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When I get around at park , I made up a question to hang out by myself.However , I stuck in it.

Question: Let $75!$ be in base of $10$ , then what is the last two digits of it in the base of $3$?

Where I stuck in :

$1-)$ I know how to write $75$ in the base of $3$ , but I do not know writing $75!$ in the base of $3$.

$2-)$ To find the last two digits, we utilize from $(\text{mod }100)$, then which modulus should I use to find its last two digits?

Thanks for your helps..

Bernard
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Not a Salmon Fish
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3 Answers3

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Hint: $75!$ is a multiple of $3^2$.

More generally, the number of trailing zeros in the representation of $n!$ in base $3$ is $$ f(n)=\sum _{{i=1}}^{\infty}\left\lfloor {\frac {n}{3^{i}}}\right\rfloor $$ but note that there are only finitely many nonzero terms.

So, $75!$ has $35$ trailing zeros in base $3$.

lhf
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Since $75!= 1\times 2\times \dots \times 9 \times \dots \times 75$, it is divisible by $9$, which means that it can be written as $$75!= 3^2(a_0 + 3a_1 +3^2a_2 +\dots +3^n a_n) =3^2 a_0 +3^3a_1 +3^4 a_2 +\dots+3^{n+2}a_n\\ =(a_n \dots a_2 \ a_1 a_0 \ 0\ 0)_3 $$

Vishu
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You can answer with Legendre's formula and calculate the $3$-valuation of $75!$: $$ v_3(75!)=\biggl\lfloor\frac{75}3\biggr\rfloor+\biggl\lfloor\frac{75}9\biggr\rfloor + \biggl\lfloor\frac{75}{27}\biggr\rfloor=25+8+2=35,$$ so you know that, in bases $3$, the last $\mathbf{35}$ digits will be $0$.

Bernard
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