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Find the number of positive numbers less than 2020, which can be written as the sum of five consecutive positive numbers.

User Mattia
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1 Answer

3 votes

Integers*?

Let
n be the smallest of these five consecutive integers. Then the sum of the five consecutive integers is
(5(2n+4))/(2)=5n+10, so we seek the largest positive integer
n such that
5n+10<2020. Solving this inequality, we find that
n<402, so there are
\boxed{401} such positive integers.

User Mohammad Kholghi
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