Let \(n\) be a three digit natural number such that \(n^{5} - 5\) is divisible by \(91\). Find the least possible value of \(n\).

I contributed this problem to Briiliant but was rejected. So everybody enjoy solving it... and suggest more methods to make this problem even More Thoughtful and Interesting....

For its solution see my reply to Sebastian's comment.

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TopNewestCould you write a valid solution, please? I found it to be 122 with a computer search.

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Yes you are correct! Its 122. And for a valid solution here it is:

The statement

\(n^{5} - 5\) is divisible by 91can be inferred as \(n^{5} \equiv 5\pmod{7}\) as well as\(\pmod{13}\). By a direct check, modulo 7, n = 3 is the only value satisfying \(n^{5} \equiv 5\). So \(n \equiv 3\pmod{7}\). Similarly, \(n \equiv 5\pmod{13}\). By the Chinese Remainder Theorem, these two conditions are equivalent to saying that \(n \equiv 31\pmod{91}\). Therefore the least 3 digit possible value of \(n = 91 \times 1 + 31 = 122\)Log in to reply

You seems to have the belief that \(x\equiv a\pmod {pq}\implies x\equiv a\pmod {p}, x\equiv a\pmod{q}\) which is not true.

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do you have a legitimate solution for it?

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Yes you are correct! Its 122. And for a valid solution here it is:

The statement n5−5 is divisible by 91 can be inferred as n5≡5(mod7) as well as(mod13). By a direct check, modulo 7, n = 3 is the only value satisfying n5≡5. So n≡3(mod7). Similarly, n≡5(mod13). By the Chinese Remainder Theorem, these two conditions are equivalent to saying that n≡31(mod91). Therefore the least 3 digit possible value of n=91×1+31=122

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I think the answer is 31. (Gotta say that I cheated--I used C++ programming) You got solution for this??

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\(n\) is a

THREEdigit number.Log in to reply

Still your cheat hasn't worked... Answer is 122.

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