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$90!$ when divided by $n$, gives an odd number. How could we find the minimum and the maximum values of $n$?

I am not sure how to approach this one, any ideas?

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    @Paul: I reposted as an answer.2011-12-26

1 Answers 1

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A result of Legendre (formula 5 in the link, and sometimes also attributed to de Polignac) states that the largest power of a prime $p$ dividing $n!$ is given by

$\sum_{k=1}^{\lfloor\log_p n\rfloor}\left\lfloor\frac{n}{p^k}\right\rfloor$

The highest power of $2$ that divides $90!$ is thus given by

$\left\lfloor\frac{90}{2}\right\rfloor+\left\lfloor\frac{90}{4}\right\rfloor+\left\lfloor\frac{90}{8}\right\rfloor+\left\lfloor\frac{90}{16}\right\rfloor+\left\lfloor\frac{90}{32}\right\rfloor+\left\lfloor\frac{90}{64}\right\rfloor=86$

and thus $\dfrac{90!}{2^{86}}$ is odd. As Paul mentions, $\dfrac{90!}{90!}=1$ is also odd.

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    It's maybe worth mentioning that this result is really easy. The meaning of the formula is the following: out of the numbers 1,...,90, there will be $\lfloor 90/2 \rfloor$ even numbers, and for each of these a factor of 2 in $90!$. Moreover, there will be $\lfloor 90/4 \rfloor$ which are also divisible by four, and each of these contributes an additional factor of two to $90!$. There will be $\lfloor 90/8 \rfloor$ ones divisible by eight, and for each of these *yet another* factor of two. And so on.2011-12-26