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number theory

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by priyankagumber » Wed Jul 22, 2009 9:01 am
Find the number of possible values of integer ‘x’ such that 41^n when divided by ‘x’ leaves 1 as the remainder. (‘n’ is any prime number greater than 5)
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Re: number theory

by real2008 » Thu Jul 23, 2009 10:44 am
priyankagumber wrote:Find the number of possible values of integer ‘x’ such that 41^n when divided by ‘x’ leaves 1 as the remainder. (‘n’ is any prime number greater than 5)
Is the answer 8? I have one doubt.... Is it needed that 'n' to be prime number greater than 5 but not any positive integer?
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by PussInBoots » Fri Jul 24, 2009 11:12 pm
This cannot be on GMAT. 41 ^ 7 -1 vs 41 ^ 11 - 1 have different divisors. Only answers of some f-on of n would be correct. That'd take more than 3 minutes to solve.
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by prindaroy » Sat Jul 25, 2009 12:00 am
That's certainly not a GMAT question. I don't know where you're pulling these questions from. Anyways, the number of divisors of 41^n = n+1, so for numbers that have a remainder of 1, the answer is n .
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by PussInBoots » Sat Jul 25, 2009 11:47 am
prindaroy wrote:That's certainly not a GMAT question. I don't know where you're pulling these questions from. Anyways, the number of divisors of 41^n = n+1, so for numbers that have a remainder of 1, the answer is n .
Wrong

41^ 2 = 1681

10, 20, 40, 80, 2, 5...
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