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 ?

Let the first of the numbers be x. Then the other two are x+1 and x+2.
Let Q1, Q2, Q3 be the three quotients.

Q1 = (x*(x+1)*(x+2))/x = (x+1)*(x+2) = x^2 + 3x + 2
Q2 = (x*(x+1)*(x+2))/(x+1) = x*(x+2) = x^2 + 2x
Q3 = (x*(x+1)*(x+2))/(x+2) = x*(x+1) = x^2 + x

Therefore:

x^2 + x + x^2 + 2x + x^2 + 3x + 2 = 74
-> 3x^2 + 6x - 72 = 0
-> x^2 + 2x - 24 = 0
-> (x + 6)*(x - 4) = 0
-> x = -6 or 4

So the numbers are either 4,5,6 or -6,-5,-4

Posted by fwaff on 2003-03-12 02:27:32