What is the positive integer n? (1) For every integer m,

Question

What is the positive integer n?

(1) For every integer m, (m+n)!/(m – 1)! is divisible by 16
(2) n^2- 9n + 20 = 0

GMAT TIGER · Answer

1: (m+n)!/(m – 1)! >= 6!m could be anything so does n.
if m = 10 and n = 5, (m+n)!/(m – 1)! >= 6!
if m = 5 and n = 10, (m+n)!/(m – 1)! >= 6!.so nsf.

2: n^2- 9n + 20 = 0(n - 4) (n - 5) = 0so n = 4 or 5. nsf.

E. togather also nsf.

marcodonzelli · Answer

OA is C. your reasoning is very good. anyway you forgot to think about that:

m(m+1)(m+2)....has to be a multiple of 16, thus a multiple of 6! as you said. this actually means that we must have at least 6 terms, in order to have at least 3 even numbers (i.e. 2,4 and 6). obviously n can be >=5, so that we have at least 6 terms. Cond 2 says that n can be either 4 or 5. only 5 works, so OA is C

GMAT TIGER · Answer

m or n not necessarily be 6!. only (m+n)!/(m – 1)! has to be equal or greater than 6!.
ok, from ii, n = either 4 or 5. but it doesnot mean n has to be 5. n could be 4 if m is 2 or greater. 
so imo, it is still E.
correct me if i missed anything.

AlbertNTN · Answer

"n could be 4 if m is 2 or greater"
->i think there is a constraint "for every integer m". Choose m=1 -> min(n)=5
but comes another prob, can m<=0 ???

What is the positive integer n? (1) For every integer m,

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