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Let n be a positive integer.

When n is divided by 2, the remainder is 1.

When n is divided by 3, the remainder is 2.

When n is divided by 4, the remainder is 3.

When n is divided by 5, the remainder is 4.

When n is divided by 6, the remainder is 5.

What is the smallest possible value of n?

May 11, 2020

#1
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When n is divided by 2, the remainder is 1.

When n is divided by 3, the remainder is 2.

When n is divided by 4, the remainder is 3.

When n is divided by 5, the remainder is 4.

When n is divided by 6, the remainder is 5.

LCM[2, 3, 4, 5, 6] = 60

n =60m + 59, where m =0, 1, 2, 3, 4......etc.

So, the smallest n = 59.

May 11, 2020
#2
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I accidentally reposted that question, but I think that answer is 59 from brute forcing it. I think there is a more logical way of solving it using mod however my math skills aren't great enough.

May 11, 2020
#3
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Since the mods are in consecutive order and the remainders are also in consecutive order, and are separated in every case by 1, then all you have to do is to take their common LCM and subtract 1 from it. And that would be the smallest n.

Guest May 11, 2020