Cut Out Anything Unnecessary

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As we know

$\large 4!=24 , 5!=24\times5,6!=24\times6\times5 ..........$

Means every number starting from $4!$ will give us remainder as $0$ on dividing by $24$

Means we will get remainder from $\large 1!+2!+3!=9$

and $\large 24 \mod 9=9$

$\therefore$ the answer is $\Huge\Rightarrow\color{royalblue}{\boxed{9}}$

$4!$ is $24$ , then, for $n! \geq 4, 24|n!$ that means its remainder is $0$

So

$(1!+2!+3!+……+100!)\equiv(1!+2!+3!+0+...+0) \mod24$

$(1!+2!+3!+……+100!)\equiv 9\mod24$

$\therefore$ remainder is $\boxed{9}$

By careful observation, we get that $\frac { 1!+2!+3!+....100! }{ 24 } =\frac { 1!+2!+3! }{ 24 } +\frac { 4!+5!+....+100! }{ 24 }$

We know that any number above $4!$ is divisible by $24$ so $\frac { 4!+5!+....+100! }{ 24 }$ has a remainder of $0$

We need to take care of the first fraction: $frac { 1!+2!+3! }{ 24 }$ , which leaves a remainder of $\boxed{9}$

LOL the title of the question is a clue HAHAAH

24 is a factor of 4!, 5!, 6!.......100!. Therefore remainder is 1! +2! +3! = 1+2+6 = 9 Even if the series continued to infinity answer would have been the same.