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2021 AMC 10A Problem 16

Problem 16 of 25IntermediateAlgebra

In the following list of numbers, the integer nn appears nn times in the list for 1≤n≤200.1\leq n\leq200. 1,2,2,3,3,3,4,4,4,4,…,200,200,…,200 \begin{aligned} &1,2,2,3,3,3,4,4,4,4,\ldots,\\ &200,200,\ldots,200 \end{aligned} What is the median of the numbers in this list?

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Solution

The list contains 1+2+⋯+200=200⋅2012=20100 \begin{aligned} 1+2+\cdots+200&=\frac{200\cdot201}{2}\\ &=20100 \end{aligned} entries, so its two middle positions are 1005010050 and 10051.10051. There are 141⋅1422=10011\frac{141\cdot142}{2}=10011 entries through the last 141,141, and 142⋅1432=10153\frac{142\cdot143}{2}=10153 entries through the last 142.142. Thus both middle entries are 142,142, so the median is 142.142. Thus, C is the correct answer.

More practice

Concepts: median (data) · triangular number

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.