Question

The sum of the reciprocals of three consecutive positive integers is equal to 47 divided by the product of the integers. What is the smallest of the three integers?

Answer 1

1/(x-1) + 1/x + 1/(x+1) = 47/[(x-1)x(x+1)]
put all over a common denominator of (x-1)x(x+1). The numerators then satisfy

x(x+1) + (x-1)(x+1) + (x-1)x = 47
3x^2 - 1 = 47
3x^2 = 48
x = 4

check:
1/3 + 1/4 + 1/5 = 47/60