Question

Find the remainder when 1!+2!+3!+4!+5!+...+100! is divided by 24. Plz write the full solution and not any recommended website to this question .

Answer 1

Let's call this remainder x the result of a%b
means a is divided by b and remainder is x.

Now some properties of modulo...

$(a*b)$%$c=((a$%$c) * (b$%$c))$%$c$

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

these two properties are used in this question.

so,
1! % 24 = 1
2! % 24 = ((1! %24) * (2 % 24)) % 24 = 2
3! % 24 = ((2! %24) * (3 % 24)) % 24 = 6
4! % 24 = ((3! %24) * (4 % 24)) % 24 = 0
5! % 24 = ((4! %24) * (5 % 24)) % 24 = 0
. = 0
. = 0
. = 0
.
.
.
rest all will be zero.
hence answer is
(1! + 2! + 3! ...... + 100!) % 24 = (1! % 24 + 2! %24 + 3! %24 + 4! %24 + .......100! % 24) % 24
= (1 + 2 + 6 + 0 + 0 + 0 .........) % 24
= 9 % 24 = 9

Is that correct answer? Please excuse if there is any calculation error. :)