Question

a,b,c,d are four positive prime numbers such that the product of these four prime numbers is equal to to the sum of 55 consecutive positive integers. Find smallest possible value of a plus a + b + c+ d, where a,b,c,d, are not necessarily distinct.

Answer 1

Answer:

Step-by-step explanation:

Sum of first fifty-five consecutive integers = 55*56/2 = 55*28=1540

1540  = 4*5*7*11

Sum = 4+5+7+11 = 27