tarek99 wrote:

Well guys, the OA is C. Here is the OE, which I never understood clearly. Maybe someone could use that to come up with a better explanation:

(1) Note that (m+n)!/(n-1)! = m(m+1)(m+2).....(m+n), the product of the n+1 integers from m to m+n, inclusive. The product is a multiple of 16 for every integer n if and only if three of the terms are even, i.e. if the product consists of at least 6 terms. Thus n+1 is greater than or equal to 6, and n is at least 5. NOT SUFF.

(2) n is either 4 or 5 NOT SUFF.

(1&2) n=5 SUFF.

We can do that!

Let us extend our earlier analysis

n= (m^2+20) / 9

for n to be a positive integer m^2+20 should be multiple of 9

Try values 9, 18,27 , 36, 45 for m^2+20

m^2+20 = 36 => m=+4 or -4

m^2+20 = 45 => m=+5 or -5

substitute m value to get n

==> n can be 4 or 5

let us see which value suits best.

(m+n)!/(n-1)! = m(m+1)(m+2).....(m+n)

for n=4 => it becomes m(m+1)(m+2)(m+3)(m+4)

for a product of consecutive numbers series to have 16 as a factor , there should be atleast three even numbers in the series.

let us see 2*3*4*5*6 = 720 ,divisible by 16

so above n=4 does not suffice.

try n=5 ==> it contains 6 terms..

m can be either even or odd it does not matter..it is divisible by 16.

truly tough one..