The product of three consecutive numbers when divided by each of them in turn gives three quotients. The sum of these three quotients is equal to 74.

What are the numbers ?

(In reply to

answer by K Sengupta)

Let the numbers be x, x+1 and x+2

Then, by the problem:

x(x+1) + (x+1)(x+2) + x(x+2) = 74

Or, 3*x^2 + 6x - 72 = 0

Or, x^2 + 2x - 24 = 0

Or, (x+6)(x-4) = 0

Or, x = -6, 4

x= 4 gives: (x+1, x+2) = (5, 6), while:

x = -6 gives: (x+1, x+2) = (-5, -4)

Thus there exists two sets of triplets satisfying the conditions of the problem, and these are:

(4, 5, 6) and (-6, -5, -4).