2019 AMC 10A Problem 5

Below is the professionally curated solution for Problem 5 of the 2019 AMC 10A, 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.

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Concepts:arithmetic sequenceextremal argument

Difficulty rating: 1070

5.

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

99

2525

4545

9090

120120

Solution:

Note that negative integers are allowed to be in the sequence.

This means that we could form the sequence 44,43,,44,45, -44, -43, \cdots, 44, 45, which clearly adds up to 45.45. There are 9090 terms in this sequence.

Adding another negative term wouldn't work, since that would require having to go up to 4646 in the positive numbers, which puts the sum over 45.45.

Thus, D is the correct answer.

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