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular}$

Solution: Let $a_n$ be the interior angle of the nth vertex, and let $b_n$ be the exterior angle of the nth vertex.  From the conditions in the problem, \[a_n = 7.5b_n\] That means \[a_1 + a_2 \cdots a_n = 7.5(b_1 + b_2 \cdots b_n)\] Since the sum of the exterior angles of a polygon is $360^{\circ}$ , the equation can be simplified as \[a_1 + a_2 \cdots a_n = 7.5 \cdot 360\] \[a_1 + a_2 \cdots a_n = 2700\] The sum of the interior angles is $2700$ degrees.  However, there is no other constraint on what angle the interior angles can be, so we can not for sure claim that the polygon is regular or not.  Thus, the answer is $\boxed{2700}$

Answer: E 2700
Problem: You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$ , and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$ . 
Let $N$ be the number of primes appearing in this sequence. Then $N$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$

Solution: First, note that $n$ does not have a prime number larger than $61$ as one of its factors.  Also, note that $n$ does not equal $1$
Therefore, since the prime factorization of $n$ only has primes from $2$ to $59$ $n$ and $P$ share at least one common factor other than $1$ .  Therefore $P+n$ is not prime for any $n$ , so the answer is $\Rightarrow{\boxed{0}$

Answer: A 0
Problem: Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool, 
one at the rate of $3$ feet per second, the other at $2$ feet per second. 
They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$

Solution: First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side.  Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.
[asy] draw((0,0)--(0,105),EndArrow); draw((0,0)--(105,0),EndArrow); for (int i=0; i<6;++i) { dot((0,18i)); } for (int j=0;j<7;++j) { dot((15j,0)); } label("0",(0,0),SW); label("90",(0,90),W); label("180",(90,0),S); draw((0,0)--(15,90)--(30,0)--(45,90)--(60,0)--(75,90)--(90,0),red); draw((0,90)--(22.5,0)--(45,90)--(67.5,0)--(90,90),blue); [/asy]
At this point, find the number of meeting points in the first $3$ minutes, then multiply by four to get the answer.  From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other $20$ times, which is answer choice $\boxed{20}$

Answer: C 20
Problem: Each edge of a cube is increased by $50$ %. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$

Solution: Note that increasing the length of each edge by $50$ % with result in a cube that is similar to the original cube with scale factor $1.5$ . Therefore, the surface area will increase by a factor of $1.5^2$ , or $2.25$ . Converting this back into a percent, the percent increase will be $125$ %. Therefore, the answer is $\boxed{125}$

Answer: B 125
Problem: The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$

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

Answer: A 4
Problem: A farmer divides his herd of $n$ cows among his four sons so that one son gets one-half the herd, a second son, one-fourth, a third son, one-fifth, and the fourth son, $7$ cows. Then $n$ is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$

Solution: The first three sons get $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}$ of the herd, so that the fourth son should get $\frac{1}{20}$ of it. But the fourth son gets $7$ cows, so the size of the herd is $n=\frac{7}{\frac{1}{20}} = 140$ . Then our answer is $\boxed{140}$ , and we are done.

Answer: C 140
Problem: In $\triangle ABC$ with $\overline{AB}=\overline{AC}=3.6$ , a point $D$ is taken on $AB$ at a distance $1.2$ from $A$ . Point $D$ is joined to $E$ in the prolongation of $AC$ so that $\triangle AED$ is equal in area to $ABC$ . Then $\overline{AE}$ is: $\textbf{(A)}\ 4.8 \qquad\textbf{(B)}\ 5.4\qquad\textbf{(C)}\ 7.2\qquad\textbf{(D)}\ 10.8\qquad\textbf{(E)}\ 12.6$

Solution: Note that $\frac{1}{2}AB * AC *\sin\angle BAC = \frac{1}{2}AD * AE *\sin\angle DAE$ . Since $\angle BAC = \angle DAE$ , we have $AB*AC = AD*AE$ , so that $3.6*3.6 = 1.2*AE$ . Therefore, $AE = \frac{3.6^2}{1.2} = 10.8$ . Thusly, our answer is $\boxed{10.8}$ , and we are done.

Answer: D 10.8
Problem: The arithmetic mean (average) of a set of $50$ numbers is $38$ . If two numbers, namely, $45$ and $55$ , are discarded, the mean of the remaining set of numbers is: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$

Solution: Since the arithmetic mean of the $50$ numbers is $38$ , their sum must be $50*38 = 1900$ . After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$ , so it must become $1900 - 100 = 1800$ . 
But this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\frac{1800}{48} = 37.5$ . Thusly, our answer is $\boxed{37.5}$ , and we are done.

Answer: D 37.5
Problem: 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.

Answer: C 45
Problem: 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: 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: C 45
Problem: With the use of three different weights, namely $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale? $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7$

Solution: The heaviest object that could be weighed with this set weighs $1 + 3 + 9 = 13$ lb., and we can weigh any positive integer weight at most that. This means that $13$ different objects could be weighed, so our answer is $\boxed{13}$ and we are done.

Answer: B 13
Problem: Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ , where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$ 's are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$ . [See example 25 for the meaning of $|x|$ .] The number of polynomials with $h=3$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9$

Solution: We perform casework by the value of $n$ , the degree of our polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$
Case $n = 0$ : In this case we are forced to set $a_0 = 3$ . This contributes $1$ possibility.
Case $n = 1$ : In this case we must have $a_0 + |a_1| = 2$ , so our polynomial could be $1 + 1x, 0 + 2x, -1 + 1x$ . This contributes $3$ possibilities.
Case $n = 2$ : In this case we must have $a_0 + |a_1| + |a_2| = 1$ . However, because $a_0$ must be positive, it has to be $1$ , so our polynomial can only be $0 + 0x + 1x^2$ . This contributes $1$ possibility.
Case $n\geq 3$ : This case is impossible because $h = n+a_0+|a_1|+|a_2|+\cdots+|a_n|\geq n + a_0\geq 3 + 1 = 4 > 3$ , so it contributes $0$ possibilities.
Adding the results from all four cases, we find that there are $1 + 3 + 1 + 0 = 5$ possibilities in total, so our answer is $\boxed{5}$

Answer: B 5
Problem: The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \not = 0$ , is:
$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$

Solution: We have $\frac{1}{2}\cdot \left(\frac{x + a}{x} + \frac{x - a}{x}\right) = \frac{2}{2} = \boxed{1}$

Answer: B 1
Problem: If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad  \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad  \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad  \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$

Solution: \[P=\frac{s}{(1+k)^n}\]
\[(1+k)^n=\frac{s}{P}\]
Take the $\log$ of each side.
\[n \log(1+k) = \log\left(\frac{s}{P}\right)\]
\[n = \frac{\log\left(\frac{s}{P}\right)}{\log(1+k)} \to \boxed{1}\]

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

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

Answer: B 12
Problem: The sides of a triangle are $30$ $70$ , and $80$ units. If an altitude is dropped upon the side of length $80$ , the larger segment cut off on this side is:
$\textbf{(A)}\ 62\qquad  \textbf{(B)}\ 63\qquad  \textbf{(C)}\ 64\qquad  \textbf{(D)}\ 65\qquad  \textbf{(E)}\ 66$

Solution: Let the shorter segment be $x$ and the altitude be $y$ . The larger segment is then $80-x$ .  By the Pythagorean Theorem \[30^2-y^2=x^2 \qquad(1)\] and \[(80-x)^2=70^2-y^2 \qquad(2)\] Adding $(1)$ and $(2)$ and simplifying gives $x=15$ . Therefore, the answer is $80-15=\boxed{65}$

Answer: D 65
Problem: In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:
$\textbf{(A)}\ 11\qquad  \textbf{(B)}\ 33\qquad  \textbf{(C)}\ 55\qquad  \textbf{(D)}\ 66\qquad  \textbf{(E)}\ 132$

Solution: Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:
\[\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}\]
So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:
\[\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{66}\]

Answer: D 66
Problem: The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$

Solution: [asy] size(2cm); draw((-3,0)--(0,4)--(3,0)--cycle); draw((0,0)--(0,4), red);  draw((-3,0)--(0.84, 2.88), green); draw((-3,0)--(1.5, 2), green); draw((-3,0)--(1.636, 1.818), green);  draw((3,0)--(-0.84, 2.88), blue); draw((3,0)--(-1.5, 2), blue); draw((3,0)--(-1.636, 1.818), blue); [/asy]
As shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\boxed{7}$ , and we are done.

Answer: B 7
Problem: The area of a circle inscribed in an equilateral triangle is $48\pi$ . The perimeter of this triangle is:
$\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$

Solution: [asy] draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); draw(circle((0,0),sqrt(3))); dot((0,0)); draw((0,0)--(0,-sqrt(3))); [/asy] We can see that the radius of the circle is $4\sqrt{3}$ . We know that the radius is $\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\sqrt{3}$ . Since the median line completes a $30$ $60$ $90$ triangle, we can conclude that one of the sides of the triangle is $24$ . Triple the side length and we get our answer, $\boxed{72}$

Answer: E 72
Problem: The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$ . The sum of $x, y$ , and $z$ is $100$ . The number y is given by the equation $y = ax - 10$ . Then a is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$

Solution: In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$ , and $z$ would have $5$ ). One part, after some algebra, equals $10$ , so $x$ $y$ , and $z$ are $20$ $30$ , and $50$ , respectively.
We can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$ , and then solve to get $a = \boxed{2}$

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

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

Answer: B 14
Problem: The table below shows the distance $s$ in feet a ball rolls down an inclined plane in $t$ seconds.
\[\begin{tabular}{|c|c|c|c|c|c|c|}\hline t & 0 & 1 & 2 & 3 & 4 & 5\\ \hline s & 0 & 10 & 40 & 90 & 160 & 250\\ \hline\end{tabular}\] The distance $s$ for $t = 2.5$ is:
$\textbf{(A)}\ 45\qquad \textbf{(B)}\ 62.5\qquad \textbf{(C)}\ 70\qquad \textbf{(D)}\ 75\qquad \textbf{(E)}\ 82.5$

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

Answer: B 62.5
Problem: The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$ . In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$ . The numeral $10011$ in the binary system would then be written in the decimal system as:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 10011\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 7$

Solution: Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$
Using this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{19}$

Answer: A 19
Problem: The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:
$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$

Solution: We can merge the two equations to create $x^2+y=x+y$ . Using either the quadratic equation or factoring, we get two solutions with $x$ -coordinates $0$ and $1$
Plugging this into either of the original equations, we get $(0,10)$ and $(1,9)$ . The distance between those two points is $\boxed{2}$

Answer: C 2
Problem: If $9^{x + 2} = 240 + 9^x$ , then the value of $x$ is:
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2\qquad \textbf{(C)}\ 0.3\qquad \textbf{(D)}\ 0.4\qquad \textbf{(E)}\ 0.5$

Solution: $9^{x+2}$ can be rewritten as $9^x*9^2=9^x*81$ , which means the equation can be rewritten as $81(9^x)=240+9^x$ , or $80(9^x)=240$ , or $9^x=3$ . Therefore, $x=\boxed{0.5}$

Answer: E 0.5
Problem: If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$

Solution: We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$ . Note that $i^4=1$ and $(-i)^4=1$ , so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$
For $n=0$ , we have $i^0+(-i)^0=1+1=2$
For $n=1$ , we have $i^1+(-i)^1=i-i=0$
For $n=2$ , we have $i^2+(-i)^2=-1-1=-2$
For $n=3$ , we have $i^3+(-i)^3=-i+i=0$
Hence, the answer is $\boxed{3}$

Answer: C 3
Problem: If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$

Solution: Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.
For $n=0$ : we have $2$
For $n=1$ : we have $0$
For $n=2$ : we have $-2$
For $n=3$ : we have $0$
For $n=4$ : we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\boxed{3}$

Answer: C 3