\(N = \dfrac{a!\cdot b!\cdot c!\cdot d!}{e!}\) 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 \(\dfrac{a+b+c+d}{e},\) if \(N\) is divisible by cube of product of the three smallest odd prime numbers?

A. \(\dfrac{26}3\)
B. \(9\)
C. \(\dfrac{21}2\)
D. \(13\)
E. \(\dfrac{27}2\)

Answer: A

Source: e-GMAT