Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5));  for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num)  { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); }  draw_num((0,0), 17); draw_num((4, 0), 81);  draw_num((0, 4), 1);  draw_num((4,4), 25);   void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); }  foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " ");   [/asy]

Solution: The middle term of the first row is $\frac{25+1}{2}=13$ , since the middle number is just the average in an arithmetic sequence. Similarly, the middle of the bottom row is $\frac{17+81}{2}=49$ . Applying this again for the middle column, the answer is $\frac{49+13}{2}=\boxed{31}$

Answer: B 31
Problem: An arithmetic sequence is a sequence in which each term after the first is obtained by adding a constant to the previous term. For example, $2,5,8,11,14$ is an arithmetic sequence with five terms, in which the first term is $2$ and the constant added is $3$ . Each row and each column in this $5\times5$ array is an arithmetic sequence with five terms. The square in the center is labelled $X$ as shown. What is the value of $X$
$\textbf{(A) }21\qquad\textbf{(B) }31\qquad\textbf{(C) }36\qquad\textbf{(D) }40\qquad \textbf{(E) }42$
[asy] size(3.85cm); label("$X$",(2.5,2.1),N); for (int i=0; i<=5; ++i) draw((i,0)--(i,5), linewidth(.5));  for (int j=0; j<=5; ++j) draw((0,j)--(5,j), linewidth(.5)); void draw_num(pair ll_corner, int num)  { label(string(num), ll_corner + (0.5, 0.5), p = fontsize(19pt)); }  draw_num((0,0), 17); draw_num((4, 0), 81);  draw_num((0, 4), 1);  draw_num((4,4), 25);   void foo(int x, int y, string n) { label(n, (x+0.5,y+0.5), p = fontsize(19pt)); }  foo(2, 4, " "); foo(3, 4, " "); foo(0, 3, " "); foo(2, 3, " "); foo(1, 2, " "); foo(3, 2, " "); foo(1, 1, " "); foo(2, 1, " "); foo(3, 1, " "); foo(4, 1, " "); foo(2, 0, " "); foo(3, 0, " "); foo(0, 1, " "); foo(0, 2, " "); foo(1, 0, " "); foo(1, 3, " "); foo(1, 4, " "); foo(3, 3, " "); foo(4, 2, " "); foo(4, 3, " ");   [/asy]

Solution: The value of $X$ is simply the average of the average values of both diagonals that contain $X$ . This is $\frac{\frac{1+81}{2}+\frac{17+25}{2}}{2} =\frac{\frac{82}{2}+\frac{42}{2}}{2} = \frac{41+21}{2} = \boxed{31}$

Answer: B 31
Problem: Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: So, let there be $x$ pairs of $$1$ socks, $y$ pairs of $$3$ socks, and $z$ pairs of $$4$ socks.
We have $x+y+z=12$ $x+3y+4z=24$ , and $x,y,z \ge 1$
Now, we subtract to find $2y+3z=12$ , and $y,z \ge 1$ .
It follows that $2y$ is a multiple of $3$ and $3z$ is a multiple of $3$ . Since sum of 2 multiples of 3 = multiple of 3, so we must have $2y=6$
Therefore, $y=3$ , and it follows that $z=2$ . Now, $x=12-y-z=12-3-2=\boxed{7}$ , as desired.

Answer: D 7
Problem: Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: Since the total cost of the socks was $$24$ and Ralph bought $12$ pairs, the average cost of each pair of socks is $\frac{$24}{12} = $2$
There are two ways to make packages of socks that average to $$2$ . You can have:
$\bullet$ Two $$1$ pairs and one $$4$ pair (package adds up to $$6$
$\bullet$ One $$1$ pair and one $$3$ pair (package adds up to $$4$
Now, we need to solve \[6a+4b=24,\] where $a$ is the number of $$6$ packages and $b$ is the number of $$4$ packages. We see our only solution (that has at least one of each pair of sock) is $a=2, b=3$ , which yields the answer of $2\times2+3\times1 = \boxed{7}$

Answer: D 7
Problem: Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: Since there are 12 pairs of socks, and Ralph bought at least one pair of each, there are $12-3=9$ pairs of socks left. Also, the sum of the three pairs of socks is $1+3+4=8$ . This means that there are $24-8=16$ dollars left. If there are only $1$ dollar socks left, then we would have $9\cdot1=9$ dollars wasted, which leaves $7$ more dollars. If we replace one pair with a $3$ dollar pair, then we would waste an additional $2$ dollars. If we replace one pair with a $4$ dollar pair, then we would waste an additional $3$ dollars. The only way $7$ can be represented as a sum of $2$ s and $3$ s is $2+2+3$ . If we change $3$ pairs, we would have $6$ pairs left. Adding the one pair from previously, we have $\boxed{7}$ pairs.

Answer: D 7
Problem: Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: Let the amount of $1$ dollar socks be $a$ $3$ dollar socks be $b$ , and $4$ dollar socks be $c$ . We then know that $a+b+c=12$ and $a+3b+4c=24$ . We can make $a+b+c=12$ into $a=12-b-c$ and then plug that into the other equation, producing $12-b-c+3b+4c=24$ which simplifies to $2b+3c=12$ . It's not hard to see $b=3$ and $c=2$ . Now that we know $b$ and $c$ , we know that $a=7$ , meaning the number of $1$ dollar socks Ralph bought is $\boxed{7}$

Answer: D 7
Problem: Ralph went to the store and bought 12 pairs of socks for a total of $$24$ . Some of the socks he bought cost $$1$ a pair, some of the socks he bought cost $$3$ a pair, and some of the socks he bought cost $$4$ a pair. If he bought at least one pair of each type, how many pairs of $$1$ socks did Ralph buy?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$

Solution: If Ralph bought one sock of each kind, he already used $$8$ , so there are $$16$ left and 9 socks. If we split the $$16$ into four $$4$ sections, (as it is the smallest possible number that 1, 3, 4, can make in different ways that in all use at least each of the numbers once,) if Ralph bought a $$3$ pair, he would need to buy a $$1$ pair in order for it to add up to a multiple of four. Similarly, if Ralph bought a $$1$ pair, he would either need to buy three $$1$ pairs or a $$3$ pair. If Ralph bought a $$4$ pair, it would already make a group. Now, the problem is just how we can split 9 into 4 groups of 1, 2, or 4. We clearly see that $1 + 2 + 2 + 4 = 9$ , or a $$4$ pair, two $$3$ pairs, and six $$1$ pairs. Because we subtracted the necessary one of each kind, there are two $$4$ pairs, three $$3$ pairs, and seven $$1$ pairs. Therefore, the number of $$1$ pairs Ralph bought is $\boxed{7}$ .
~strongstephen

Answer: D 7
Problem: In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$

Solution: Clearly, since $\overline{FE}$ is a side of a square with area $32$ $\overline{FE} = \sqrt{32} = 4 \sqrt{2}$ . Now, since $\overline{FE} = \overline{BC}$ , we have $\overline{BC} = 4 \sqrt{2}$
Now, $\overline{JB}$ is a side of a square with area $18$ , so $\overline{JB} = \sqrt{18} = 3 \sqrt{2}$ . Since $\Delta JBK$ is equilateral, $\overline{BK} = 3 \sqrt{2}$
Lastly, $\Delta KBC$ is a right triangle. We see that $\angle JBA + \angle ABC + \angle CBK + \angle KBJ = 360^{\circ} \rightarrow 90^{\circ} + 120^{\circ} + \angle CBK + 60^{\circ} = 360^{\circ} \rightarrow \angle CBK = 90^{\circ}$ , so $\Delta KBC$ is a right triangle with legs $3 \sqrt{2}$ and $4 \sqrt{2}$ . Now, its area is $\dfrac{3 \sqrt{2} \cdot 4 \sqrt{2}}{2} = \dfrac{24}{2} = \boxed{12}$

Answer: C 12
Problem: In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$

Solution: Since $\overline{FE} = \sqrt{32}$ , and $\overline{FE} = \overline{BC}$ $\overline{BC} = 4\sqrt{2}$ . Meanwhile, $\overline{JB} = 3\sqrt{2}$ , and since $\triangle JBK$ is equilateral, $\overline{BK} = 3\sqrt{2}$ . If $ABCDEF$ is equiangular, $\angle ABC = \frac{180 \cdot (n-2)}{n} = 120^{\circ}$ , where $n$ is the number of sides of the shape. Adding all the angles around $B$ gives $270^{\circ}$ , so $\angle KBC = 360 - 90 = 270^{\circ}$ . Because $\triangle KBC$ is right, the area of $\triangle KBC = \frac{4\sqrt{2} \cdot 3\sqrt{2}} {2} = 12$ . Therefore, the answer is $\boxed{12}$ . ~strongstephen

Answer: C 12
Problem: On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

Solution: The text suggests that the number of students in the group has $12$ factors, since each arrangement is a factor. The smallest integer with $12$ factors is $2^2\cdot3\cdot5=\boxed{60}$

Answer: C 60
Problem: On June 1, a group of students is standing in rows, with 15 students in each row. On June 2, the same group is standing with all of the students in one long row. On June 3, the same group is standing with just one student in each row. On June 4, the same group is standing with 6 students in each row. This process continues through June 12 with a different number of students per row each day. However, on June 13, they cannot find a new way of organizing the students. What is the smallest possible number of students in the group?
$\textbf{(A) } 21 \qquad \textbf{(B) } 30 \qquad \textbf{(C) } 60 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 1080$

Solution: Since we know the number must be a multiple of $15$ , we can eliminate $A$ . We also know that after $12$ days, the students can't find any more arrangements, meaning the number has $12$ factors. Now, we just list the factors of every number, starting with $30$ \[30=1\cdot30, 2\cdot15, 3\cdot10, 5\cdot6\] \[60=1\cdot60, 2\cdot30, 3\cdot20, 4\cdot15, 5\cdot12, 6\cdot10\] $60$ has $12$ factors, so the answer is $\boxed{60}$

Answer: C 60
Problem: A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solution: On one team they play $3N$ games in their division and $4M$ games in the other.  This gives $3N+4M=76$
Since $M>4$ we start by trying M=5. This doesn't work because $56$ is not divisible by $3$
Next, $M=6$ does not work because $52$ is not divisible by $3$
We try $M=7$ does work by giving $N=16$ $~M=7$ and thus $3\times 16=\boxed{48}$ games in their division.

Answer: B 48
Problem: A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solution: $76=3N+4M > 10M$ , giving $M \le 7$ .
Since $M>4$ , we have $M=5,6,7$ .
Since $4M$ is $1$ $\pmod{3}$ , we must have $M$ equal to $1$ $\pmod{3}$ , so $M=7$
This gives $3N=48$ , as desired. The answer is $\boxed{48}$

Answer: B 48
Problem: A baseball league consists of two four-team divisions. Each team plays every other team in its division $N$ games. Each team plays every team in the other division $M$ games with $N>2M$ and $M>4$ . Each team plays a $76$ game schedule. How many games does a team play within its own division?
$\textbf{(A) } 36 \qquad \textbf{(B) } 48 \qquad \textbf{(C) } 54 \qquad \textbf{(D) } 60 \qquad \textbf{(E) } 72$

Solution: Notice that each team plays $N$ games against each of the three other teams in its division. So that's $3N$
Since each team plays $M$ games against each of the four other teams in the other division, that's $4M$
So $3N+4M=76$ , with $M>4, N>2M$
Let's start by solving this Diophantine equation. In other words, $N=\frac{76-4M}{3}$
So $76-4M\equiv0 \pmod{3}$ (remember: $M$ must be divisible by 3 for $N$ to be an integer!). Therefore, after reducing $76$ to $1$ and $-4M$ to $2M$ (we are doing things in $\pmod{3}$ ), we find that $M\equiv1 \pmod{3}$
Since $M>4$ , so the minimum possible value of $M$ is $7$ . However, remember that $N>2M$ ! To find the greatest possible value of M, we assume that $N=2M$ and that is the upper limit of $M$ (excluding that value because $N>2M$ ). Plugging $N=2M$ in, $10M=76$ . So $M<7.6$ . Since you can't have $7.6$ games, we know that we can only check $M=7$ since we know that since $M>4, M<7.6, M\equiv1 \pmod{3}$ . After checking $M=7$ , we find that it works.
So $M=7, N=16$ . So each team plays 16 games against each team in its division. Since they are asking for games in it division, which equals $3n = 48$ . Select $\boxed{48}$

Answer: B 48
Problem: One-inch squares are cut from the corners of this 5 inch square.  What is the area in square inches of the largest square that can be fitted into the remaining space?
[asy] size(75); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]
$\textbf{(A)  } 9\qquad \textbf{(B)  } 12\frac{1}{2}\qquad \textbf{(C)  } 15\qquad \textbf{(D)  } 15\frac{1}{2}\qquad \textbf{(E)  } 17$

Solution: Proceed as in solution 1 to get $x^2 - 5x + 5 = 0$
Notice that the side length of the square is $\sqrt{x^2+(5-x)^2}$ by the Pythagorean Theorem. Thus, the area of the square is \[\left(\sqrt{x^2+(5-x)^2}\right)^2\] \[= x^2 + x^2 - 10x + 25\] \[= (2x^2 - 10x + 10) + 15\] \[= 2(x^2 - 5x + 5) + 15\] \[= 2(0) + 15\] \[= \boxed{15}\]

Answer: C 15
Problem: Harry and Terry are each told to calculate $8-(2+5)$ . Harry gets the correct answer. Terry ignores the parentheses and calculates $8-2+5$ . If Harry's answer is $H$ and Terry's answer is $T$ , what is $H-T$
$\textbf{(A) }-10\qquad\textbf{(B) }-6\qquad\textbf{(C) }0\qquad\textbf{(D) }6\qquad \textbf{(E) }10$

Solution: We have $H=8-7=1$ and $T=8-2+5=11$ . Clearly $1-11=-10$ , so our answer is $\boxed{10}$

Answer: A 10
Problem: Paul owes Paula $35$ cents and has a pocket full of $5$ -cent coins, $10$ -cent coins, and $25$ -cent coins that he can use to pay her. What is the difference between the largest and the smallest number of coins he can use to pay her?
$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution: The fewest amount of coins that can be used is $2$ (a quarter and a dime). The greatest amount is $7$ , if he only uses nickels. Therefore we have $7-2=\boxed{5}$

Answer: E 5
Problem: Isabella had a week to read a book for a school assignment. She read an average of $36$ pages per day for the first three days and an average of $44$ pages per day for the next three days. She then finished the book by reading $10$ pages on the last day. How many pages were in the book?
$\textbf{(A) }240\qquad\textbf{(B) }250\qquad\textbf{(C) }260\qquad\textbf{(D) }270\qquad \textbf{(E) }280$

Solution: Isabella read $3\cdot 36+3\cdot 44$ pages in the first 6 days. Although this can be calculated directly, it is simpler to calculate it as $3\cdot (36+44)=3\cdot 80$ , which gives that she read $240$ pages. On her last day, she read $10$ more pages for a total of $240+10=\boxed{250}$ pages.

Answer: B 250
Problem: The sum of two prime numbers is $85$ . What is the product of these two prime numbers?
$\textbf{(A) }85\qquad\textbf{(B) }91\qquad\textbf{(C) }115\qquad\textbf{(D) }133\qquad \textbf{(E) }166$

Solution: Since the two prime numbers sum to an odd number, one of them must be even. The only even prime number is $2$ . The other prime number is $85-2=83$ , and the product of these two numbers is $83\cdot2=\boxed{166}$

Answer: E 166
Problem: Margie's car can go $32$ miles on a gallon of gas, and gas currently costs $ $4$ per gallon. How many miles can Margie drive on $\textdollar 20$ worth of gas?
$\textbf{(A) }64\qquad\textbf{(B) }128\qquad\textbf{(C) }160\qquad\textbf{(D) }320\qquad \textbf{(E) }640$

Solution: Margie can afford $20/4=5$ gallons of gas. She can go $32\cdot5=\boxed{160}$ miles on this amount of gas.

Answer: C 160
Problem: Six rectangles each with a common base width of $2$ have lengths of $1, 4, 9, 16, 25$ , and $36$ . What is the sum of the areas of the six rectangles?
$\textbf{(A) }91\qquad\textbf{(B) }93\qquad\textbf{(C) }162\qquad\textbf{(D) }182\qquad \textbf{(E) }202$

Solution: The sum of the areas is equal to $2\cdot1+2\cdot4+2\cdot9+2\cdot16+2\cdot25+2\cdot36$ . This is equal to $2(1+4+9+16+25+36)$ , which is equal to $2\cdot91$ . This is equal to our final answer of $\boxed{182}$

Answer: D 182
Problem: Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

Solution: Since all the eleven members paid the same amount, that means that the total must be divisible by $11$ . We can do some trial-and-error to get $A=3$ , so our answer is $\boxed{3}$ ~SparklyFlowers

Answer: D 3
Problem: Eleven members of the Middle School Math Club each paid the same amount for a guest speaker to talk about problem solving at their math club meeting. They paid their guest speaker $\textdollar\underline{1} \underline{A} \underline{2}$ . What is the missing digit $A$ of this $3$ -digit number?
$\textbf{(A) }0\qquad\textbf{(B) }1\qquad\textbf{(C) }2\qquad\textbf{(D) }3\qquad \textbf{(E) }4$

Solution: We know that a number is divisible by $11$ if the odd digits added together minus the even digits added together (or vice versa) is a multiple of $11$ . Thus, we have $1+2-A$ = a multiple of $11$ . The only multiple that works here is $0$ , as $11 \cdot 0 = 0$ . Thus, $A = \boxed{3}$ ~fn106068

Answer: D 3
Problem: The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$

Solution: The seventh AMC 8 would have been given in $1991$ . If Samantha was 12 then, that means she was born 12 years ago, so she was born in $1991-12=1979$
Our answer is $\boxed{1979}$ corrections made by DrDominic

Answer: A 1979
Problem: The first AMC $8$ was given in $1985$ and it has been given annually since that time. Samantha turned $12$ years old the year that she took the seventh AMC $8$ . In what year was Samantha born?
$\textbf{(A) }1979\qquad\textbf{(B) }1980\qquad\textbf{(C) }1981\qquad\textbf{(D) }1982\qquad \textbf{(E) }1983$

Solution: Since she was 12 when she took the seventh AMC 8, she should be $12-6=6$ years old when the first AMC 8 occurred. Therefore, she was born or was 'age 0' in $1985-6=\boxed{1979}$ ~SweetMango77
corrections made by DrDominic

Answer: A 1979