math
posted by alex on .
the sum of the reciprocals of 3 consecutive positive integers is 47 divided by the product of the integers. what is the smallest of the 3 integers

1/(x1) + 1/x + 1/(x+1) = 47/((x1)x(x+1))
x(x+1) + (x1)(x+1) + x(x1) = 47
x^2 + x + x^2  1 + x^2  x
3x^2  1 = 47
3x^2 = 48
x^2 = 16
x = 4
so, the numbers are 3,4,5