# I don't understand one thing in a solution of a number theory task

I'm going through a solution of a number theory task and I don't understand one thing.

$d_1,d_2,...,d_n$ are all natural factors of $10!$. Find the following sum

This sum can be rewritten as

Here's what I don't understand, in the solution it's stated that the following sum is equal to the number of all the factors of 10!. Why? Can someone explain please? I understand all the rest.

$10! = 2^83^45^27^1$. So the the amount of all the natural factors is $(8+1)(4+1)(2+1)(1+1) = 270$.

Hence the answer is $S=270/(2\sqrt{10!})$.

Thanks.

• Use the original definition of $S$: $$S=\sum_{i=1}^n\frac1{d_i+\sqrt{10!}}.$$ So $\sqrt{10!}S=\sum_{j=1}^n\frac{\sqrt{10!}}{d_j+\sqrt{10!}}$. – awllower Jun 26 '17 at 7:34
• @awllower, thanks, I understand it, but I don't get why 2√(10!)S=n. – Snate Jun 26 '17 at 7:38
• Notice that $$\frac{\sqrt{10!}}{d_j+\sqrt{10!}}+\frac{d_j}{d_j+\sqrt{10!}}=1.$$ – awllower Jun 26 '17 at 7:42

First, if $d_1,...,d_n$ are the positive factors of $10!$, then for each $i$ we have $10!/d_i$ is also a factor of $10!$ (since if you multiply by $d_i$ you get $10!$), and all of these are different. So $10!/d_1,...,10!/d_n$ are just $d_1,...,d_n$ in a different order. Therefore
$$\frac{1}{d_1+\sqrt{10!}}+\cdots+\frac{1}{d_n+\sqrt{10!}}=\frac{1}{10!/d_1+\sqrt{10!}}+\cdots+\frac{1}{10!/d_n+\sqrt{10!}}=S,$$ since these sums have the same terms in a different order.
Now $$\frac{1}{10!/d_i+\sqrt{10!}}=\frac{d_i}{10!+d_i\sqrt{10!}}=\frac1{\sqrt{10!}}\frac{d_i}{\sqrt{10!}+d_i},$$ and so $$S=\sum_i\frac{1}{10!/d_i+\sqrt{10!}}=\frac1{\sqrt{10!}}\sum_i\frac{d_i}{\sqrt{10!}+d_i}.$$ Therefore we have $$\sqrt{10!}S=\sum_i\frac{d_i}{\sqrt{10!}+d_i}.$$ But also, using the original sum $\frac{1}{d_1+\sqrt{10!}}+\cdots+\frac{1}{d_n+\sqrt{10!}}$, $$\sqrt{10!}S=\sqrt{10!}\sum_i\frac{1}{d_i+\sqrt{10!}}=\sum_i\frac{\sqrt{10!}}{d_i+\sqrt{10!}}.$$ Adding these two expressions together we get $$2\sqrt{10!}S=\sum_i\frac{d_i+\sqrt{10!}}{d_i+\sqrt{10!}}=n.$$