2019 AMC 10A Problem 5

Attempt Problem 5 of the 2019 AMC 10A 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 10A 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 extremal argument

Difficulty rating: 1070

Solution:

Suppose there are $k$ consecutive integers with first term $a.$ Their sum is $\dfrac{k(2a+k-1)}{2}=45,$ so $k$ must divide $90.$ Therefore the number of terms cannot exceed $90.$

This bound is attained by $-44, -43, \cdots, 44, 45$ which has $90$ terms and sum $45.$

Thus the greatest possible number of consecutive integers is $90.$

Thus, D is the correct answer.

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