1
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$1! + 2! + 3!+\cdots + 99! + 100!$

I am not getting any idea on how to solve this problem. I know that modular arithmetic should be used but not getting how to start off with the solution. Please give me some hint on how to approach this question. Thanks in advance.

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  • $\begingroup$ HINT: Think of the first factorial that is divisible by $12$ - lets call this $x$. So we know $x!$ is divisible by $12$. Then you know that all others after this will also be divisible by $12$ because for example $(x+1)!=(x+1)*x!$ and $x!$ is divisivble by $12$. That should drastically reduce the set of numbers you need to consider. $\endgroup$ – Mufasa Aug 26 '14 at 16:11
  • $\begingroup$ See math.stackexchange.com/questions/618992/… $\endgroup$ – Yiorgos S. Smyrlis Aug 26 '14 at 16:32
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    $\begingroup$ $1!+2!+3!+\cdots+99!+100!=(1+2+6)+12k$ $\endgroup$ – BigM Aug 26 '14 at 16:41
3
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Hint: Note that $4!\equiv 0\pmod{12}$.

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3
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Since $4!$ is congruent to $0$ (mod $12$) then any multiple of $4!$ is congruent to $0$ (mod $12$). So we need only look at the first 3 terms, and since each of the first 3 terms is congruent to themselves (mod $12$), then the addition of all the terms in (mod $12$) is:

$(1! + 2! + 3! + 0 + 0 +...+ 0)$(mod 12). So the remainder should be 9.

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1
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Hint: Most elements in the sum are divisable by $12$

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0
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Hint $\ n(n\!-\!1)\mid n!\mid (n\!+\!1)!\mid (n\!+\!2)!\mid \cdots\ $ for $\, n>1$

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    $\begingroup$ Generally $\ a_n\! > \cdots\! > a_2 > a_1 \ge 1\,\Rightarrow\, a_n\! \cdots a_2 a_1\mid a_n!\,\ $ i.e. $\ S \subset T\,\Rightarrow\, \prod S\, \mid\, \prod T\ \ \ $ $\endgroup$ – Bill Dubuque Aug 26 '14 at 16:20

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