2004 AMC 12A Problem 10

Attempt Problem 10 of the 2004 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 12A solutions, or check the answer key.

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10.

The sum of

49

consecutive integers is

7^5.

What is their median?

7

7^2

7^3

7^4

7^5

For consecutive integers the median equals the mean

The median is the sum divided by the number of terms,

\dfrac{7^5}{49}

The sum of a set of consecutive integers equals the number of terms times their mean, and for consecutive integers the mean equals the median.

So the median is

\dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343.

Thus, the correct answer is C.

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