Solution

Given:

• The product of 3 consecutive positive integers is 8 times their sum.

To find:

• The sum of the squares of the 3 integers.

Approach and Working:
Let us assume the 3 consecutive numbers are n – 1, n and n + 1.

As per the given condition,

• (n – 1) n (n + 1) = 8 (n – 1 + n + n + 1)
Or, \(n (n^2 – 1) = 24n\)
Or, \(n^2 – 1 = 24\) (as n ≠ 0)
Or, \(n^2 = 25\)
Or, \(n = 5\) (as n is positive)

Hence, the integers are 4, 5 and 6.

Therefore, the sum of their squares = \(4^2 + 5^2 + 6^2 = 16 + 25 + 36 = 77\)

Hence, the correct answer is option B.

Answer: B


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I tried the same approach with x, x+1, x+2 but it seems more complicated.

How did you know that we should use x-1,x,x+1?

Is it generally easier to use n-1,n,n+1 for three consecutive integers or is it just for this case?

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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