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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 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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