2018 AMC 12B Problem 10

Attempt Problem 10 of the 2018 AMC 12B 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 2018 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

10.

A list of 20182018 positive integers has a unique mode, which occurs exactly 1010 times. What is the least number of distinct values that can occur in the list?
202202
223223
224224
225225
234234
Answer: D
Concepts:modepigeonhole principle
Small hint:
The 2018−10=20082018-10=2008 non-mode entries can each appear at most 99 times
Big hint:
Find the fewest distinct non-mode values needed, then add 11 for the mode
Solution:
The mode uses 1010 of the entries, leaving 2008.2008. Because the mode is unique, every other value appears at most 99 times, so at least ⌈20089⌉=224\left\lceil\tfrac{2008}{9}\right\rceil=224 distinct non-mode values are needed.
Adding the mode gives 224+1=225.224+1=225. This is achievable: use 99 copies each of 11 through 223,223, ten copies of 224,224, and one copy of 225.225.
Thus, the correct answer is D.
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