Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Note that $\left(a_1+a_2+\cdots+a_{2018}\right)^3=a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2\\ +\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k$
Note that $a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2\left(a_1+a_2+\cdots+a_{2018}-a_1\right)+3a_2^2\left(a_1+a_2+\cdots+a_{2018}-a_2\right)+\cdots+3a_{2018}^2\left(a_1+a_2+\cdots+a_{2018}-a_{2018}\right)+6\sum_{i\neq j\neq k}^{2018} a_ia_ja_k\equiv a_1^3+a_2^3+\cdots+a_{2018}^3+3a_1^2({2018}^{2018}-a_1)+3a_2^2({2018}^{2018}-a_2)+\cdots+3a_{2018}^2({2018}^{2018}-a_{2018}) \equiv -2(a_1^3+a_2^3+\cdots+a_{2018}^3)\pmod 6$ Therefore, $-2(a_1^3+a_2^3+\cdots+a_{2018}^3)\equiv \left(2018^{2018}\right)^3\equiv\left( 2^{2018}\right)^3\equiv 4^3\equiv 4\pmod{6}$
Thus, $a_1^3+a_2^3+\cdots+a_{2018}^3\equiv 1\pmod 3$ . However, since cubing preserves parity, and the sum of the individual terms is even, the sum of the cubes is also even, and our answer is $\boxed{4}$

Answer: E 4
Problem: Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, we will assume that $a_1 = 1$
We first note that $1^3+2^3+...+n^3 = (1+2+...+n)^2$ . So what we are trying to find is what $\left(2018^{2018}\right)^2=\left(2018^{4036}\right)$ mod $6$ . We start by noting that $2018$ is congruent to $2 \pmod{6}$ . So we are trying to find $\left(2^{4036}\right) \pmod{6}$ . Instead of trying to do this with some number theory skills, we could just look for a pattern. We start with small powers of $2$ and see that $2^1$ is $2$ mod $6$ $2^2$ is $4$ mod $6$ $2^3$ is $2$ mod $6$ $2^4$ is $4$ mod $6$ , and so on... So we see that since $\left(2^{4036}\right)$ has an even power, it must be congruent to $4 \pmod{6}$ , thus giving our answer $\boxed{4}$ . You can prove this pattern using mods. But I thought this was easier.

Answer: E 4
Problem: Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: First, we can assume that the problem will have a consistent answer for all possible values of $a_1$ . For the purpose of this solution, assume $a_1, a_2, ... a_{2017}$ are multiples of 6 and find $2018^{2018} \pmod{6}$ (which happens to be $4$ ). Then ${a_1}^3 + ... + {a_{2018}}^3$ is congruent to $64 \pmod{6}$ or just $\boxed{4}$

Answer: E 4
Problem: Let $a_1,a_2,\dots,a_{2018}$ be a strictly increasing sequence of positive integers such that \[a_1+a_2+\cdots+a_{2018}=2018^{2018}.\] What is the remainder when $a_1^3+a_2^3+\cdots+a_{2018}^3$ is divided by $6$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution: Due to the large amounts of variables in the problem, and the fact that the test is only 75 minutes, you can assume that the answer is probably just $2018^{2018} \pmod{6}$ , which is $\boxed{4}$

Answer: E 4
Problem: In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$
$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution: Let $AP=BQ=x$ . Then $AB=8-2x$
Now notice that since $CD=8-2x$ we have $QC=DR=x-1$
Thus by the Pythagorean Theorem we have $x^2+(x-1)^2=(8-2x)^2$ which becomes $2x^2-30x+63=0\implies x=\dfrac{15-3\sqrt{11}}{2}$
Our answer is $8-(15-3\sqrt{11})=3\sqrt{11}-7\implies \boxed{7}$ . (Mudkipswims42)

Answer: B 7
Problem: In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$
$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution: Denote the length of the equilateral octagon as $x$ . The length of $\overline{BQ}$ can be expressed as $\frac{8-x}{2}$ . By the Pythagorean Theorem, we find that: \[\left(\frac{8-x}{2}\right)^2+\overline{CQ}^2=x^2\implies \overline{CQ}=\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}\] Since $\overline{CQ}=\overline{DR}$ , we can say that $x+2\sqrt{x^2-\left(\frac{8-x}{2}\right)^2}=6\implies x=-7\pm3\sqrt{11}$ . We can discard the negative solution, so $k+m+n=-7+3+11=\boxed{7}$ ~ blitzkrieg21

Answer: B 7
Problem: In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$
$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution: Let the octagon's side length be $x$ . Then $PH = \frac{6 - x}{2}$ and $PA = \frac{8 - x}{2}$ . By the Pythagorean theorem, $PH^2 + PA^2 = HA^2$ , so $\left(\frac{6 - x}{2} \right)^2 + \left(\frac{8 - x}{2} \right)^2 = x^2$ . By expanding the left side and  combining the like terms, we get $\frac{x^2}{2} - 7x + 25 = x^2 \implies -\frac{x^2}{2} - 7x + 25 = 0$ . Solving this using the quadratic formula, $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ , we use $a = -\frac{1}{2}$ $b = -7$ , and $c = 25$ , to get one positive solution, $x=-7+3\sqrt{11}$ , so $k+m+n=-7+3+11=\boxed{7}$

Answer: B 7
Problem: In rectangle $PQRS$ $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$
$\textbf{(A) } 1 \qquad \textbf{(B) } 7 \qquad \textbf{(C) } 21 \qquad \textbf{(D) } 92 \qquad \textbf{(E) } 106$

Solution: Let $AB$ , or the side of the octagon, be $x$ . Then, $BQ = \left(\frac{8-x}{2}\right)$ and $CQ = \left(\frac{6-x}{2}\right)$ . By the Pythagorean Theorem $BQ^2+CQ^2=x^2$ , or $\left(\frac{8-x}{2}\right)^2+\left(\frac{6-x}{2}\right)^2 = x^2$ . Multiplying this out, we have $x^2 = \frac{64-16x+x^2+36-12x+x^2}{4}$ . Simplifying, $-2x^2-28x+100=0$ . Dividing both sides by $-2$ gives $x^2+14x-50=0$ . Therefore, using the quadratic formula , we have $x=-7 \pm 3\sqrt{11}$ . Since lengths are always positive, then $x=-7+3\sqrt{11} \Rightarrow k+m+n=-7+3+11=\boxed{7}$

Answer: B 7
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: We can begin to put this into cases. Let's call the pairs $a$ $b$ and $c$ , and assume that a member of pair $a$ is sitting in the leftmost seat of the second row. We can have the following cases then.
Case $1$ : 
Second Row: a b c
Third Row: b c a
Case $2$ :
Second Row: a c b
Third Row: c b a
Case $3$ : 
Second Row: a b c
Third Row: c a b
Case $4$ : 
Second Row: a c b
Third Row: b a c
For each of the four cases, we can flip the siblings, as they are distinct. So, each of the cases has $2 \cdot 2 \cdot 2 = 8$ possibilities. Since there are four cases, when pair $a$ has someone in the leftmost seat of the second row, there are $32$ ways to rearrange it. However, someone from either pair $a$ $b$ , or $c$ could be sitting in the leftmost seat of the second row. So, we have to multiply it by $3$ to get our answer of $32 \cdot 3 = 96$ . So, the correct answer is $\boxed{96}$

Answer: D 96
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: Lets call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$ . We can split our problem into two cases:
There is a child of each family in each row (There is an A, B, C in each row ) or There are two children of the same family in a row.
Starting off with the first case, we see that there are $3!=6$ ways to arrange the A,B,C. Then, we have to choose which sibling sits. There are $2$ choices for each set of siblings meaning we have $2^3=8$ ways to arrange that. So, there are $48$ ways to arrange the siblings in the first row. The second row is a bit easier. We see that there are $2$ ways to place the A sibling and each placement yields only $1$ possibility. So, our first case has $48\cdot2=96$ possibilities.
In our second case, there are $3$ ways to choose which set of siblings will be in the same row and $4$ ways to choose the person in between. So, there $3*4 = 12$ ways to arrange the first row. In the second row, however, we see that it is impossible to make everything work out. So, there are $0$ possibilities for this case.
Thus, there are $96+0 = \boxed{96}$ possibilities for this trip.

Answer: D 96
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: Call the siblings $A_1$ $A_2$ $B_1$ $B_2$ $C_1$ , and $C_2$
There are 6 choices for the child in the first seat, and it doesn't matter which one takes it, so suppose Without loss of generality that $A_1$ takes it ( $\circ$ denotes an empty seat):
\[A_1 \circ \circ\] \[\circ \ \circ \ \circ\]
Then there are 4 choices for the second seat ( $B_1$ $B_2$ $C_1$ , or $C_2$ ). Like before, it doesn't matter who takes the seat, so WLOG suppose it is $B_1$
\[A_1 B_1 \circ \\\] \[\circ \ \circ \ \circ\]
The last seat in the first row cannot be $A_2$ because it would be impossible to create a second row that satisfies the conditions. Therefore, it must be $C_1$ or $C_2$ . Let's say WLOG that it is $C_1$ . There are two ways to create a second row:
\[A_1 B_1 C_1 \\\] \[B_2 C_2 A_2\]
and
\[A_1 B_1 C_1 \\\] \[C_2 A_2 B_2\]
Therefore, there are $6 \cdot 4 \cdot 2 \cdot 2= \boxed{96}$ possible seating arrangements.

Answer: D 96
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: WLOG, define the three pairs of siblings to be: $A$ $B$ , and $C$ . Now, notice that you can only form a correct grouping either like this:
\[A B C\]
\[B C A\]
or this:
\[C B A\]
\[A C B\]
However, we need to consider the different orders. There are $3!$ ways to order each pair (eg. the same letters) and $2^3$ ways to order the people each of the three pairs. Now, we can just multiply everything together, yielding:
\[2\cdot3!\cdot2^3\]
Which is $\boxed{96}$

Answer: D 96
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: Let the families be $A$ $B$ $C$ . In any given possible arrangement, there are $3! = 6$ ways to arrange the families and $2 \cdot 2 \cdot 2 = 8$ ways to arrange the siblings. This means the answer has to be divisble by $6 \cdot 8 = 48$ . The only answer choice that satisfies this is $\boxed{96}$

Answer: D 96
Problem: Three young brother-sister pairs from different families need to take a trip in a van. These six children will occupy the second and third rows in the van, each of which has three seats. To avoid disruptions, siblings may not sit right next to each other in the same row, and no child may sit directly in front of his or her sibling. How many seating arrangements are possible for this trip?
$\textbf{(A)} \text{ 60} \qquad \textbf{(B)} \text{ 72} \qquad \textbf{(C)} \text{ 92} \qquad \textbf{(D)} \text{ 96} \qquad \textbf{(E)} \text{ 120}$

Solution: If a pair of siblings are in the same row, the other $2$ pairs of siblings cannot fit in the remaining $4$ seats to meet the requirements. Therefore, the siblings must be in different rows.
Let the first pair of siblings be $a_1$ $a_2$ , the second pair of siblings be $b_1$ $b_2$ , and the third pair of siblings be $c_1$ $c_2$ . For each row there must be $1$ child from each family. WLOG, let $a_1$ $b_1$ $c_1$ be in the first row. There are $3!$ arrangements. $a_2$ $b_2$ $c_2$ are in the second row, and they cannot be in the same column as their sibling. Now the problem becomes a $3$ element Derangement problem.
An $n$ element derangement problem is to find the number of permutations of $n$ elements where each element has a specified location, and no element is in it's specified location. For example, there are $3$ elements $1$ $2$ $3$ $1$ cannot be in the first location, $2$ cannot be in the second location, and $3$ cannot be in the third location. The derangements are: $(2,3,1)$ $(3,1,2)$ $D_3=2$
Each pair of siblings can swap their position. There are $2^3$ swaps. The total arrangements are: $3! \cdot D_3 \cdot 2^3=6 \cdot 2 \cdot 8 =\boxed{96}$

Answer: D 96
Problem: Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution: Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.
Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{11}.$

Answer: E 11
Problem: Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution: Let Joey's age be $j$ , Chloe's age be $c$ , and we know that Zoe's age is $1$
We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.
Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$ . Therefore, we know that, as there are $9$ solutions for $k$ , there must be $9$ solutions for $c-1$ . We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$ , so $c=37$ . Therefore, $j=38$ . Now, since $j-1=37$ , by similar logic, $37=(1+k)(a-1)$ , so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{11}$

Answer: E 11
Problem: Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution: Here's a different way of stating Solution 2:
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$ , so Chloe is $37$ , and Joey is $38$ . The common factor that will divide both of their ages is $37$ , so Joey will be $74$ . The answer is $7 + 4 =  \boxed{11}$

Answer: E 11
Problem: Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?
$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution: Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$ , it will hit every natural number less than $C-Z$ , including every factor of $C-Z$ . For $C+n$ to be an integral multiple of $Z+n$ , the difference $C-Z$ must also be a multiple of $Z$ , which happens if $Z$ is a factor of $C-Z$ . Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$ . Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$ . Because $37$ is prime, the next time this happens is at $Z+n=37$ , when $J+n=74$ . The answer is $7+4=\boxed{11}$

Answer: E 11
Problem: A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$

Solution: For all integers $n \geq 7,$ note that \begin{align*} f(n)&=f(n-1)-f(n-2)+n \\ &=[f(n-2)-f(n-3)+n-1]-f(n-2)+n \\ &=-f(n-3)+2n-1 \\ &=-[f(n-4)-f(n-5)+n-3]+2n-1 \\ &=-f(n-4)+f(n-5)+n+2 \\ &=-[f(n-5)-f(n-6)+n-4]+f(n-5)+n+2 \\ &=f(n-6)+6. \end{align*} It follows that \begin{align*} f(2018)&=f(2012)+6 \\ &=f(2006)+12 \\ &=f(2000)+18 \\ & \ \vdots \\ &=f(2)+2016 \\ &=\boxed{2017} ~MRENTHUSIASM

Answer: B 2017
Problem: A function $f$ is defined recursively by $f(1)=f(2)=1$ and \[f(n)=f(n-1)-f(n-2)+n\] for all integers $n \geq 3$ . What is $f(2018)$
$\textbf{(A) } 2016 \qquad \textbf{(B) } 2017 \qquad \textbf{(C) } 2018 \qquad \textbf{(D) } 2019 \qquad \textbf{(E) } 2020$

Solution: For all integers $n\geq3,$ we rearrange the given equation: \[f(n)-f(n-1)+f(n-2)=n. \hspace{28.25mm}(1)\] For all integers $n\geq4,$ it follows that \[f(n-1)-f(n-2)+f(n-3)=n-1. \hspace{15mm}(2)\] For all integers $n\geq4,$ we add $(1)$ and $(2):$ \[f(n)+f(n-3)=2n-1. \hspace{38.625mm}(3)\] For all integers $n\geq7,$ it follows that \[f(n-3)+f(n-6)=2n-7. \hspace{32mm}(4)\] For all integers $n\geq7,$ we subtract $(4)$ from $(3):$ \[f(n)-f(n-6)=6. \hspace{47.5mm}(5)\] From $(5),$ we have the following system of $336$ equations: \begin{align*} f(2018)-f(2012)&=6, \\ f(2012)-f(2006)&=6, \\ f(2006)-f(2000)&=6, \\ & \ \vdots \\ f(8)-f(2)&=6. \end{align*} We add these equations up to get \[f(2018)-f(2)=6\cdot336=2016,\] from which $f(2018)=f(2)+2016=\boxed{2017}.$

Answer: B 2017