2004 AMC 12A Problem 10

Below is the professionally curated solution for Problem 10 of the 2004 AMC 12A, 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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Concepts:arithmetic sequencemean

Difficulty rating: 1370

10.

The sum of 4949 consecutive integers is 75.7^5. What is their median?

77

727^2

737^3

747^4

757^5

Solution:

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 7549=7572=73=343. \dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343.

Thus, the correct answer is C.

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