# 2013 AMC 12B Problems/Problem 7

The following problem is from both the 2013 AMC 12B #7 and 2013 AMC 10B #13, so both problems redirect to this page.

## Problem

Jo and Blair take turns counting from $1$ to one more than the last number said by the other person. Jo starts by saying $1"$, so Blair follows by saying $1, 2"$. Jo then says $1, 2, 3"$, and so on. What is the $53^{\text{rd}}$ number said? $\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 8$

## Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. Thus, after nine "turns", $1+2+3+4+5+6+7+8+9=45$ numbers have been said. In the tenth turn, the eighth number will be the 53rd number said, because $53-45=8$. Since we are starting from 1 every turn, the 53rd number said will be $\boxed{\textbf{(E) }8}$.

## Faster Solution

We notice that the number of numbers said is incremented by one each time; that is, Jo says one number, then Blair says two numbers, then Jo says three numbers, etc. We notice that the number of numbers is $1 + 2 + 3 + 4 ...$ every time we finish a "turn" we notice the sum of these would be the largest number $\frac{n(n+1)}{2}$ under 53, we can easily see that if we double this it's $n^2 + n \simeq 106$, and we immediately note that 10 is too high, but 9 is perfect, meaning that at 9, 45 numbers have been said so far, $\frac{9(9+1)}{2} = 45$ and $53 - 45 = \boxed{\textbf{(E) }8}$

## Solution 3

Let $T(n)$ denote the $n$th triangle number. Then, observe that the $T(n)$th number said is $n$. It follows that the $55$th number is $10$ (as $55 = T(10)$). Thus, the $53$rd number is $10 - 2 = 8$, which is answer choice $\boxed{\textbf{(E)}}$.

~Mathavi

## Video Solution

~no one

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