# 2004 AMC 12A Problems/Problem 10

## Problem

The sum of $49$ consecutive integers is $7^5$. What is their median? $\text {(A)}\ 7 \qquad \text {(B)}\ 7^2\qquad \text {(C)}\ 7^3\qquad \text {(D)}\ 7^4\qquad \text {(E)}\ 7^5$

## Solutions

### Solution 1

The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$.

### Solution 2

Notice that $49\cdot7^3=7^5$. So, our middle number (median) must be $7^3\ \mathrm{(C)}$ since all the other terms can be grouped to form an additional $48$ copies $7^3$. Adding them would give $7^5$.

Solution by franzliszt

## See also

 2004 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 9 Followed byProblem 11 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

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