Question

Sum of three consecutive prime numbers have product of their multiplication 190747. Find the value of their sum

Answer 1

Answer:

Consider a, b and c as consecutive numbers their product is

(b-1)b(b+1) = 190747

b(b^2–1) = 190747

If b = 57, LHS = 185136

If b = 58, LHS = 195054.

The middle term should be a prime number around 58. The prime number nearest to 58 is 59. So the three consecutive prime numbers are

53, 59, 61.

Check: 53*59*61 = 190747. Correct.