Can you tell me what power is greater?

If the equation $36! ≡ 0 (mod$ $n)$ becomes true only when we achieve the number 36! , this means that any divisors of $n$ are being "taken away" at each factor of this number, and the last divisor (or divisors) is (are) taken on $36=2^{2} \times{3^{2}}$ . The divisor taken on 36! only can be 2 or 3 . The number of times which the factor 2 appears between 1 and 35 , can be counted by the intervals. Hence, we have:

$\frac{[2, 34]}{2}=17$

$\frac{[4, 32]}{4}=8$

$\frac{[8, 32]}{8}=4$

$\frac{[16, 32]}{16}=2$

$\frac{32}{32}=1$

$17+8+4+2+1=32$

$\frac{[3, 33]}{3}=11$

$\frac{[9, 27]}{9}=3$

$\frac{27}{27}=1$

$11+3+1=15$

When 36! takes away any divisor, n becomes $2^{33}$ or $3^{16}$ . Now, we only need to find which power is smaller. Two different numbers can only be smaller than another, equal to another or greater than another. We can use a question mark as operator, as follows:

$2^{33}$ ? $3^{16}$

$2^{\frac{33}{16}}$ ? $3$

$2^{2} \times {2^{\frac{1}{16}}}$ ? $3$

$2^{\frac{1}{16}}$ ? $\frac{3}{4}$

As we know, any power of a number greater than one, is greater than one too. So, the left side of the comparation is greater than the right side.

$2^{33}$ > $3^{16}$

$n=a^{b}=3^{16}$

And the sum is $a+b=3+16=\boxed{19}$ .