Homework Help: smallest of 3 integers

Posted by Anonymous on Wednesday, June 27, 2012 at 1:15pm.

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?

smallest of 3 integers - Steve, Wednesday, June 27, 2012 at 2:26pm
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

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