jeeteshsingh wrote:

Is there a specific approach to tackle this?

No specific approach.

We have the sum of many numbers: \((1!)^3+ (2!)^3 + (3!)^3 +...+(1152!)^3\) and want to determine the remainder when this sum is divided by 1152.

First we should do the prime factorization of 1152: \(1152=2^7*3^2\).

Consider the third and fourth terms:

\((3!)^3=2^3*3^3\) not divisible by 1152;

\((4!)^3=2^9*3^3=2^2*3*(2^7*3^2)=12*1152\) divisible by 1152, and all the other terms after will be divisible by 1152.

We'll get \(\{(1!)^3+ (2!)^3 + (3!)^3\} +\{(4!)^3+...+(1152!)^3\}=225+1152k\) and this sum divided by 1152 will result remainder of 225.