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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.

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..

From choice 1, we can deduce that n should atleast be 5 because, the numeration will only yield a product of 16, it (m+n) is 6!. This is because, 6! = 6*5*4*3*2*1 and this is the smallest number that has 4 multiples of 2s, which will give 16. So, n>=5. However, this is insufficient.

Choice 2: Here the equation should be n^2-9n+20 = 0 ; The question cannot relate choice 1 and choice 2. There is nowhere else about "m" except in choice 1. Hence, when we solve for the above equation, we get, (n-5)*(n-4)=0, which will give n = 5 or n = 4; So, this is insufficient.

Combining both the choices, we get n = 5 and hence the correct answer is C.

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

.........a very simple question ...........u got the core right