# 2010 AMC 12B Problems/Problem 10

The following problem is from both the 2010 AMC 12B #10 and 2010 AMC 10B #14, so both problems redirect to this page.

## Problem

The average of the numbers $1, 2, 3,\cdots, 98, 99,$ and $x$ is $100x$. What is $x$? $\textbf{(A)}\ \dfrac{49}{101} \qquad \textbf{(B)}\ \dfrac{50}{101} \qquad \textbf{(C)}\ \dfrac{1}{2} \qquad \textbf{(D)}\ \dfrac{51}{101} \qquad \textbf{(E)}\ \dfrac{50}{99}$

## Solution

We first sum the first $99$ numbers: $\frac{99(100)}{2}=99\cdot50$. Then, we know that the sum of the series is $99\cdot50+x$. There are $100$ terms, so we can divide this sum by $100$ and set it equal to $100x$: $$\frac{99\cdot50+x}{100}=100x \Rightarrow 99\cdot50=100^2x-x\Rightarrow \frac{99\cdot50}{100^2-1}=x$$ Using difference of squares: $$x=\frac{99\cdot50}{101\cdot99}=\frac{50}{101}$$ Thus, the answer is $\boxed{\text{B}}$.

## Video Solution

~IceMatrix

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