2021 AMC 12A Spring Problem 16

Below is the professionally curated solution for Problem 16 of the 2021 AMC 12A Spring, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12A Spring solutions, or check the answer key.

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Concepts:triangular numbermedian (data)

Difficulty rating: 1730

16.

In the following list of numbers, the integer nn appears nn times in the list for 1n200.1 \le n \le 200. 1,2,2,3,3,3,4,4,4,4,,200,200,,200 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots, 200

What is the median of the numbers in this list?

100.5100.5

134134

142142

150.5150.5

167167

Solution:

The list has 1+2++200=2002012=201001 + 2 + \cdots + 200 = \dfrac{200\cdot 201}{2} = 20100 terms, so the median is the average of the 1005010050th and 1005110051st terms.

The value nn occupies positions up to n(n+1)2.\dfrac{n(n+1)}{2}. Since 1411422=10011\dfrac{141\cdot 142}{2} = 10011 and 1421432=10153,\dfrac{142\cdot 143}{2} = 10153, positions 1001210012 through 1015310153 all equal 142.142. Both middle positions fall in this block, so the median is 142.142.

Thus, the correct answer is C.

Problem 16 in Other Years