AoPS - Problem

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Problem

Source: 2019 Irish Mathematical Olympiad paper 2 p6

Tags: number theory, Sum of powers, consecutive

parmenides51

05.10.2020 17:11

The number $2019$ has the following nice properties: (a) It is the sum of the fourth powers of fuve distinct positive integers. (b) It is the sum of six consecutive positive integers. In fact, $2019 = 1^4 + 2^4 + 3^4 + 5^4 + 6^4$ (1) $2019 = 334 + 335 + 336 + 337 + 338 + 339$ (2) Prove that $2019$ is the smallest number that satises both (a) and (b). (You may assume that (1) and (2) are correct!)

i3435

05.10.2020 17:19

In order for a number to be a sum of

$6$ consecutive integers, it must be

$3\mod 6$ . The only smaller sums of 5 distinct

$4$ th powers are

$1^4+2^4+3^4+4^4+5^4$ and

$1^4+2^4+3^4+4^4+6^4$ (since

$1^4+2^4+3^4+4^4+7^4>1^4+2^4+3^4+5^4+6^4$ ). Checking both of these the first one is

$1+2+3+4+1\equiv 5\mod 6$ and the second one is

$1+2+3+4+0\equiv 4\mod 6$ .

parmenides51

05.10.2020 17:22

2 days , 5 problems each day , so p6 stands for 1st problem of day 2