BTGmoderatorDC wrote:N = a!×b!×c!×d!/e! where a, b, c, and d, are four distinct positive integers, which are greater than 1, and a, e are two consecutive numbers, which are prime. What is the least possible value of a+b+c+d/e , if N is divisible by cube of product of the three smallest odd prime numbers?

A.26/3
B. 9
C. 21/2
D. 13
E. 27/2

OA A

Source: e-GMAT

So, we have N = a!*b!*c!*d!/e!, where a, b, c, and d, are four distinct positive integers, which are greater than 1, and a, e are two consecutive numbers, which are prime.

Since the only two consecutive numbers that are prime are 2 and 3, we have a = 2 and e = 3.

Thus, N = a!*b!*c!*d!/e! = 2!*b!*c!*d!/3! = 2*b!*c!*d!/3

Given that N = b!*c!*d!/3 is divisible by cube of product of the three smallest odd prime numbers, 2*b!*c!*d!/3 is divisible by (3*5*7)^3. Note that 3, 5 and 7 are the three smallest odd prime numbers

Thus, [2*b!*c!*d!] / [3*(3*5*7)^3] is an integer.

Since the greatest number in the denominator is 7, the numerator 2*b!*c!*d! must have three 7s. Least possible values of b, c, and d are possible if b = 7, c = 8 and d = 9. Each of b, c and d will render one 7.

Thus, we have a = 2, b = 7, c = 8, d = 9 and e = 3. Thus, the least possibel value of (a + b + c + d)/e = (2 + 7 + 8 + 9)/3 = 26/3

The correct answer: A

Hope this helps!

-Jay
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