Question

What is the remainder when $(1!{)}^3+(2!{)}^3+(3!{)}^3+(4!{)}^3+\dots \dots (1152!{)}^3$ is divided by 1152?

• A. 125
• B. 225
• C. 325
• D. 205

Answer 1

The correct option is B 225

$1152=2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3=2^7\times 3^2$
Essentially every number starting from $4{!}^3$ would be divisible perfectly by 1152 since each number after that would have at least 7 twos and 2 threes.
Thus, the required remainder is got by the first three terms:
$\frac{(1+8+216)}{1152}=\frac{225}{1152}$ gives us 225 as the required remainder.