Prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000.

Question

Amanda Benjamin

Mathematics

4 March, 02:22

Prove that among 502 positive integers, there are always two integers so that either their sum or their difference is divisible by 1000.

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Answers (1)

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Jamal Hull · Answer 1

We conlude that always exists two integers so that either their sum or their difference is divisible by 1000.

Step-by-step explanation:

We have a 502 positive integers. Therefore we have a numbers 1, 2, 3, ..., 500, 501, 502.

For the sum, we can choose the numbers 502 and 488. We get

(502+488) / 1000=1000/1000=1.

For the difference, we can choose the numbers 1 and 501. We get

(1-501) / 1000=-500/1000=-0.5.

We conlude that always exists two integers so that either their sum or their difference is divisible by 1000.