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