what is remainder of 1!+2!+3!+4!+......+151! is divisible by 5

Question

Pls solv the que.. n Giv ne the . Methods of solution...

Answer 1

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Answer 2

1!+2!+3!+4!+......+151!=1!+2!+3!+4!+5(4!)+....

the remainder is 1!+2!+3!+4!=33

Answer 3

In this series, each term which is greater than or equal to the 5th, 5!, includes 5 in its factorial.

Therefore, each term is divisible by 5, so that the sum of them is divisible by 5 as well.

That is: 5!+6!+ … +150!+151! ≡ 0 (mod 5) ··· Ex.1

While, 1!+2!+3!+4!=33=5x6+3 so, the sum of 1!+2!+3!+4! is not divisible by 5.

That is: 1!+2!+3!+4! ≡ 3 (mod 5) ··· Ex.2 

Add Ex.1 and Ex.2.   We have: 1!+2!+3!+4!+ …+150!+151! ≡ 3 (mod 5) 

Therefore, the given series is not divisible by 5.   If the series is divided by 5, the remainder is 3.