# $90!$ when divided by $n$, gives an odd number. What can be the minimum and the maximum values of $n$?

$90!$ 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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maximum value of $n$ is $90!$ because $90!/90!=1$ which is odd, right? –  Paul Dec 26 '11 at 12:58
$\dfrac{90!}{2^{86}}$ is an odd number. –  Ｊ. Ｍ. Dec 26 '11 at 12:59
What is the 2-adic valuation of $90$? There's a nice formula for this, try some small examples. –  jspecter Dec 26 '11 at 13:00
–  Martin Sleziak Dec 26 '11 at 13:03
@Neal. That's incorrect. $2|90$ and $4\not|90$, but $4|90!$ so $\frac{90!}{2}$ is not even. –  jspecter Dec 26 '11 at 13:04

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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btw how we know for sure that $\dfrac{90!}{2^{86}}$ is odd? –  Quixotic Dec 26 '11 at 13:16
Because dividing by $2^{86}$ leaves only the odd factors? –  Ｊ. Ｍ. Dec 26 '11 at 13:18
$$\begin{split}\frac{90!}{2^{86}}=&3^{44}\times 5^{21}\times 7^{13}\times 11^8\times 13^6\times 17^5\times 19^4\times 23^3\times 29^3\times 31^2\times 37^2\times 41^2\times 43^2\times\\&47\times 53\times 59\times 61\times 67\times 71\times 73\times 79\times 83\times 89\end{split}$$ –  Ｊ. Ｍ. Dec 26 '11 at 13:24
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. –  Dan Petersen Dec 26 '11 at 13:42