Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $S$ be a square of side length 1. Two points are chosen independently at random on the sides of $S$ . The probability that the straight-line distance between the points is at least $\frac12$ is $\frac{a-b\pi}{c}$ , where $a,b,$ and $c$ are positive integers and $\text{gcd}(a,b,c) = 1$ . What is $a+b+c$
$\textbf{(A)}\ 59 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 61 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 63$

Solution: Choose a certain side for one of the points to be on. Let the distance from the point to the vertex on its left be $x$
We split this into two cases:
Case 1: $0\leq x\leq \frac12$
The total length of the segments for which the other point can be on such that the straight-line distance between the points is less than $\frac12$ is \[\sqrt{\frac14-x^2}+x+\frac12.\] We can graph this in the Cartesian plane and find the area of the region below the curve and above the line $y=0$
Case 2: $\frac12< x\leq 1$
This is basically Case 1 but flipped over the line $x=\frac12$
So our total probability is 1 minus the area of the graph over the total area (4, perimeter of square). Notice that the desired area of the region below the curve we found earlier is the sum of a quarter circle with radius $\frac12$ and centered at $(0,0)$ and a trapezoid with height $\frac12$ and bases of length $\frac12$ and $\frac32$ . Adding this all up then multiplying by 2, we have \[\frac{\pi}{8}+\frac34\] and then the probability of the desired result would be \[1-\frac{\pi+6}{32}=\frac{26-\pi}{32}\] and our answer is $26+1+32=\boxed{59}$ . ~caroline2023

Answer:  59
Problem: Isaac has written down one integer two times and another integer three times. The sum of the five numbers is 100, and one of the numbers is 28. What is the other number?
$\textbf{(A)}\; 8 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 14 \qquad\textbf{(D)}\; 15 \qquad\textbf{(E)}\; 18$

Solution: Let $a$ be the number written two times, and $b$ the number written three times. Then $2a + 3b = 100$ . Plugging in $a = 28$ doesn't yield an integer for $b$ , so it must be that $b = 28$ , and we get $2a + 84 = 100$ . Solving for $a$ , we obtain $a = \boxed{8}$

Answer: A 8
Problem: Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?
$\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75$

Solution: There are a total of $(12+1) \times (12+1) = 169$ products, and a product is odd if and only if both its factors are odd. There are $6$ odd numbers between $0$ and $12$ , namely $1, 3, 5, 7, 9, 11,$ hence the number of odd products is $6 \times 6 = 36$ . Therefore the answer is $36/169 \doteq \boxed{0.21}$

Answer: A 0.21
Problem: Back in 1930, Tillie had to memorize her multiplication facts from $0 \times 0$ to $12 \times 12$ . The multiplication table she was given had rows and columns labeled with the factors, and the products formed the body of the table. To the nearest hundredth, what fraction of the numbers in the body of the table are odd?
$\textbf{(A)}\; 0.21 \qquad\textbf{(B)}\; 0.25 \qquad\textbf{(C)}\; 0.46 \qquad\textbf{(D)}\; 0.50 \qquad\textbf{(E)}\; 0.75$

Solution: Note that if we had an $11$ by $11$ multiplication table, the fraction of odd products becomes $0.25$ . If we add the $12$ by $12$ , the fraction of odd products decreases. Because $0.21$ is the only option less than $0.25$ , our answer is $\boxed{0.21}$

Answer: A 0.21
Problem: A regular 15-gon has $L$ lines of symmetry, and the smallest positive angle for which it has rotational symmetry is $R$ degrees. What is $L+R$
$\textbf{(A)}\; 24 \qquad\textbf{(B)}\; 27 \qquad\textbf{(C)}\; 32 \qquad\textbf{(D)}\; 39 \qquad\textbf{(E)}\; 54$

Solution: From consideration of a smaller regular polygon with an odd number of sides (e.g. a pentagon), we see that the lines of symmetry go through a vertex of the polygon and bisect the opposite side. Hence $L=15$ , the number of sides / vertices. The smallest angle for a rotational symmetry transforms one side into an adjacent side, hence $R = 360^\circ / 15 = 24^\circ$ , the number of degrees between adjacent sides. Therefore the answer is $L + R = 15 + 24 = \boxed{39}$

Answer: D 39
Problem: What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$

Solution: $(625^{\log_5 2015})^\frac{1}{4} = ((5^4)^{\log_5 2015})^\frac{1}{4} = (5^{4 \cdot \log_5 2015})^\frac{1}{4} = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} = ((5^{\log_5 2015})^4)^\frac{1}{4} = (2015^4)^\frac{1}{4} = \boxed{2015}$

Answer: D 2015
Problem: What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$

Solution: We can rewrite $\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$

Answer:  2015
Problem: What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$

Solution: $(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{2015}$

Answer: D 2015
Problem: What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$
$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$

Solution: We note that the year number is just $2015$ , so just guess $\boxed{2015}$

Answer: D 2015
Problem: How many noncongruent integer-sided triangles with positive area and perimeter less than 15 are neither equilateral, isosceles, nor right triangles?
$\textbf{(A)}\; 3 \qquad\textbf{(B)}\; 4 \qquad\textbf{(C)}\; 5 \qquad\textbf{(D)}\; 6 \qquad\textbf{(E)}\; 7$

Solution: Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$ . Furthermore, "positive area" tells us that $c < a + b$ and the perimeter constraints means $a+b+c < 15$
There are no triangles when $a = 1$ because then $c$ must be less than $b+1$ , implying that $b \geq c$ , contrary to $b < c$
When $a=2$ , similar to above, $c$ must be less than $b+2$ , so this leaves the only possibility $c = b+1$ . This gives 3 triangles $(2,3,4), (2,4,5), (2,5,6)$ within our perimeter constraint.
When $a=3$ $c$ can be $b+1$ or $b+2$ , which gives triangles $(3,4,5), (3,4,6), (3,5,6)$ . Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles.
All in all, this gives us $3+2 = \boxed{5}$ triangles.

Answer: C 5
Problem: Quadrilateral $ABCD$ is inscribed in a circle with $\angle BAC=70^{\circ}, \angle ADB=40^{\circ}, AD=4,$ and $BC=6$ . What is $AC$
$\textbf{(A)}\; 3+\sqrt{5} \qquad\textbf{(B)}\; 6 \qquad\textbf{(C)}\; \dfrac{9}{2}\sqrt{2} \qquad\textbf{(D)}\; 8-\sqrt{2} \qquad\textbf{(E)}\; 7$

Solution: $\angle ADB$ and $\angle ACB$ are both subtended by segment $AB$ , hence $\angle ACB = \angle ADB = 40^\circ$ . By considering $\triangle ABC$ , it follows that $\angle ABC = 180^\circ - (70^\circ + 40^\circ) = 70^\circ$ . Hence $\triangle ABC$ is isosceles, and $AC = BC = \boxed{6}.$

Answer: B 6
Problem: For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$

Solution: This problem becomes simple once we recognize that the domain of the function is $\{4, 6, 8, 9, 10, 12, 14, 15, \dots\}$ . By evaluating $r(4)$ to be $4$ , we can see that $\textbf{(E)}$ is incorrect. Evaluating $r(6)$ to be $5$ , we see that both $\textbf{(B)}$ and $\textbf{(C)}$ are incorrect. Since our domain consists of composite numbers, which, by definition, are a product of at least two positive primes, the minimum value of $r(n)$ is $4$ , so $\textbf{(A)}$ is incorrect. That leaves us with $\boxed{3}$

Answer: D 3
Problem: For every composite positive integer $n$ , define $r(n)$ to be the sum of the factors in the prime factorization of $n$ . For example, $r(50) = 12$ because the prime factorization of $50$ is $2 \times 5^{2}$ , and $2 + 5 + 5 = 12$ . What is the range of the function $r$ $\{r(n): n \text{ is a composite positive integer}\}$
$\textbf{(A)}\; \text{the set of positive integers} \\ \textbf{(B)}\; \text{the set of composite positive integers} \\ \textbf{(C)}\; \text{the set of even positive integers} \\ \textbf{(D)}\; \text{the set of integers greater than 3} \\ \textbf{(E)}\; \text{the set of integers greater than 4}$

Solution: Think backwards. The range is the same as the numbers $y$ that can be expressed as the sum of two or more prime positive integers.
The lowest number we can get is $y = 2+2 = 4$ . For any number greater than 4, we can get to it by adding some amount of 2's and then possibly a 3 if that number is odd. For example, 23 can be obtained by adding 2 ten times and adding a 3; this corresponds to the argument $n = 2^{10} \times 3$ .  Thus our answer is $\boxed{3}$

Answer: D 3
Problem: For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$

Solution: Simply take some time to draw a table of values of $f(i,j)$ for the first few values of $i$
\[\begin{array}{|c || c | c | c | c | c |} \hline i \text{\ \textbackslash\ } j & 0 & 1 & 2 & 3 & 4\\ \hline\hline 0 & 1 & 2 & 3 & 4 & 0\\ \hline 1 & 2 & 3 & 4 & 0 & 1\\ \hline 2 & 3 & 0 & 2 & 4 & 1\\ \hline 3 & 0 & 3 & 4 & 1 & 0\\ \hline 4 & 3 & 1 & 3 & 1 & 3\\ \hline 5 & 1 & 1 & 1 & 1 & 1\\ \hline \end{array}\]
Now we claim that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$ .  We will prove this by induction on $i$ and $j$ .  The base cases for $i = 5$ , have already been proven.
For our inductive step, we must show that for all valid values of $j$ $f(i, j) = 1$ if for all valid values of $j$ $f(i - 1, j) = 1$
We prove this itself by induction on $j$ .  For the base case, $j=0$ $f(i, 0) = f(i-1, 1) = 1$ .  For the inductive step, we need $f(i, j) = 1$ if $f(i, j-1) = 1$ .  Then, $f(i, j) = f(i-1, f(i, j-1)).$ $f(i, j-1) = 1$ by our inductive hypothesis from our inner induction and $f(i-1, 1) = 1$ from our outer inductive hypothesis.  Thus, $f(i, j) = 1$ , completing the proof.
It is now clear that for $i \ge 5$ $f(i,j) = 1$ for all values $0 \le j \le 4$
Thus, $f(2015,2) = \boxed{1}$

Answer: B 1
Problem: For every positive integer $n$ , let $\text{mod}_5 (n)$ be the remainder obtained when $n$ is divided by 5. Define a function $f: \{0,1,2,3,\dots\} \times \{0,1,2,3,4\} \to \{0,1,2,3,4\}$ recursively as follows:
\[f(i,j) = \begin{cases}\text{mod}_5 (j+1) & \text{ if } i = 0 \text{ and } 0 \le j \le 4 \text\\ f(i-1,1) & \text{ if } i \ge 1 \text{ and } j = 0 \text{, and} \\ f(i-1, f(i,j-1)) & \text{ if } i \ge 1 \text{ and } 1 \le j \le 4. \end{cases}\]
What is $f(2015,2)$
$\textbf{(A)}\; 0 \qquad\textbf{(B)}\; 1 \qquad\textbf{(C)}\; 2 \qquad\textbf{(D)}\; 3 \qquad\textbf{(E)}\; 4$

Solution: We are given that \[f(0,n) \equiv n+1\pmod{5} .\] Then, $f(1,n) = f(0, f(1, n-1)) \equiv f(1,n-1) + 1\pmod{5}$ . Thus $f(1,n) \equiv n+f(1,0)\pmod{5}$ . Since $f(1,0)=f(0,1)=2$ , we get \[f(1,n) \equiv n+2\pmod{5} .\] Then, $f(2,n) = f(1, f(2, n-1)) \equiv f(2,n-1) + 2\pmod{5}$ . Thus $f(2,n) \equiv 2n+f(2,0)\pmod{5}$ . Since $f(2,0)=f(1,1)=3$ , we get \[f(2,n) \equiv 2n+3\pmod{5} .\] Now $f(3,n) = f(2, f(3, n-1)) \equiv 2f(3,n-1) + 3\pmod{5}$ . Thus \begin{align*}     f(3,n) &\equiv 2f(3,n-1) + 3 &\pmod{5} \\     2f(3,n-1) &\equiv 2^2f(3,n-2) + 3\cdot 2 &\pmod{5} \\     \vdots \qquad &\quad\vdots \quad\qquad \vdots \qquad\qquad \vdots &\\     2^{n-1}f(3,1) &\equiv 2^nf(3,0) + 3\cdot 2^{n-1} &\pmod{5}  \end{align*} Adding them all up we get \[f(3,n) \equiv 3(2^n-1)\pmod{5} .\] This means that $f(3,0)=0$ $f(3,1)=3$ $f(3,2)=4$ $f(3,3)=1$ , and $f(3,4)=0$ . Thus $f(3,n)$ never takes the value 2.
Since $f(4,n)=f(3,f(4,n-1))$ , this implies that $f(4,n) \neq 2$ for any $n$ . By induction, $f(3,n) \neq f(3,2) = 4$ for any $n$ . It follows that $f(3,n) \neq f(3,4) = 0$ for any $n$ . Thus $f(4,n)$ only takes values in $\{1,3\}$ . In fact, it alternates between 1 and 3: $f(4,0)=f(3,1)=3$ , then $f(4,1)=f(3,f(4,0))=f(3,3)=1$ , then $f(4,2)=f(3,f(4,1))=f(3,1)=3$ , and so on.
Repeating the argument above, we see that $f(5,n) = f(4, f(5,n-1))$ can only take values in $\{1,3\}$ . However, $f(5,n-1)\neq 0$ for any $n$ implies that $f(5,n)\neq f(4,0)=3$ for any $n$ . Thus $f(5,n)=1$ for all $n$ . We can easily verify this: $f(5,0)=f(4,1)=1$ , then $f(5,1)=f(4,f(5,0))=f(4,1)=1$ , then $f(5,2)=f(4,f(5,1))=f(4,1)=1$ , and so on.
Then $f(6,0)=f(5,1)=1$ . Moreover, $f(6,n) = f(5,f(6,n-1)) = 1$ for all $n$ . Continuing in this manner we see that $f(m,n)=1$ for all $m\ge 5$
In particular, $f(2015,2) = \boxed{1}$

Answer: B 1
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution: We can translate this wordy problem into this simple equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil\]
We will proceed to solve this equation via casework.
Case 1: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}$
Our equation becomes $\frac{s}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=1$ and $j=4$ yield $s=64$ and $s=66$ , respectively.
Case 2: $\left\lceil \frac{s}{2} \right\rceil = \frac{s}{2}+\frac{1}{2}$
Our equation becomes $\frac{s}{2} +\frac{1}{2} - 19 = \frac{s}{5} + \frac{j}{5}$ , where $j \in \{0,1,2,3,4\}$ Using the fact that $s$ is an integer, we quickly find that $j=2$ yields $s=63$
Summing up we get $63+64+66=193$ . The sum of the digits is $\boxed{13}$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution: It can easily be seen that the problem can be expressed by the equation: \[\left\lceil \frac{s}{2} \right\rceil - \left\lceil \frac{s}{5} \right\rceil = 19\]
However, because the ceiling function is difficult to work with, we can rewrite the previous equation as:
\[\frac{s+a}{2} - \frac{s+b}{5} = 19\] Where $a \in \{0,1\}$ and $b \in \{0,1,2,3,4\}$ Multiplying both sides by ten and simplifying, we get: \[5s+5a-2s-2b=190\] \[3s = 190+2b-5a\] \[s = 63 + \frac{1+2b-5a}{3}\]
Because s must be an integer, we need to find the values of $a$ and $b$ such that $2b-5a \equiv 2 \mod 3$ . We solve using casework.
Case 1: $a = 0$
If $a = 0$ , we have $2b \equiv 2 \mod 3$ . We can easily see that $b = 1$ or $b = 4$ , which when plugged into our original equation lead to $s = 64$ and $s=66$ respectively.
Case 2: $a = 1$
If $a = 1$ , we have $2b-5 \equiv 2 \mod 3$ , which can be rewritten as $2b \equiv 1 \mod 3$ . We can again easily see that $b = 2$ is the only solution, which when plugged into our original equation lead to $s = 63$
Adding these together we get $64+66+63=193$ . The sum of the digits is $\boxed{13}$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution: As before, we write the equation:
\[\left\lceil \frac{s}{2} \right\rceil - 19 = \left\lceil \frac{s}{5} \right\rceil.\]
To get a ballpark estimate of where $s$ might lie, we remove the ceiling functions to find:
\[\frac{s}{2} - 19 = \frac{s}{5}.\]
This gives $\frac{3s}{10} = 19$ , and thus values for $s$ will be around $\frac{190}{3} = 63.\overline3$
Now, to establish some bounds around this estimated working value, we note that if $s=60$ , Cozy takes 30 steps while Dash takes 12, a difference of 18.  If $s=70$ , Cozy takes 35 steps while Dash takes 14, a difference of 21.  When $s$ increases from a multiple of ten, the difference will never decrease beyond what it is at the multiple of ten, and likewise, when it decreases, it never becomes greater than at the multiple of ten, so any working values of $s$ will be between $60$ and $70$
Then, by inspection, $s=63, 64,$ or $66$ , so $\sum s = 193 \implies \boxed{13}.$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution: Notice that the possible number of steps in the staircase is around 60 to 70. By testing all of the values between 60 and 70, we see that 63, 64 and 66 work. Adding those up gives 193, so the answer is $1+9+3=\boxed{13}.$

Answer: D 13
Problem: Cozy the Cat and Dash the Dog are going up a staircase with a certain number of steps. However, instead of walking up the steps one at a time, both Cozy and Dash jump. Cozy goes two steps up with each jump (though if necessary, he will just jump the last step). Dash goes five steps up with each jump (though if necessary, he will just jump the last steps if there are fewer than 5 steps left). Suppose that Dash takes 19 fewer jumps than Cozy to reach the top of the staircase. Let $s$ denote the sum of all possible numbers of steps this staircase can have. What is the sum of the digits of $s$
$\textbf{(A)}\; 9 \qquad\textbf{(B)}\; 11 \qquad\textbf{(C)}\; 12 \qquad\textbf{(D)}\; 13 \qquad\textbf{(E)}\; 15$

Solution: We represent C's steps with $2c + a = n$ and D's steps with $5d + b = n$ , where $a \in \{1,2\}$ and $b \in \{1,2,3,4,5\}$ , where $n$ is the number of steps, $c$ is the number of jumps C takes bar the last one, and $d$ is the number of jumps D takes bar the last.
The reason for starting at 1 and ending at 5 instead of 0 through 4 is that the last step can be quite problematic to deal with, especially if it is possible to make it in one go, so we treat it as a different jump that can take all possible jump values. We know that D makes it in 19 fewer than C, so $c+1-(d+1) = 19 \implies c = 19+d \implies 5d + b = 2(19+d) + a \implies 3d = 38 + a - b$
Now that we have this nice equivalence, we can do the thing and it works.
If we take both sides mod 3 and rearrange, we get $b \equiv 2+a \pmod{3}$ This gives us the following satisfactory $(a,b)$ relational pairs: $\{(1,3),(2,1),(2,4)\}$ . We can now just find the corresponding $d$ value for each pair, sum it all up using $5d + b = n$ and sum the digits to reveal $\boxed{13}$ as the answer.

Answer: D 13
Problem: Six chairs are evenly spaced around a circular table. One person is seated in each chair. Each person gets up and sits down in a chair that is not the same and is not adjacent to the chair he or she originally occupied, so that again one person is seated in each chair. In how many ways can this be done?
$\textbf{(A)}\; 14 \qquad\textbf{(B)}\; 16 \qquad\textbf{(C)}\; 18 \qquad\textbf{(D)}\; 20 \qquad\textbf{(E)}\; 24$

Solution: Consider shifting every person over three seats left after each person has gotten up and sat back down again. Now, instead of each person being seated not in the same chair and not in an adjacent chair, each person will be seated either in the same chair or an adjacent chair. The problem now becomes the number of ways in which six people can sit down in a chair that is either the same chair or an adjacent chair in a circle.
Consider the similar problem of $n$ people sitting in a chair that is either the same chair or an adjacent chair in a row. Call the number of possibilities for this $F_n$ . Then if the leftmost person stays put, the problem is reduced to a row of $n-1$ chairs, and if the leftmost person shifts one seat to the right, the new person sitting in the leftmost seat must be the person originally second from the left, reducing the problem to a row of $n-2$ chairs. Thus, $F_n = F_{n-1} + F_{n-2}$ for $n \geq 3$ . Clearly $F_1 = 1$ and $F_2 = 2$ , so $F_3 = 3$ $F_4 = 5$ , and $F_5 = 8$
Now consider the six people in a circle and focus on one person. If that person stays put, the problem is reduced to a row of five chairs, for which there are $F_5 = 8$ possibilities. If that person moves one seat to the left, then the person who replaces him in his original seat will either be the person originally to the right of him, which will force everyone to simply shift over one seat to the left, or the person originally to the left of him, which reduces the problem to a row of four chairs, for which there are $F_4 = 5$ possibilities, giving $1 + 5 = 6$ possibilities in all. By symmetry, if that person moves one seat to the right, there are another $6$ possibilities, so we have a total of $8+6+6 = \boxed{20}$ possibilities.

Answer: D 20