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vcbabu
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remainder 28 May 2009, 09:09
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what is the remainder when 40! / 41? and when 39! / 41 ?



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remainder 28 May 2009, 12:28
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40! mod 41 = 40 (any prime number - 1) mod prime number = prime number - 1
39! mod 41 = 1 (any prime number - 2) mod Prime number = 1

Think about it.

Proof to follow....
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remainder 28 May 2009, 12:31
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Good to see you Akamai!
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remainder 28 May 2009, 13:03
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AkamaiBrah
40! mod 41 = 40 (any prime number - 1) mod prime number = prime number - 1
39! mod 41 = 1 (any prime number - 2) mod Prime number = 1

Think about it.

Proof to follow....
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Actually, I don't think this is a fair GMAT question. The problem is an application of "Wilson's Theorem" which is far beyond the knowledge needed for the GMAT. I would be happy to see an answer that has a solution which falls under GMAT-required knowledge.
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remainder 31 May 2009, 00:50
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40! =40! +1-1= (40! +1)+40-41

when 40! / 41 remainder is 40 since except 40 other terms are divisble by 41 and hence the remainder is 40/41 i.e. 40 .
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remainder 01 Jun 2009, 15:50
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vcbabu
40! =40! +1-1= (40! +1)+40-41

when 40! / 41 remainder is 40 since except 40 other terms are divisble by 41 and hence the remainder is 40/41 i.e. 40 .
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Hmmmm. Please demonstrate to me some proof or method to determine that (40! + 1) is divisible by 41. This is the same as saying 40! mod 41 = 40. Essentially, you are using the answer to the problem to find the answer -- circular logic.

If you are claiming that (n! + 1) is always divisible by n + 1, THAT IS WRONG; here is a simple counter example: (3! + 1) = 7 which is certainly NOT divisible by 4.

The ONLY time (n! + 1) is divisible by n + 1 is when n + 1 is PRIME, which is the essense of Wilson's formula and which no GMAT student should be expected to know.

Nuff said.



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