Question

If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

• A. 1777
• B. 1785
• C. 1875
• D. 1877

Answer 1

The correct option is D

1877

Let the three consecutive numbers be (n-1), n, and (n+1)
Then, according to question:
(n-1) x n x (n+1) = 15600
$\Rightarrow$ $n^3$ – n = 15600.
We have to try for a number whose cube is very close to but also little more than 15600.
We get, n = 25
So, the required sum = $(n-1{)}^2+(n{)}^2+(n+1{)}^2=(25-1{)}^2+(25{)}^2+(25+1{)}^2={24}^2+{25}^2+{26}^2$
$\Rightarrow$ 576 + 625 + 676 = 1877