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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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    maximum value of $n$ is $90!$ because $90!/90!=1$ which is odd, right?2011-12-26
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    $\dfrac{90!}{2^{86}}$ is an odd number.2011-12-26
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    What is the 2-adic valuation of $90$? There's a nice formula for this, try some small examples.2011-12-26
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    @J.M. I think that's an answer.2011-12-26
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    You know that $n$ has to contain every factor of $2$ in $90$. That's the minimum $n$ can be.2011-12-26
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    See here: http://math.stackexchange.com/questions/20737/prime-powers-that-divide-a-factorial2011-12-26
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    @Neal. That's incorrect. $2|90$ and $4\not|90$, but $4|90!$ so $\frac{90!}{2}$ is not even.2011-12-26
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    @J.M.: There is some problem with your equation.2011-12-26
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    @Paul: I reposted as an answer.2011-12-26

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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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    btw how we know for sure that $ \dfrac{90!}{2^{86}}$ is odd?2011-12-26
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    Because dividing by $2^{86}$ leaves only the odd factors?2011-12-26
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    $$\begin{split}\frac{90!}{2^{86}}=&3^{44}\times 5^{21}\times 7^{13}\times 11^8\times 13^6\times 17^5\times 19^4\times 23^3\times 29^3\times 31^2\times 37^2\times 41^2\times 43^2\times\\&47\times 53\times 59\times 61\times 67\times 71\times 73\times 79\times 83\times 89\end{split}$$2011-12-26
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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