Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Find the number of ordered pairs of real numbers $(a,b)$ such that $(a+bi)^{2002} = a-bi$
$\text{(A) }1001 \qquad \text{(B) }1002 \qquad \text{(C) }2001 \qquad \text{(D) }2002 \qquad \text{(E) }2004$

Solution: Let $z = a + bi = re^{i\theta}$ where $r = |z|$ and $0\leq \theta < 2\pi.$ We want to solve $z^{2002} = \overline{z}.$ Since $z\overline{z} = |z|^2,$ we multiply by $z$ on both sides to get \[z^{2003} = |z|^2 = r^{2003}e^{2023i\theta} = r^2,\] from which we get $r\in \{0, 1\}$ (since $r\in \mathbb{R}$ ), immediately yielding $z = 0$ as an answer. If $r=1$ then we have $z^{2003} = 1$ , which each of the $2003$ rd roots of unity satisfies. Altogether, there are $1 + 2003 = \boxed{2004}$ values of $z$ , each of which correspond to a unique ordered pair $(a, b)$

Answer: E 2004
Problem: How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: Subtracting 10 from each number in the set, and dividing the results by 3, we obtain the set $\{-3, -2, -1, 0, 1, 2, 3\}$ .  It is easy to see that we can get any integer between $-6$ and $6$ inclusive as the sum of three elements from this set, for the total of $\boxed{13}$ integers.

Answer: A 13
Problem: How many different integers can be expressed as the sum of three distinct members of the set $\{1,4,7,10,13,16,19\}$
$\text{(A)}\ 13 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 24 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 35$

Solution: The set is an arithmetic sequence of numbers each $1$ more than a multiple of $3$ . Thus the sum of any three numbers will be a multiple of $3$ . All the multiples of $3$ from $1+4+7=12$ to $13+16+19=48$ are possible, totaling to $\boxed{13}$ integers.

Answer: A 13
Problem: How many four-digit numbers $N$ have the property that the three-digit number obtained by removing the leftmost digit is one ninth of $N$
$\mathrm{(A)}\ 4 \qquad\mathrm{(B)}\ 5 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7 \qquad\mathrm{(E)}\ 8$

Solution: Since N is a four digit number, assume WLOG that $N = 1000a + 100b + 10c + d$ , where a is the thousands digit, b is the hundreds digit, c is the tens digit, and d is the ones digit.
Then, $\frac{1}{9}N = 100b + 10c + d$ , so $N = 900b + 90c + 9d$ Set these equal to each other: \[1000a + 100b + 10c + d = 900b + 90c + 9d\] \[1000a = 800b + 80c + 8d\] \[1000a = 8(100b + 10c + d)\] Notice that $100b + 10c + d = N - 1000a$ , thus: \[1000a = 8(N - 1000a)\] \[1000a = 8N - 8000a\] \[9000a = 8N\] \[N = 1125a\]
Go back to our first equation, in which we set $N = 1000a + 100b + 10c + d$ ,
Then: \[1125a = 1000a + 100b + 10c + d\] \[125a = 100b + 10c + d\] The upper limit for the right hand side (RHS) is $999$ (when $b = 9$ $c = 9$ , and $d = 9$ ).
It's easy to prove that for an $a$ there is only one combination of $b, c,$ and $d$ that can make the equation equal. Just think about the RHS as a three digit number $bcd$ . There's one and only one way to create every three digit number with a certain combination of digits.
Thus, we test for how many as are in the domain set by the RHS. Since $125\cdot7 = 875$ which is the largest $a$ value, then $a$ can be $1$ through $7$ , giving us the answer of $\boxed{7}$

Answer: D 7
Problem: If $a,b,$ and $c$ are positive real numbers such that $a(b+c) = 152, b(c+a) = 162,$ and $c(a+b) = 170$ , then $abc$ is
$\mathrm{(A)}\ 672 \qquad\mathrm{(B)}\ 688 \qquad\mathrm{(C)}\ 704 \qquad\mathrm{(D)}\ 720 \qquad\mathrm{(E)}\ 750$

Solution: Adding up the three equations gives $2(ab + bc + ca) = 152 + 162 + 170 = 484 \Longrightarrow ab + bc + ca = 242$ . Subtracting each of the above equations from this yields, respectively, $bc = 90, ca = 80, ab = 72$ . Taking their product, $ab \cdot bc \cdot ca = a^2b^2c^2 = 90 \cdot 80 \cdot 72 = 720^2 \Longrightarrow abc = \boxed{720}$

Answer: D 720
Problem: For all positive integers $n$ less than $2002$ , let
\begin{eqnarray*} a_n =\left\{ \begin{array}{lr} 11, & \text{if\ }n\ \text{is\ divisible\ by\ }13\ \text{and\ }14;\\ 13, & \text{if\ }n\ \text{is\ divisible\ by\ }14\ \text{and\ }11;\\ 14, & \text{if\ }n\ \text{is\ divisible\ by\ }11\ \text{and\ }13;\\ 0, & \text{otherwise}. \end{array} \right. \end{eqnarray*}
Calculate $\sum_{n=1}^{2001} a_n$
$\mathrm{(A)}\ 448 \qquad\mathrm{(B)}\ 486 \qquad\mathrm{(C)}\ 1560 \qquad\mathrm{(D)}\ 2001 \qquad\mathrm{(E)}\ 2002$

Solution: Find the LCMs of the groups of the numbers.
Notice that the groups are relatively prime.
So $a_n=$
11 if $n$ is a multiple of 182.
13 if $n$ is a multiple of 154.
14 if $n$ is a multiple of 143.
When do we see ambiguities (for example: $n$ is a multiple of 11, 13, and 14)? This is only done when $n$ is a multiple of $\operatorname{lcm}(11,13,14)=2002$ . However, since $n<2002$ , this can never happen.
So we have 10 multiples of 182 we have to count (1 to 10 $*182$ ), and similarly, 12 multiples of 154, and 13 multiples of 143. The sum is $10*11+12*13+13*14=448$ . Select $\boxed{448}$

Answer: A 448
Problem: For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1  \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$

Solution: Note that $\frac{1}{\log_a b}=\log_b a$ . Thus $a_n=\log_{2002} n$ . Also notice that if we have a log sum, we multiply, and if we have a log product, we divide. Using these properties, we get that the result is the following:
\[\log_{2002}\left(\frac{2*3*4*5}{10*11*12*13*14}=\frac{1}{11*13*14}=\frac{1}{2002}\right)=\boxed{1}\]

Answer: B 1
Problem: For all integers $n$ greater than $1$ , define $a_n = \frac{1}{\log_n 2002}$ . Let $b = a_2 + a_3 + a_4 + a_5$ and $c = a_{10} + a_{11} + a_{12} + a_{13} + a_{14}$ . Then $b- c$ equals
$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1  \qquad\mathrm{(C)}\ \frac{1}{2002} \qquad\mathrm{(D)}\ \frac{1}{1001} \qquad\mathrm{(E)}\ \frac 12$

Solution: Note that $a_2 = \frac{1}{\log_2 2002}$ $1$ is also equal to $\log_2 2$ . So $a_2 = \frac{\log_2 2}{\log_2 2002}$ . By the change of bases formula, $a_2 = \log_{2002} 2$ . Following the same reasoning, $a_3 = \log_{2002} 3$ $a_4 = \log_{2002} 4$ and so on.
\[b = \log_{2002} 2 + \log_{2002} 3 + .....+ \log_{2002} 5 = \log_{2002} 5! = \log_{2002} 120\] Now solving for $c$ , we see that it equals $\log_{2002} (10\cdot 11 \cdot 12 \cdot 13 \cdot 14)$ \[b-c = \log_{2002} 120 - \log_{2002} 240240 \rightarrow \log_{2002} \frac{1}{2002} = \boxed{1}\]

Answer:  1
Problem: In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$

Solution: From Stewart's Theorem , we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{2}.$ - awu2014

Answer:  2
Problem: In $\triangle ABC$ , we have $AB = 1$ and $AC = 2$ . Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$

Solution: There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$ , and you draw a median from point $A$ to side $BC$ (label this as $M$ ), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$ . Note that $b$ is the length of side $\overline{AC}$ $c$ is the length of side $\overline{AB}$ , and $a$ is length of side $\overline{BC}$ . Let $MB = MC = x$ . Then $AM = 2x$ . Now, we can plug into the formula given above: $AM = 2x$ $b = 2$ $c = 1$ , and $a = 2x$ . After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$ . Then, $BC = \boxed{2}$

Answer: C 2
Problem: Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$

Solution: The first condition gives us that \[x^2 + 6x + 1 + y^2 + 6y + 1 \le 0 \Longrightarrow (x+3)^2 + (y+3)^2 \le 16\]
which is a circle centered at $(-3,-3)$ with radius $4$ . The second condition gives us that
\[x^2 + 6x + 1 - y^2 - 6y - 1 \le 0 \Longrightarrow (x^2 - y^2) + 6(x-y) \le 0 \Longrightarrow (x-y)(x+y+6) \le 0\]
Thus either
\[x - y \ge 0,\quad x+y+6 \le 0\]
or
\[x - y \le 0,\quad x+y+6 \ge 0\]
Each of those lines passes through $(-3,-3)$ and has slope $\pm 1$ , as shown above. Therefore, the area of $R$ is half of the area of the circle, which is $\frac{1}{2} (\pi \cdot 4^2) = 8\pi \approx \boxed{25}$

Answer: E 25
Problem: Let $f(x) = x^2 + 6x + 1$ , and let $R$ denote the set of points $(x,y)$ in the coordinate plane such that \[f(x) + f(y) \le 0 \qquad \text{and} \qquad f(x)-f(y) \le 0\] The area of $R$ is closest to
$\textbf{(A) } 21 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 23 \qquad\textbf{(D)}\ 24 \qquad\textbf{(E)}\ 25$

Solution: Similar to Solution 1, we proceed to get the area of the circle satisfying $f(x)+f(y) \le 0$ , or $16 \pi$
Since $f(x)-f(y) \le 0 \implies f(x) \le f(y)$ , we have that by symmetry, if $(x,y)$ is in $R$ , then $(y,x)$ is not, and vice versa. Therefore, the shaded part of the circle above the line $y=x$ has the same area as the unshaded part below $y=x$ , and the unshaded part above $y=x$ has the same area as the shaded part below $y=x$ . This means that exactly half the circle is shaded, allowing us to divide by two to get $\frac{16 \pi }{2} = 8\pi \approx \boxed{25}$ . ~samrocksnature + ddot1

Answer: E 25
Problem: The function $f$ is given by the table
\[\begin{tabular}{|c||c|c|c|c|c|}  \hline   x & 1 & 2 & 3 & 4 & 5 \\   \hline  f(x) & 4 & 1 & 3 & 5 & 2 \\  \hline \end{tabular}\]
If $u_0=4$ and $u_{n+1} = f(u_n)$ for $n \ge 0$ , find $u_{2002}$
$\text{(A) }1 \qquad \text{(B) }2 \qquad \text{(C) }3 \qquad \text{(D) }4 \qquad \text{(E) }5$

Solution: We can guess that the series given by the problem is periodic in some way. Starting off, $u_0=4$ is given. $u_1=u_{0+1}=f(u_0)=f(4)=5,$ so $u_1=5.$ $u_2=u_{1+1}=f(u_1)=f(5)=2,$ so $u_2=2.$ $u_3=u_{2+1}=f(u_2)=f(2)=1,$ so $u_3=1.$ $u_4=u_{3+1}=f(u_3)=f(1)=4,$ so $u_4=4.$ Plugging in $4$ will give us $5$ as found before, and plugging in $5$ will give $2$ and so on. This means that our original guess of the series being periodic was correct. Summing up our findings in a nice table,
\[\begin{tabular}{|c||c|c|c|c|c|c|}  \hline   n & 0 & 1 & 2 & 3 & 4 & ...\\   \hline  un & 4 & 5 & 2 & 1 & 4 & ...\\  \hline \end{tabular}\]
in which the next $u_n$ is found by simply plugging in the number from the last box into $f(x).$ The function is periodic every $4$ terms. $2002 \equiv 2\pmod{4}$ , and counting $4$ starting from $u_1$ will give us our answer of $\boxed{2}$

Answer: B 2
Problem: The dimensions of a rectangular box in inches are all positive integers and the volume of the box is $2002$ in $^3$ . Find the minimum possible sum of the three dimensions.
$\text{(A) }36 \qquad \text{(B) }38  \qquad \text{(C) }42 \qquad \text{(D) }44 \qquad \text{(E) }92$

Solution: Given an arbitrary product and an arbitrary amount of terms to multiply to get that product, to maximize the sum, make all of the terms $1$ with the last one being the number. To minimize the sum, make all of the terms equal to each other. (This is a corollary that follows from the $AM-GM$ proof.) Since $2002$ is not a perfect cube, we have to make the terms as close to each other as possible. A good rule of thumb is the memorize the prime factorization of the AMC year that you are doing, which is $2002= 2 \cdot 7 \cdot 11 \cdot 13$ . The three terms that are closest to each other that multiply to $2002$ are $11, 13,$ and $14$ , so our answer is $11+13+14=\boxed{38}$

Answer: B 38
Problem: Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5  \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$

Solution: For sake of speed, WLOG, let $b=1$ . This means that the ratio $\frac{a}{b}$ will simply be $a$ because $\frac{a}{b}=\frac{a}{1}=a.$ Solving for $a$ with some very simple algebra gives us a quadratic which is $5a^2 -9a +4=0.$ Factoring the quadratic gives us $(5a-4)(a-1)=0$ . Therefore, $a=1$ or $a=\frac{4}{5}=0.8.$ However, since $a$ and $b$ must be distinct, $a$ cannot be $1$ so the latter option is correct, giving us our answer of $\boxed{0.8}.$

Answer: E 0.8
Problem: Let $a$ and $b$ be distinct real numbers for which \[\frac{a}{b} + \frac{a+10b}{b+10a} = 2.\]
Find $\frac{a}{b}$
$\text{(A) }0.4 \qquad \text{(B) }0.5  \qquad \text{(C) }0.6 \qquad \text{(D) }0.7 \qquad \text{(E) }0.8$

Solution: The only tricky part about this equation is the fact that the left-hand side has fractions. Multiplying both sides by $b(b+10a)$ gives us $2ab+10a^2+10b^2=2b^2+20ab.$ Moving everything to the left-hand side and dividing by $2$ gives $5a^2-4b^2 -9ab,$ which factors into $(5a-4b)(a-b)=0.$ Because $a \neq b, 5a=4b \implies \frac{a}{b}=0.8$ giving us our answer of $\boxed{0.8}.$

Answer: E 0.8
Problem: How many three-digit numbers have at least one $2$ and at least one $3$
$\text{(A) }52 \qquad \text{(B) }54  \qquad \text{(C) }56 \qquad \text{(D) }58 \qquad \text{(E) }60$

Solution: We can do this problem with some simple case work.
Case 1: The hundreds place is not $2$ or $3.$ This means that the tens place and ones place must be $2$ and $3$ respectively or $3$ and $2$ respectively. This case covers $1, 4, 5, 6, 7, 8,$ and $9,$ so it gives us $2 \cdot 7 = 14$ cases.
Case 2: The hundreds place is $2.$ This means that $3$ must be in the tens place or ones place. Starting with cases in which the tens place is not $3$ , we get $203, 213, 223, 243, 253, 263, 273, 283,$ and $293.$ With cases in which the tens place is $3$ , we have $230-239$ , or $10$ more cases. This gives us $9 + 10=19$ cases.
Case 3: The hundreds place is $3.$ This case is almost identical to the second case, just swap the $2$ s with $3$ s and $3$ s with $2$ s in the reasoning and its the same, giving us an additional $19$ cases.
Addition up all of these cases gives $14+19+19=52$ cases, or $\boxed{52}.$

Answer: A 52
Problem: Two walls and the ceiling of a room meet at right angles at point $P.$ A fly is in the air one meter from one wall, eight meters from the other wall, and nine meters from point $P$ . How many meters is the fly from the ceiling?
$\text{(A) }\sqrt{13} \qquad \text{(B) }\sqrt{14} \qquad \text{(C) }\sqrt{15} \qquad \text{(D) }4 \qquad \text{(E) }\sqrt{17}$

Solution: We can use the formula for the diagonal of the rectangle, or $d=\sqrt{a^2+b^2+c^2}$ The problem gives us $a=1, b=8,$ and $c=9.$ Solving gives us $9=\sqrt{1^2 + 8^2 + c^2} \implies c^2=9^2-8^2-1^2 \implies c^2=16  \implies c=\boxed{4}.$

Answer: D 4
Problem: Let $f_n (x) = \text{sin}^n x + \text{cos}^n x.$ For how many $x$ in $[0,\pi]$ is it true that
\[6f_{4}(x)-4f_{6}(x)=2f_{2}(x)?\]
$\text{(A) }2 \qquad \text{(B) }4  \qquad \text{(C) }6 \qquad \text{(D) }8 \qquad \text{(E) more than }8$

Solution: Divide by 2 on both sides to get \[3f_{4}(x)-2f_{6}(x)=f_{2}(x)\] Substituting the definitions of $f_{2}(x)$ $f_{4}(x)$ , and $f_{6}(x)$ , we may rewrite the expression as \[3(\text{sin}^4{x} + \text{cos}^4{x}) - 2(\text{sin}^6{x} + \text{cos}^6{x}) = 1\] We now simplify each term separately using some algebraic manipulation and the Pythagorean identity.
We can rewrite $3(\text{sin}^4 x + \text{cos}^4 x)$ as $3(\text{sin}^2 x + \text{cos}^2 x)^2 - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $3 - 6\text{sin}^2 x \text{cos}^2 x$
As for $2(\text{sin}^6 x + \text{cos}^6 x)$ , we may factor it as $2(\text{sin}^2 x + \text{cos}^2 x)(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ which can be rewritten as $2(\text{sin}^4 x + \text{cos}^4 x - \text{sin}^2 x \text{cos}^2 x)$ , and then as $2(\text{sin}^2 x + \text{cos}^2)^2) - 6\text{sin}^2 x \text{cos}^2 x$ , which is equivalent to $2 - 6\text{sin}^2 x \text{cos}^2 x$
Putting everything together, we have $(3 - 6\text{sin}^2 {x} \text{cos}^2 {x}) - (2 - 6\text{sin}^2 {x} \text{cos}^2 {x}) = 1$ or $1 = 1$ . Therefore, the given equation $3f_{4}(x)-2f_{6}(x)=f_{2}(x)$ is true for all real $x$ , meaning that there are more than 8 values of $x$ that satisfy the given equation and so the answer is $\boxed{8}$

Answer: E 8
Problem: For how many positive integers $n$ is $n^3 - 8n^2 + 20n - 13$ a prime number?
$\text{(A) 1} \qquad \text{(B) 2} \qquad \text{(C) 3} \qquad \text{(D) 4} \qquad \text{(E) more than 4}$

Solution: Since this is a number theory question, it is clear that the main challenge here is factoring the given cubic. In general, the rational root theorem will be very useful for these situations.
The rational root theorem states that all rational roots of $n^3 - 8n^2 + 20n - 13$ will be among $1, 13, -1$ , and $-13$ . Evaluating the cubic at these values will give $n = 1$ as a root. Doing some synthetic division gives \[n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13)\]
Since $n > 0$ $n-1$ must be nonnegative. Since $(n-1)(n^2 - 7n + 13)$ evaluates to a prime, it is clear that exactly one of $n-1$ and $n^2 - 7n - 13$ is $1$ . We proceed by splitting the problem into 2 cases.
Case 1: $n-1 = 1$ It is clear that $n = 2$ .  We have $2^2 - 7(2) + 13 = 3$ , so this case yields $n = 2$ as a solution.
Case 2: $n^2 - 7n + 13 = 1$ Solving for $n$ gives $n^2 - 7n + 12 = 0$ or $(n-3)(n-4) = 0$ . Therefore, $n = 3$ or $n = 4$ .
Since both $3-1 = 2$ and $4-1 = 3$ are prime, both $n = 3$ and $n = 4$ work, yielding 2 solutions.
Putting everything together, the answer is $1 + 2 = \boxed{3}$

Answer: C 3
Problem: What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1, k_2, ... k_n$ for which
\[k^2_1 + k^2_2 + ... + k^2_n = 2002?\]
$\text{(A) }14 \qquad \text{(B) }15 \qquad \text{(C) }16 \qquad \text{(D) }17 \qquad \text{(E) }18$

Solution: Note that $k^2_1 + k^2_2 + ... + k^2_n = 2002 \leq \frac{n(n+1)(2n+1)}{6}$
When $n = 17$ $\frac{n(n+1)(2n+1)}{6} = \frac{(17)(18)(35)}{6} = 1785 < 2002$
When $n = 18$ $\frac{n(n+1)(2n+1)}{6} = 1785 + 18^2 = 2109 > 2002$
Therefore, we know $n \leq 17$
Now we must show that $n = 17$ works. We replace one of $1, 2, ... 17$ with an integer $a > 17$ to account for the amount under $2002$ , which is $2002-1785 = 217$
Essentially, this boils down to writing $217$ as a difference of squares. We know $217 = (7)(31)$ , so we assume there exist positive integers $a$ and $b$ where $a > 17$ and $b \leq 17$ such that $a^2 - b^2 = 217$
We can rewrite this as $(a+b)(a-b) = 217$ , so either $a+b = 217$ and $a-b = 1$ or $a+b = 31$ and $a-b = 7$ . We analyze each case separately.
Case 1: $a+b = 217$ and $a-b = 1$
Solving this system of equations gives $a = 109$ and $b = 108$ . However, $108 > 17$ , so this case does not yield a solution.
Case 2: $a+b = 31$ and $a-b = 7$
Solving this system of equations gives $a = 19$ and $b = 12$ . This satisfies all the requirements of the problem.
The list $1, 2 ... 11, 13, 14 ... 17, 19$ has $17$ terms whose sum of squares equals $2002$ . Therefore, the answer is $\boxed{17}$

Answer: D 17
Problem: The altitudes of a triangle are $12, 15,$ and $20.$ The largest angle in this triangle is
$\text{(A) }72^\circ \qquad \text{(B) }75^\circ \qquad \text{(C) }90^\circ \qquad \text{(D) }108^\circ \qquad \text{(E) }120^\circ$

Solution: Let $a, b,$ and $c$ denote the bases of altitudes $12, 15,$ and $20,$ respectively. Since they are all altitudes and bases of the same triangle, they have the same area, so $\frac{12a}{2}=\frac{15b}{2}=\frac{20c}{2}.$ Multiplying by $2$ , we get $12a=15b=20c.$ Notice that a simple solution to the equation is if all of them equal $12 \cdot 15 \cdot 20.$ That means $a=15 \cdot 20, b=12 \cdot 20,$ and $c=12 \cdot 15.$ Simplifying our solution to check for Pythagorean triples we see that this is just a Pythagorean triple, namely a $3-4-5$ triangle. Since the other two angles of a right triangle must be acute, the right angle must be the greatest angle. Therefore, our answer is $\boxed{90}.$

Answer: C 90
Problem: Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution

Solution: By the Pythagorean identity we can rewrite the given expression as follows. \[\sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}} = \sqrt{\sin^4{x} + 4(1 - \sin^2{x})} - \sqrt{\cos^4{x} + 4(1 - \cos^2{x})}\]
Expanding each bracket gives \[\sqrt{\sin^4{x} - 4\sin^2{x} + 4} - \sqrt{\cos^4{x} - 4\cos^2{x} + 4}\]
The expressions under the square roots can be factored to get \[\sqrt{(\sin^2{x} - 2)^2} - \sqrt{(\cos^2{x} - 2)^2}\]
Since $\sin^2{x} < 2$ and $\cos^2{x} < 2$ for all real $x$ , the expression must evaluate to $(2 - \sin^2{x}) - (2 - \cos^2{x})$ , which simplifies to $\cos^2{x} - \sin^2{x} = \boxed{2}$

Answer: E 2
Problem: Let $f(x) = \sqrt{\sin^4{x} + 4 \cos^2{x}} - \sqrt{\cos^4{x} + 4 \sin^2{x}}.$ An equivalent form of $f(x)$ is
$\text{(A) }1-\sqrt{2}\sin{x} \qquad \text{(B) }-1+\sqrt{2}\cos{x} \qquad \text{(C) }\cos{\frac{x}{2}} - \sin{\frac{x}{2}} \qquad \text{(D) }\cos{x} - \sin{x} \qquad \text{(E) }\cos{2x}$
Solution

Solution: We don't actually have to solve the question. Just let $x$ equal some easy value to calculate $\cos {x}, \cos {2x}, \sin {x}, \sin {\frac{x}{2}},$ and $\cos {\frac{x}{2}}.$ For this solution, let $x=60^\circ.$ This means that the expression in the problem will give $\sqrt{\sin^4{60^\circ} + 4 \cos^2{60^\circ}} - \sqrt{\cos^4{60^\circ} + 4 \sin^2{60^\circ}}=\sqrt{(\frac{\sqrt{3}}{2})^4 + 4(\frac{1}{2})^2} - \sqrt{(\frac{1}{2})^4 + 4(\frac{\sqrt{3}}{2})^2}=\sqrt{\frac{9}{16} +1} - \sqrt{\frac{1}{16} + 3} = \frac{-1}{2}.$ Plugging in $x=60^\circ$ for the rest of the expressions, we get
$\text{(A) }1-\sqrt{2}\sin{60^\circ}=1-\frac{\sqrt{6}}{2} \neq \frac{-1}{2}.$
$\text{(B) }1+\sqrt{2}\cos{60^\circ}=1+\frac{\sqrt{2}}{2} \neq \frac{-1}{2}.$
$\text{(C) }\cos{\frac{60^\circ}{2}} - \sin{\frac{60^\circ}{2}}=\frac{\sqrt{3}}{2}-1 \neq \frac{-1}{2}.$
$\text{(D) }\cos{60^\circ} - \sin{60^\circ}=\frac{1-\sqrt{3}}{2} \neq \frac{-1}{2}.$
$\text{(E) }\cos {120^\circ}=\frac{-1}{2}.$
Therefore, our answer is $\boxed{2}$

Answer: E 2
Problem: If $a,b,c$ are real numbers such that $a^2 + 2b =7$ $b^2 + 4c= -7,$ and $c^2 + 6a= -14$ , find $a^2 + b^2 + c^2.$
$\text{(A) }14 \qquad \text{(B) }21 \qquad \text{(C) }28 \qquad \text{(D) }35 \qquad \text{(E) }49$

Solution: Adding all of the equations gives $a^2 + b^2 +c^2 + 6a + 2b + 4c=-14.$ Adding 14 on both sides gives $a^2 + b^2 +c^2 + 6a + 2b + 4c+14=0.$ Notice that 14 can split into $9, 1,$ and $4,$ which coincidentally makes $a^2 +6a, b^2+2b,$ and $c^2+4c$ into perfect squares. Therefore, $(a+3)^2 + (b+1)^2 + (c+2) ^2 =0.$ An easy solution to this equation is $a=-3, b=-1,$ and $c=-2.$ Plugging in that solution, we get $a^2+b^2+c^2=-3^2+-1^2+-2^2=\boxed{14}.$

Answer: A 14
Problem: Let $f$ be a real-valued function such that
\[f(x) + 2f(\frac{2002}{x}) = 3x\]
for all $x>0.$ Find $f(2).$
$\text{(A) }1000 \qquad \text{(B) }2000 \qquad \text{(C) }3000 \qquad \text{(D) }4000 \qquad \text{(E) }6000$

Solution: Setting $x = 2$ gives $f(2) + 2f(1001) = 6$ .
Setting $x = 1001$ gives $2f(2) + f(1001) = 3003$
Adding these 2 equations and dividing by 3 gives $f(2) + f(1001) = \frac{6+3003}{3} = 1003$
Subtracting these 2 equations gives $f(2) - f(1001) = 2997$
Therefore, $f(2) = \frac{1003+2997}{2} = \boxed{2000}$

Answer: B 2000