Bunuel wrote:
The product of three consecutive positive integers is 8 times their sum. What is the sum of their squares?
(A) 50
(B) 77
(C) 110
(D) 149
(E) 194
We let n = the middle integer; thus, we can create the equation:
(n - 1)(n)(n + 1) = 8(n - 1 + n + n + 1)
n(n^2 - 1) = 8(3n)
n^3 - n = 24n
n^3 - 25n = 0
n(n^2 - 25) = 0
n(n - 5)(n + 5) = 0
n = 0 or n = 5 or n = -5
Since we are told that the integers are positive, then n = 5 only. Therefore, the three integers are 4, 5 and 6, and the sum of their squares is 4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77.
Alternate Solution:
If we let n be the middle integer, then the sum of the squares, in terms of n, is:
(n - 1)^2 + n^2 + (n + 1)^2 = n^2 -2n + 1 + n^2 + n^2 + 2n + 1 = 3n^2 + 2
From the information given to us, we can write the equation:
(n - 1)(n)(n + 1) = 8(n - 1 + n + n + 1)
n(n^2 - 1) = 8(3n)
Since all the integers are positive, we can divide each side by n:
n^2 - 1 = 24
n^2 = 25
Thus, the sum of their squares is 3n^2 + 2 = 3(25) + 2 = 77.
Answer: B
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