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

2004 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2004 AMC 12A Problems/Problem 10

Contents

Problem

Solutions

Solution 1

Solution 2

See also