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2004 AMC 12A Problems/Problem 10

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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

Solution

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)}.

See also

2004 AMC 12A (ProblemsResources)
Preceded by
Problem 9
Followed by
Problem 11
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Looking for a challenging algebra text? Preparing for MATHCOUNTS or the AMC exams?
Check out Art of Problem Solving's Introduction to Algebra by Richard Rusczyk.
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