Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Lines $L_1,L_2,\dots,L_{100}$ are distinct. All lines $L_{4n}, n$ a positive integer, are parallel to each other. 
All lines $L_{4n-3}, n$ a positive integer, pass through a given point $A.$ The maximum number of points of intersection of pairs of lines from the complete set $\{L_1,L_2,\dots,L_{100}\}$ is
$\textbf{(A) }4350\qquad \textbf{(B) }4351\qquad \textbf{(C) }4900\qquad \textbf{(D) }4901\qquad  \textbf{(E) }9851$

Solution: We partition $\{L_1,L_2,\dots,L_{100}\}$ into three sets. Let \begin{align*} X &= \{L_n\mid n\equiv0\pmod{4}\}, \\ Y &= \{L_n\mid n\equiv1\pmod{4}\}, \\ Z &= \{L_n\mid n\equiv2,3\pmod{4}\}, \\ \end{align*} from which $|X|=|Y|=25$ and $|Z|=50.$
Any two distinct lines can intersect at most once. To maximize the number of points of intersection, note that each point must be passed by exactly two lines. If three or more lines pass through the same point, then we can create more points of intersection by translating the lines.
We construct the sets one by one:
Together, the answer is $626+3725=\boxed{4351}.$

Answer: B 4351
Problem: Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as 
Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is $44$ years, then Ann's age is
$\textbf{(A) }22\qquad \textbf{(B) }24\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad  \textbf{(E) }28$

Solution: This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$ . Let $y$ equal the difference between their ages, so $y=a-b$ . Know that $y$ is constant because the difference between their ages will always be the same.
Now, let's tackle the equation: $b=$ Ann's age when Barbara was Ann's age when Barbara was $\frac{a}{2}$ . When Barbara was $\frac{a}{2}$ years old, Ann was $\frac{a}{2}+y$ years old. So the equation becomes $b=$ Ann's age when Barbara was $\frac{a}{2}+y$ . Adding on their age difference again, we get $b = \frac{a}{2} + y + y \Rightarrow b = \frac{a}{2} + 2y$ . Substitute $a-b$ back in for $y$ to get $b = \frac{a}{2} + 2(a-b)$ . Simplify: $2b = a + 4(a-b) \Rightarrow 6b = 5a$ . Solving $b$ in terms of $a$ , we have $b = \frac{5a}{6}$ . Substitute that back into the first equation of $a+b=44$ to get $\frac{11a}{6}=44$ . Solve for $a$ , and the answer is $\boxed{24}$ . ~ jiang147369

Answer: B 24
Problem: How many distinct ordered triples $(x,y,z)$ satisfy the following equations? \begin{align*} x + 2y + 4z &= 12 \\ xy + 4yz + 2xz &= 22 \\ xyz &= 6 \end{align*} $\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$

Solution: The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac b2,\frac c4\right).$
We rewrite the given equations in terms of $a,b,$ and $c:$ \begin{align*} a + b + c &= 12, \\ \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ \frac{abc}{8} &= 6. \end{align*} We clear fractions in these equations: \begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*} By Vieta's Formulas, note that $a,b,$ and $c$ are the roots of the equation \[r^3 - 12r^2 + 44r - 48 = 0,\] which factors as \[(r - 2)(r - 4)(r - 6) = 0.\] It follows that $\{a,b,c\}=\{2,4,6\}.$ Since the substitution $(x,y,z)=\left(a,\frac b2,\frac c4\right)$ is not symmetric with respect to $x,y,$ and $z,$ we conclude that different ordered triples $(a,b,c)$ generate different ordered triples $(x,y,z),$ as shown below: \[\begin{array}{c|c|c||c|c|c} & & & & & \\ [-2.5ex] \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] \hline & & & & & \\ [-2ex] 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array}\] So, there are $\boxed{6}$ such ordered triples $(x,y,z).$

Answer: E 6
Problem: For which real values of m are the simultaneous equations
\begin{align*}y &= mx + 3 \\  y& = (2m - 1)x + 4\end{align*}
satisfied by at least one pair of real numbers $(x,y)$
$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\  \text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m$

Solution: Solution by e_power_pi_times_i
Solving the systems of equations, we find that $mx+3 = (2m-1)x+4$ , which simplifies to $(m-1)x+1 = 0$ . Therefore $x = \dfrac{1}{1-m}$ $x$ is only a real number if $\boxed{1}$

Answer: D 1
Problem: If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad  \textbf{(B)}\ \sqrt2 \qquad  \textbf{(C)}\ 1/2 \qquad  \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$

Solution: Solution by e_power_pi_times_i
Denote the side of one square as $s$ . Then the diagonal of the second square is $s$ , so the side of the second square is $\dfrac{s\sqrt{2}}{2}$ . The area of the second square is $\dfrac{1}{2}s^2$ , so the ratio of the areas is $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{2}$

Answer: A 2
Problem: The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad  \textbf{(B)}\ 20 \qquad  \textbf{(C)}\ 40 \qquad  \textbf{(D)}\ 60 \qquad  \textbf{(E)}\ 80$

Solution: Solution by e_power_pi_times_i
When the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{80}$

Answer: E 80
Problem: Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$ , and $a_{100} + b_{100} = 100$ . 
Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$
$\textbf{(A)}\ 0 \qquad  \textbf{(B)}\ 100 \qquad  \textbf{(C)}\ 10,000 \qquad  \textbf{(D)}\ 505,000 \qquad \\ \textbf{(E)}\ \text{not enough information given to solve the problem}$

Solution: Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$ , respectively. Therefore $k_2 = -k_1$ . Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$ . The sum of the first $100$ terms is $100\cdot100 = \boxed{10,000}$

Answer: C 10,000
Problem: The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$ , where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad  \textbf{(B)}\ 4n \qquad  \textbf{(C)}\ 2+2n \qquad  \textbf{(D)}\ 4n^2 \qquad  \textbf{(E)}\ n^2+n+2$

Solution: We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: \[110\text{ some number of zeros }011.\] This works because of the way they are multiplied. Therefore, the answer is $\boxed{4}$

Answer: A 4
Problem: In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad  \textbf{(B)}\ 4 \qquad  \textbf{(C)}\ 2 \qquad  \textbf{(D)}\ 1 \qquad  \textbf{(E)}\ -1$

Solution: First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$ th term is $-2$ and the sum of the all those previous terms is $0$ . Then, write the 97th to the 100th terms down: $1, 3, 2,-1$ and add them up to get the sum of $\boxed{5}$

Answer: A 5
Problem: A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of $x$ working days, the man took the bus to work in the morning $8$ times, came home by bus in the afternoon $15$ times, and commuted by train (either morning or afternoon) $9$ times. Find $x$
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 16 \qquad \\ \textbf{(E)}\ \text{ not enough information given to solve the problem}$

Solution: The man has three possible combinations of transportation: \[\text{Morning train, Afternoon bus (m.t., a.b.)}\] \[\text{Morning bus, Afternoon train (m.b., a.t.)}\] \[\text{Morning bus, Afternoon bus (m.b, a.b.)}\]
Let $y$ be the number of times the man takes the $\text{a.t.}$ . Then, $9-y$ is the number of times he takes the $\text{m.t.}$ . Keep in mind that $\text{m.b.}=y$ and $\text{a.b.}=9-y$
Let $z$ be the number of times the man takes the $\text{m.b.}$ and $\text{a.b.}$ . Now, we get the two equations \[y+z=8\] and \[9-y+z=15.\]
Solving the system of equations, we get $y=1$ and $z=7$
So during the $x$ working days, the man took the $\text{(m.t., a.b.)}$ on $9-1=8$ days, the $\text{(m.b., a.t.)}$ on $1$ day, and the $\text{(m.b., a.b.)}$ on $7$ days.
Therefore, $x=8+1+7= \boxed{16}$ . ~ jiang147369

Answer: D 16
Problem: Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$
$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad  \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qquad \\ \textbf{(D)}\ \text{all positive } x \neq 1 \qquad  \textbf{(E)}\ \text{none of these}$

Solution: By the change-of-base formula, we can simplify the left side of the equation: $(\log_3x)(\log_x5) = (\frac{\log_x}{\log_3})(\frac{\log_5}{\log_x}) = \frac{\log_5}{\log_3}$
We see that this in fact simplifies to $\log_35$ , which will always equal the right side of the equation, since they are the same exact expressions.
But we have to be careful because $x \neq 1$ . Plugging in $x=1$ , the left side would equal $(\log_31)(\log_x5) = 0 \cdot \log_x5 = 0$ , and $\log_35$ definitely does not equal $0$
Besides $1$ $x$ can take on any positive value, and the equation would work. Therefore, the answer is $\boxed{1}$ jiang147369

Answer: D 1
Problem: If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$

Solution: If $p$ is a root of $x^3 - x^2 + x - 2 = 0$ , then $p^3 - p^2 + p - 2 = 0$ , or \[p^3 = p^2 - p + 2.\] Similarly, $q^3 = q^2 - q + 2$ , and $r^3 = r^2 - r + 2$ , so \[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]
By Vieta's formulas $p + q + r = 1$ $pq + pr + qr = 1$ , and $pqr = 2$ . Squaring the equation $p + q + r = 1$ , we get \[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\] Subtracting $2pq + 2pr + 2qr = 2$ , we get \[p^2 + q^2 + r^2 = -1.\]
Therefore, $p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1  + 6 = \boxed{4}$

Answer: E 4
Problem: If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$

Solution: We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$ . By Vieta's formulas, $p+q+r=1$ $pqr=2$ , and $pq+qr+pr=1$ .
So if we can find $p^2+q^2+r^2$ , we are done. Notice that $(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr$ , so $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1$ , which means that $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{4}$

Answer: E 4
Problem: If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$

Solution: Use Vieta's formulas to get $p+q+r=1$ $pq+qr+pr=1$ , and $pqr=2$
Square $p+q+r=1$ , and get $p^2+q^2+r^2+2pq+2pr+2qr=1$
Substitute $pq+qr+pr=1$ and simplify to get $-1=p^2+q^2+r^2$
After that, multiply both sides by $1=p+q+r$ , to get $-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q$
Then, factor out $pq$ $qr$ , and $pr$ $-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)$
Then, substitute the first equation into $p+q$ $q+r$ , and $p+r$ $-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)$
Then, multiply it out: $-1=p^3+q^3+r^3+pq+qr+pr-3pqr$
After that, substitute the equations $pq+qr+pr=1$ and $pqr=2$ $-1=p^3+q^3+r^3+1-6$
Solving that, you get $p^3+q^3+r^3=\boxed{4}$ ~EZ PZ Ms.Lemon SQUEEZY

Answer: E 4
Problem: What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$

Solution: $(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$ . Finally, multiply and add to find that the smallest integer higher is $\boxed{970}$

Answer: C 970
Problem: What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$

Solution: Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6$ . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\sqrt{3}-\sqrt{2})^6) < 1$ , the greatest integer greater than our original expression is $\boxed{970}$

Answer: C 970
Problem: Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$ $i=1, 2$ . Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \  } \frac{b}{3} \qquad \mathrm{(D) \  } 2b \qquad \mathrm{(E) \  }-\frac{b}{3}$

Solution: Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$ . By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{3}$

Answer: B 3
Problem: What is the remainder when $x^{51}+51$ is divided by $x+1$
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 49 \qquad \mathrm{(D) \  } 50 \qquad \mathrm{(E) \  }51$

Solution: From the Remainder Theorem , the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \boxed{50}$

Answer: D 50
Problem: Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$ , if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$ , find $\measuredangle EBC$
$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \  } 70^\circ \qquad \mathrm{(D) \  } 88^\circ \qquad \mathrm{(E) \  }92^\circ$

Solution: Since $ABCD$ is cyclic, opposite angles must sum to $180^\circ$ . Therefore, $\angle ADC+\angle ABC=180^\circ$ , and $\angle ABC=180^\circ-\angle ADC=180^\circ-68^\circ=112^\circ$ . Notice also that $\angle ABC$ and $\angle CBE$ form a linear pair, and so they sum to $180^\circ$ . Therefore, $\angle EBC=180^\circ-\angle ABC=180^\circ-112^\circ=68^\circ, \boxed{68}$ . Notice that the answer didn't even depend on $\angle BAD$

Answer: B 68
Problem: What is the smallest integral value of $k$ such that \[2x(kx-4)-x^2+6=0\] has no real roots?
$\mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } 4 \qquad \mathrm{(E) \  }5$

Solution: Expanding, we have $2kx^2-8x-x^2+6=0$ , or $(2k-1)x^2-8x+6=0$ . For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6}$ . From here, we can easily see that the smallest integral value of $k$ is $2, \boxed{2}$

Answer: B 2
Problem: If $i^2=-1$ , then $(1+i)^{20}-(1-i)^{20}$ equals
$\mathrm{(A)\ } -1024 \qquad \mathrm{(B) \ }-1024i \qquad \mathrm{(C) \  } 0 \qquad \mathrm{(D) \  } 1024 \qquad \mathrm{(E) \  }1024i$

Solution: Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,
\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \boxed{0}.\]

Answer: C 0
Problem: For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \  } 80400 \qquad \mathrm{(D) \  } 80600 \qquad \mathrm{(E) \  }80800$

Solution: The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$
We now want to evaluate $\sum_{p=1}^{10}(1600p-780)$ \[\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)\] \[=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{80200}.\]

Answer: B 80200
Problem: A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If $R$ is the ratio of the lesser part to the greater part, then the value of
\[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}\]
is
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }2R \qquad \mathrm{(C) \  } R^{-1} \qquad \mathrm{(D) \  } 2+R^{-1} \qquad \mathrm{(E) \  }2+R$

Solution: Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\frac{w}{l}=\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0$ . Therefore, $R^2+R-1=0$
From this, we have $R^2=-R+1$ . Dividing both sides by $R$ , we get $R=-1+\frac{1}{R}\implies R^{-1}=R+1$ . Therefore, $R^2+R^{-1}=-R+1+R+1=2$ . Finally, we have \[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \boxed{2}.\]

Answer: A 2
Problem: One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$

Solution: The total number of cubes is $10^3$ or $1000$ . Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$ , since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\boxed{488}$ cubes that have at least one face painted.

Answer: C 488
Problem: One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$

Solution: Each face has $100$ cubes, so multiply by six to get $600$ .  However, we overcounted each small cube on the edge but not on corner of the big cube once and each small cube on the corner of the big cube twice.  Thus, there are $600 - (12 \cdot 8 + 2 \cdot 8) = \boxed{488}$ cubes that have at least one face painted.

Answer: C 488
Problem: The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is
$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80$

Solution: We can guess and check small primes, subtract it from $126$ , and see if the result is a prime because the further away the two numbers are, the greater the difference will be.  Since $126 = 2 \cdot 3^2 \cdot 7$ , we can eliminate $2$ $3$ , and $7$ as an option because subtracting these would result in a composite number.
If we subtract $5$ , then the resulting number is $121$ , which is not prime.  If we subtract $11$ , then the resulting number is $115$ , which is also not prime.  But when we subtract $13$ , the resulting number is $113$ , a prime number.  The largest possible difference is $113-13=\boxed{100}$

Answer: B 100
Problem: If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$ , when written in base 10, equals
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$

Solution: Write out the numbers using the definition of base numbers \[554_b = 5b^2 + 5b + 4\] \[24_b = 2b+4\] Since $554_b = (24_b)^2$ , we can write an equation. \[5b^2 + 5b + 4 = (2b+4)^2\] \[5b^2 + 5b + 4 = 4b^2 + 16b + 16\] \[b^2 - 11b - 12 = 0\] \[(b-12)(b+1) = 0\] Since base numbers must be positive, $b$ in base 10 equals $\boxed{12}$

Answer: C 12
Problem: The sum of all integers between 50 and 350 which end in 1 is
$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$

Solution: The numbers that we are adding are $51,61,71 \cdots 341$ .  The numbers are part of an arithmetic series with first term $51$ , last term $341$ , common difference $10$ , and $30$ terms.  Using the arithmetic series formula, the sum of the terms is $\tfrac{30 \cdot 392}{2} = \boxed{5880}$

Answer: A 5880
Problem: If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high is
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$

Solution: The statues are similar, and since the height if the smaller statue is $\tfrac{1}{6}$ of the original statue, the surface area is $\tfrac{1}{36}$ of the original statue.  Thus, $\tfrac{1}{36}$ pints of paint is needed for one 1 ft. statue, so painting 540 of these statues requires $\tfrac{540}{36} = \boxed{15}$ pints of paint.

Answer: E 15
Problem: In $\triangle ABC$ with right angle at $C$ , altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$ , then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$

Solution: [asy] pair A=(-6,0),B=(6,0),C=(-3,5.196),M=(0,0),H=(-3,0); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-3,5.196)--(0,0)); draw(C--H); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); dot(M); label("$M$",M,S); dot(H); label("$H$",H,S);  markscalefactor=0.1; draw(anglemark((-6,0),C,(6,0))); draw((-3,0.5)--(-2.5,0.5)--(-2.5,0));  [/asy]
Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).
Note that because $CM$ is a median, $AM = BM$ .  Also, by ASA Congruency, $\triangle CHA = \triangle CHM$ , so $AH = HM$ .  That means $HM = \tfrac{1}{4} \cdot AB$ , and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{4}$

Answer: E 4