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E
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Difficulty:
65%
(hard)
Question Stats:
53% (01:57) correct
47%
(01:51)
wrong
based on 137
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History
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If n and k are integers greater than 0. What is the remainder when (n - 1)! is divided by kn ?
(1) n is not a prime number
(2) k < 2
(1) n is not a prime number
(2) k < 2
17 Sep 2010, 07:33
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shrouded1
How is the answer C?
Well, clearly each statement alone is not sufficient.
(1)+(2) \(0<k=integer<2\) --> \(k=1\), so we are looking for the remainder when \((n-1)!\) is divided by \(n\neq{prime}\). If \(n=4\neq{prime}\) then \((n-1)!=6\) divided by 4 yields remainder of 2 but if \(n=6\neq{prime}\) then \((n-1)!=120\) divided by 6 yields remainder of 0. Two different answers. Not sufficient.
Answer: E.
I think the question intends to use the theorem which states that \((n-1)!\) is always divisible by \(n\) if \(n\) is not a prime number. But it's true only if \(n>4\).
So C to be an answer, statement (1) should state: \(n\) is greater than 4 and \(n\) is not a prime number.
General Discussion
17 Sep 2010, 07:21
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(1) obviously insufficient we don't know anything about k
(2) k < 2 so k = 1
In that case, either n is prime, or n is not prime.
If n is prime, (n-1)! is not divisible by n and will give each time a different remainder.
If n is not prime, (n-1)! is divisible by and will give a remainder = 0. If your are not convinced, use example !!
(1) and (2) : using the previous argument, the remainder will be 0 every time!
ANS: C.
Hope it's clear
(2) k < 2 so k = 1
In that case, either n is prime, or n is not prime.
If n is prime, (n-1)! is not divisible by n and will give each time a different remainder.
If n is not prime, (n-1)! is divisible by and will give a remainder = 0. If your are not convinced, use example !!
(1) and (2) : using the previous argument, the remainder will be 0 every time!
ANS: C.
Hope it's clear
