Question

Question : Find the remainder when 1! + 2! + 3! + 4! + 5! + . . . . . 100! is divided by 24.

Answer 1

Answer:

See this question can be written  as 

1+2×1+3×2×1+4×3×2×1+5×4×3×2×1 . .. .100!

Or

1+2+6+24+24×5.. .. ..+100!

You will notice that every factorial after 4 factorial is divisible by 24

So answer is 1+2+6=9

• You'll see that 4!=24. This means every no. after 4! will be having a 2 x 3 x 4 in it's prime factorization.
• Thus , all nos. after and including 4! will be divisible by 24. Therefore remainder will come from 1!+2!+3! i.e. we need to find the remainder when 1! + 2! + 3! or 1+2+6 or 9 is divided by 24. 

Therefore the remainder is 9.

Answer 2

We know that 4! =24 which is divisible by 4.

Now, after 4!,we can write each factorial term with as 4! * y where y is a positive number.

ex - 5! = 4! *5

6! = 4! * 5 * 6.

So 4! is a factor of 5!,6!,7!, ... ,100!.

So , from 4! to 100!, each term is divisible by 4!(24) as they have 4! as their factor.

So, 4! mod 24 = 0.

Similarly , 5! %24 =0 and so for rest upto 100!.

Now, Rule of modular summation states that

(a+b)%c = (a%c + b%c) %c.

Now, (1! +2! +3! +4! ...+100!) mod 24 =

(1!mod24 + 2!mod24 + 3!mod 24 + 4! mod24 + 5!mod24 + 6!mod24 +... + 100!mod24)%24

=( 1 mod 24 + 2 mod 24 + 6 mod 24 + 0 + 0 +0 .. +0)mod 24

= (1 + 2 +6) mod 24

= 9 mod 24

=9.

Hence , 9 is the required answer.

Hope that helps. :)