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not much of a pure thinker, but have come across a problem which states the following:find the number, N, which lets the numbers 1 through to 10 divide it and leave a remainder 1, but when divided by 11 leaves no remainder.tks people look forward to hearing some answers

10!+1 does the trick if you drop the condition that dividing by 1 should give remainder 1

sorry i wasn't clear, the number when divided by 2,3,4,5,6....,9,10 all have to give remainder 1, and when divided by 11 have no remainder

first 2 are:N = 25,201 and 52,921

Is it possible to characterize all N for which N+1 divides N!+1? Edit: easy to verify that necessary condition is N = even, but N=8 shows that this isn't a sufficient condition

QuoteOriginally posted by: AKYfirst 2 are:N = 25,201 and 52,921followed by.....80641,108361,136081,etc.....There are a ton of them. 16 where N<1,000,000.

QuoteOriginally posted by: BramJIs it possible to characterize all N for which N+1 divides N!+1? Edit: easy to verify that necessary condition is N = even, but N=8 shows that this isn't a sufficient conditionSuch N are characterized by the Wilson's theorem: p is prime iff p divides (p-1)!+1So the N has to be p-1, where p is a prime. i.e. N = those even numbers whose successor is a prime (in your case, 9 is not prime)

Thanks. Now you mention the theorem it vaguely rings a bell...