Question

Find the remainder when 1! + 2! + 3! + 4! + ... + n! is divided by 15, if n≥5​

Answer 1

5! = 120 is divisible by 15.

So series be like, 1! + 2! + 3! + 4! + 5! + (5!×6) + (5!×6×7) + ... + (5!×6×7×8×.....×n)

After 5!, there is always 5! Or 120 in each term. So, each term after 5! is divisible by 15 and only 1! , 2!, 3!, 4! Leaves reminder when divided by 15.

1!/15 = 1

2!/15=2

3!/15=6

4!/15=9

So total reminder when 1! + 2! + 3! + 4! + ... + n! is divided by 15 is (1+2+6+9) = 18

Answer 2

Answer: 3

Consider n = 5.

1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153.

153 leaves remainder 3 on division by 15.

Thus 153 can be taken as 15k + 3, where k = 10.

Consider n = 6.

1! + 2! + 3! + 4! + 5! + 6!

⇒ 15k + 3 + 720

⇒ 15k + 3 + 15 × 48

⇒ 15(k + 48) + 3     →   (1)

Here, remainder 3 is also obtained.

Consider n = 7.

1! + 2! + 3! + 4! + 5! + 6! + 7!

⇒ 15(k + 48) + 3+ 5040

⇒ 15(k + 48) + 3 + 15 × 336

⇒ 15(k + 48 + 336) + 3

⇒ 15(k + 384) + 3

Here, it's also.

We found that 1! + 2! + 3! + 4! + ... + n! leaves remainder 3 on division by 15 where n = 5.

On considering n greater than 5, each n! contains multiples of 3 and 5, thus n! would be multiple of 15. Hence same remainder is obtained.

Thus the answer is 3.