Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution: Let's arrange all the teams in a circle and assume that each team won against the first 10 teams clockwise of themselves, and lost against the first 10 teams counter-clockwise of themselves.
Consider a working set of $3$ teams: moving clockwise, we know that the first team beat the team clockwise of itself, that team beat the next team clockwise of itself, and the final team beat the first team, which would be clockwise of itself. However, we also must remember that if the first team beat the second team, moving clockwise, the second team cannot be more than $10$ teams away from the first team; the same applies to the second and third team, and the third and first team.
Let's say, WLOG, that the first team, team $A$ , is at position $0$ out of $20$ on the circle.
If team $A$ is to beat team $B$ , since we are assuming each team beats the 10 members clockwise of themselves, there are $10$ places on the circle that team $B$ could be: positions $1$ through $10$ . Also, if team $C$ is to beat team $A$ , team $C$ must be located from positions $11$ to $20$
If Team $B$ is in position $n$ $C$ must be located from position $n+1$ to position $n+10$ , since $B$ beats $C$ . We also just found that $C$ 's position must be between $11$ and $20$ inclusive. So, when $B$ is in position $1$ $C$ can only be in position $11$ . When $B$ is in position $2$ $C$ can be in position $11$ or $12$ . In general, when $B$ is in position $n$ , there are $n$ choices for where $C$ can be. So, there are $1+2+3+ \dots + 10 = 55$ ways to place $B$ and $C$ . There are $21$ players to choose as player $A$ , and each working set will be counted $3$ times, so our final answer is $55 \cdot 21 \div 3 = \boxed{385,}$

Answer: A 385,
Problem: A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$ $B$ beat $C$ , and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$

Solution: This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$ , there are $10$ teams that lost to $A$ and $10$ teams that won over $A$ . Call the group of teams that lost to $A$ group $L$ , and the group of teams that won over $A$ group $W$
[asy] size(7cm); label("A",(20,5)); label("Group W",(27,0)); label("Group L",(13,0)); draw((19,4)--(16,1),EndArrow); draw((16,0)--(24,0),EndArrow); draw((24,1)--(21,4),EndArrow); [/asy]
Any team from group $L$ that won a team from group $W$ will form a cycle with $A$ . Now we need to count how many teams in group $L$ won over a team from group $W$ . The total number of wins in group $L$ is $10 \cdot 10 =100$ . There are $\tbinom{10}2=45$ wins among the teams inside group $L$ . So group $L$ has $100-45=55$ wins over group $W$
\[\frac{21 \cdot 55}{3}=\boxed{385,}\]

Answer: A 385,
Problem: How many four-digit integers $abcd$ , with $a \neq 0$ , have the property that the three two-digit integers $ab<bc<cd$ form an increasing arithmetic sequence? One such number is $4692$ , where $a=4$ $b=6$ $c=9$ , and $d=2$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$

Solution: The numbers are $10a+b, 10b+c,$ and $10c+d$ . Note that only $d$ can be zero, the numbers $ab$ $bc$ , and $cd$ cannot start with a zero, and $a\le b\le c$
To form the sequence, we need $(10c+d)-(10b+c)=(10b+c)-(10a+b)$ . This can be rearranged as $10(c-2b+a)=2c-b-d$ . Notice that since the left-hand side is a multiple of $10$ , the right-hand side can only be $0$ or $10$ . (A value of $-10$ would contradict $a\le b\le c$ .) Therefore we have two cases: $a+c-2b=1$ and $a+c-2b=0$
Case 1
If $c=9$ , then $b+d=8,\ 2b-a=8$ , so $5\le b\le 8$ . This gives $2593, 4692, 6791, 8890$ .
If $c=8$ , then $b+d=6,\ 2b-a=7$ , so $4\le b\le 6$ . This gives $1482, 3581, 5680$ .
If $c=7$ , then $b+d=4,\ 2b-a=6$ , so $b=4$ , giving $2470$ .
There is no solution for $c=6$ .
Added together, this gives us $8$ answers for Case 1.
Case 2
This means that the digits themselves are in an arithmetic sequence. This gives us $9$ answers, \[1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.\] Adding the two cases together, we find the answer to be $8+9=$ $\boxed{17}$

Answer: D 17
Problem: How many four-digit integers $abcd$ , with $a \neq 0$ , have the property that the three two-digit integers $ab<bc<cd$ form an increasing arithmetic sequence? One such number is $4692$ , where $a=4$ $b=6$ $c=9$ , and $d=2$
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 20$

Solution: Looking at the answer options, all the numbers are pretty small so it is easy to make a list.
$12|23|34 \rightarrow 1234$
$13|35|57 \rightarrow 1357$
$14|48|82 \rightarrow 1482$
$23|34|45 \rightarrow 2345$
$24|46|68 \rightarrow 2468$
$24|47|70 \rightarrow 2470$
$25|59|93 \rightarrow 2593$
$34|45|56 \rightarrow 3456$
$35|57|79 \rightarrow 3579$
$35|58|81 \rightarrow 3581$
$45|56|67 \rightarrow 4567$
$46|69|92 \rightarrow 4692$
$56|67|78 \rightarrow 5678$
$56|68|80 \rightarrow 5680$
$67|78|89 \rightarrow 6789$
$67|79|91 \rightarrow 6791$
$88|89|90 \rightarrow 8890$
Counting all the cases we get our answer of $17$ which is $\boxed{17}$ -srisainandan6

Answer: D 17
Problem: A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution: Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.
Triangles have $3$ edges and squares have $4$ edges, so we have a system of equations.
We have $a + b$ tiles total, so $a + b = 25$
We have $3a + 4b$ edges total, so $3a + 4b = 84$
Multiplying the first equation by $3$ on both sides gives $3a + 3b = 3(25) = 75$
Second equation minus the first equation gives $b = 9$ , so the answer is $\boxed{9}$

Answer: D 9
Problem: A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution: If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there needs to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\boxed{9}$

Answer: D 9
Problem: A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$

Solution: Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$ . There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$ . Together, the total number of edges is $4x+3(25-x)=84$ . Solving our equation, we get that $x=9$ which means that our answer is $\boxed{9}$

Answer: D 9
Problem: Ann made a $3$ -step staircase using $18$ toothpicks as shown in the figure. How many toothpicks does she need to add to complete a $5$ -step staircase?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 18\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 22\qquad\textbf{(E)}\ 24$
[asy] size(150); defaultpen(linewidth(0.8)); path h = ellipse((0.5,0),0.45,0.015), v = ellipse((0,0.5),0.015,0.45); for(int i=0;i<=2;i=i+1) { for(int j=0;j<=3-i;j=j+1) { filldraw(shift((i,j))*h,black); filldraw(shift((j,i))*v,black); } }[/asy]

Solution: We can see that a $1$ -step staircase requires $4$ toothpicks and a $2$ -step staircase requires $10$ toothpicks. Thus, to go from a $1$ -step to $2$ -step staircase, $6$ additional toothpicks are needed and to go from a $2$ -step to $3$ -step staircase, $8$ additional toothpicks are needed. Applying this pattern, to go from a $3$ -step to $4$ -step staircase, $10$ additional toothpicks are needed and to go from a $4$ -step to $5$ -step staircase, $12$ additional toothpicks are needed. Our answer is $10+12=\boxed{22}$

Answer: D 22
Problem: Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$

Solution: If the average of the first $14$ peoples' scores was $80$ , then the sum of all of their tests is $14 \cdot 80 = 1120$ .  When Payton's score was added, the sum of all of the scores became $15 \cdot 81 = 1215$ .  So, Payton's score must be $1215-1120 = \boxed{95}$

Answer: E 95
Problem: Mr. Patrick teaches math to $15$ students. He was grading tests and found that when he graded everyone's test except Payton's, the average grade for the class was $80$ . After he graded Payton's test, the test average became $81$ . What was Payton's score on the test?
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 85\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 94\qquad\textbf{(E)}\ 95$

Solution: The average of a set of numbers is the value we get if we evenly distribute the total across all entries. So assume that the first $14$ students each scored $80$ . If Payton also scored an $80$ , the average would still be $80$ . In order to increase the overall average to $81$ , we need to add one more point to all of the scores, including Payton's. This means we need to add a total of $15$ more points, so Payton needs $80+15 = \boxed{95}$

Answer: E 95
Problem: The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$

Solution: Let $a$ be the bigger number and $b$ be the smaller.
$a + b = 5(a - b)$
Multiplying out gives $a + b = 5a - 5b$ and rearranging gives $4a = 6b$ and factorised into $2a = 3b$ and then solving gives
$\frac{a}{b} = \frac32$ , so the answer is $\boxed{32}$

Answer: B 32
Problem: The sum of two positive numbers is $5$ times their difference. What is the ratio of the larger number to the smaller number?
$\textbf{(A)}\ \frac{5}{4}\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{9}{5}\qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ \frac{5}{2}$

Solution: Without loss of generality, let the two numbers be $3$ and $2$ , as they clearly satisfy the condition of the problem. The ratio of the larger to the smaller is $\boxed{32}$

Answer: B 32
Problem: How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution: $73-13 = 60$ , so the amount of terms in the sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$ is the same as in the sequence $0$ $3$ $6$ $\dotsc$ $57$ $60$
In this sequence, the terms are the multiples of $3$ going up to $60$ , and there are $20$ multiples of $3$ in $60$
However, the number 0 must also be included, adding another multiple. So, the answer is $\boxed{21}$

Answer: B 21
Problem: How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution: Using the formula for arithmetic sequence's nth term, we see that $a + (n-1)d \Longrightarrow13 + (n-1)3 =73, \Longrightarrow n = 21$ $\boxed{21}$

Answer: B 21
Problem: How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution: Minus each of the terms by $12$ to make the the sequence $1 , 4 , 7,..., 61$ $\frac{61-1}{3}=20, 20 + 1 = 21$
$\boxed{21}$

Answer: B 21
Problem: How many terms are in the arithmetic sequence $13$ $16$ $19$ $\dotsc$ $70$ $73$
$\textbf{(A)}\ 20 \qquad\textbf{(B)} \ 21 \qquad\textbf{(C)} \ 24 \qquad\textbf{(D)} \ 60 \qquad\textbf{(E)} \ 61$

Solution: Subtract each of the terms by $10$ to make the sequence $3 , 6 , 9,..., 60, 63$ . Then divide the each term in the sequence by $3$ to get $1, 2, 3,..., 20, 21$ . Now it is clear to see that there are $21$ terms in the sequence. $\boxed{21}$

Answer: B 21
Problem: Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ $1$
$\textbf{(A)}\ 2 \qquad\textbf{(B)} \ 4 \qquad\textbf{(C)} \ 5 \qquad\textbf{(D)} \ 6 \qquad\textbf{(E)} \ 8$

Solution: This problem can be converted to a system of equations. Let $p$ be Pete's current age and $c$ be Claire's current age.
The first statement can be written as $p-2=3(c-2)$ . The second statement can be written as $p-4=4(c-4)$
To solve the system of equations:
$p=3c-4$
$p=4c-12$
$3c-4=4c-12$
$c=8$
$p=20.$
Let $x$ be the number of years until Pete is twice as old as his cousin.
$20+x=2(8+x)$
$20+x=16+2x$
$x=4$
The answer is $\boxed{4}$

Answer: B 4
Problem: Two right circular cylinders have the same volume. The radius of the second cylinder is $10\%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?
$\textbf{(A)}\  \text{The second height is } 10\% \text{ less than the first.} \\ \textbf{(B)}\  \text{The first height is } 10\% \text{ more than the second.}\\ \textbf{(C)}\ \text{The second height is } 21\% \text{ less than the first.} \\ \textbf{(D)}\ \text{The first height is } 21\% \text{ more than the second.}\\ \textbf{(E)}\ \text{The second height is } 80\% \text{ of the first.}$

Solution: Let the radius of the first cylinder be $r_1$ and the radius of the second cylinder be $r_2$ . Also, let the height of the first cylinder be $h_1$ and the height of the second cylinder be $h_2$ . We are told \[r_2=\frac{11r_1}{10}\] \[\pi r_1^2h_1=\pi r_2^2h_2\] Substituting the first equation into the second and dividing both sides by $\pi$ , we get \[r_1^2h_1=\frac{121r_1^2}{100}h_2\implies h_1=\frac{121h_2}{100}.\] Therefore, $\boxed{21}$

Answer: D 21
Problem: How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet?  For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: The first thing one would want to do is place a possible letter that works and then stem off of it.  For example, if we start with an $a$ , we can only place a $c$ or $d$ next to it.  Unfortunately, after that step, we can't do too much, since:
$acbd$ is not allowed because of the $cb$ , and $acdb$ is not allowed because of the $cd$
We get the same problem if we start with a $d$ , since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$
If we start with a $b$ , the next letter would have to be a $d$ , and since we can put an $a$ next to it and then a $c$ after that, this configuration works.  The same approach applies if we start with a $c$
So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$ , giving us:
\[1 + 1 = \boxed{2}.\]

Answer: C 2
Problem: How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet?  For example, no such rearrangements could include either $ab$ or $ba$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Case 1: the first letter is A
Subcase 1: the second letter is C
The next letter must either be B or D, both of which do not satisfy the conditions.
Subcase 2: the second letter is D
The third letter is forced to be B and the fourth is forced to be C, but this doesn't work because B and C are next to each other.
Case 2: the first letter is B
The next letter is forced to be D, the third letter forced to be A and the last forced to be C. This works.
Case 3: the first letter is C
The next letter is forced to be A, the third letter is D and the last letter is B. This works.
Case 4: the first letter is D
Subcase 1: the second letter is A
The third letter has to be C and the fourth has to be B but this doesn't work because B and C are next to each other.
Subcase 2: the second letter is B
The third letter cannot be A or C, so this doesn't work.
Summing the cases, there are two that work: $BDAC$ and $CADB$ $\Longrightarrow \boxed{2}$

Answer: C 2
Problem: Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$

Solution: Since points on the graph make the equation true, substitute $\sqrt{\pi}$ in to the equation and then solve to find $a$ and $b$
$y^2 + \sqrt{\pi}^4 = 2\sqrt{\pi}^2 y + 1$
$y^2 + \pi^2 = 2\pi y + 1$
$y^2 - 2\pi y + \pi^2 = 1$
$(y-\pi)^2 = 1$
$y-\pi = \pm 1$
$y = \pi + 1$
$y = \pi - 1$
There are only two solutions to the equation $(y-\pi)^2 = 1$ , so one of them is the value of $a$ and the other is $b$ . The order does not matter because of the absolute value sign.
$| (\pi + 1) - (\pi - 1) | = 2$
The answer is $\boxed{2}$

Answer: C 2
Problem: Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$

Solution: This solution is very closely related to Solution #1 and just simplifies the problem earlier to make it easier.
$y^2 + x^4 = 2x^2 y + 1$ can be written as $x^4-2x^2y+y^2=1$ . Recognizing that this is a binomial square, simplify this to $(x^2-y)^2=1$ . This gives us two equations:
$x^2-y=1$ and $x^2-y=-1$
One of these $y$ 's is $a$ and one is $b$ . Substituting $\sqrt{\pi}$ for $x$ , we get $a=\pi+1$ and $b=\pi-1$
So, $|a-b|=|(\pi+1)-(\pi-1)|=2$
The answer is $\boxed{2}$

Answer: C 2
Problem: Points $( \sqrt{\pi} , a)$ and $( \sqrt{\pi} , b)$ are distinct points on the graph of $y^2 + x^4 = 2x^2 y + 1$ . What is $|a-b|$
$\textbf{(A)}\ 1 \qquad\textbf{(B)} \ \frac{\pi}{2} \qquad\textbf{(C)} \ 2 \qquad\textbf{(D)} \ \sqrt{1+\pi} \qquad\textbf{(E)} \ 1 + \sqrt{\pi}$

Solution: This solution is similar to Solution #1 but uses a different way to find $y$ at the end.
Just like Solution #1, we arrive at the conclusion that $y^2 - 2\pi y + \pi^2 = 1$
Simplifying we get:
$y^2 - 2\pi y + \pi^2 -1 = 0$
We now can factor this quadratic. We must find two terms that multiply to $\pi^2 -1$ and add to $2\pi$
These terms are $\pi+1$ and $\pi-1$
Subtracting one from the other, we get $2$
Thus, the answer is $\boxed{2}$

Answer: C 2
Problem: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution: Let Claudia have $x$ 5-cent coins and $\left( 12 - x \right)$ 10-cent coins. It is easily observed that any multiple of $5$ between $5$ and $5x + 10(12 - x) = 120 - 5x$ inclusive can be obtained by a combination of coins. Thus, $24 - x = 17$ combinations can be made, so $x = 7$ . But the answer is not $7,$ because we are asked for the number of 10-cent coins, which is $12 - 7 = \boxed{5}$

Answer: C 5
Problem: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution: Since the coins are 5-cent and 10-cent, all possible values that can be made will be multiples of $5.$ To have exactly $17$ different multiples of $5,$ we will need to make up to $85$ cents. If all twelve coins were 5-cent coins, we will have $60$ cents possible. Each trade of a 5-cent coin for a 10-cent coin will gain $5$ cents, and as we need to gain $25$ cents, the answer is $\boxed{5}$

Answer: C 5
Problem: Claudia has 12 coins, each of which is a 5-cent coin or a 10-cent coin. There are exactly 17 different values that can be obtained as combinations of one or more of her coins. How many 10-cent coins does Claudia have?
$\textbf{(A) }3\qquad\textbf{(B) }4\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }7$

Solution: Notice that for every $d$ dimes, any multiple of $5$ less than or equal to $10d + 5(12-d)$ is a valid arrangement. Since there are $17$ in our case, we have $10d + 5(12-d)=17 \cdot 5 \Rightarrow d=5$ . Therefore, the answer is $\boxed{5}$

Answer: C 5
Problem: The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
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$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$

Solution: The circumference of the clock is twice that of the disk. So, a quarter way around the clock (3:00), the point halfway around the disk will be tangent. The arrow will point to the left. We can see the disk made a 75% rotation from 12 to 3, and 3 is 75% of 4, so it would make 100% rotation from 12 to 4. The answer is $\boxed{4}$ . Similarly, the arrow would be pointing downward at 6:00. It would already have completed three 180 degree turns. Therefore, two 180 degree turns would be completed at 4:00.

Answer: C 4