2019 AMC 12A Problem 4

Attempt Problem 4 of the 2019 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 2019 AMC 12A solutions, or check the answer key.

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

What is the greatest number of consecutive integers whose sum is $45?$

$9$

$25$

$45$

$90$

$120$

Answer: D

Concepts:arithmetic sequence symmetry

Difficulty rating: 1170

Solution:

Negative integers are allowed. The integers from $-44$ to $44$ sum to $0,$ so the integers from $-44$ to $45$ sum to $45.$

This run has $45 - (-44) + 1 = 90$ integers. Conversely, if $L$ consecutive integers have sum $45,$ then twice their sum is $L$ times an integer, so $L\mid90.$ Hence $L\le90,$ proving that this run is longest.

Thus, the correct answer is D.

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