The divisibility rule for 5 states that a number is divisible by 5 if its last digit is either 0 or 5.
We can determine if (1!+2!+3!+...+100!)^2 is divisible by 5 by checking if the sum of the factorials of the numbers from 1 to 100 is divisible by 5. If it is, then the square of that sum will also be divisible by 5.
No, if the sum of the factorials of the numbers from 1 to 100 is not divisible by 5, then the square of that sum will also not be divisible by 5.
The sum of the factorials of the numbers from 1 to 100 is 1!+2!+3!+...+100! = 44,899,027,554,640,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000