2019 AMC 10A Problem 5

Question

What is the greatest number of consecutive integers whose sum is

45?

9

25

45

90

120

A long consecutive block can be centered around zero

Use divisibility of

90

by the number of terms

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.

Problem 5 in other years