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

Question

$90!$ 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?

Answer 1

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.