A special sum

Evaluate the sum:

$$\mathbf{ \sum_{d \mid 10!} \frac{1}{d+\sqrt{10!}}}$$

Solution



We note that $10! = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7$  hence:

a) $10!$ cannot be a perfect square and
b) it has $270$ divisors since $(8+1)(4+1)(2+1)(1+1)=270$

If ${\rm d} \mid 10! $ , then there exists ${\rm p}$ such ${\rm p d}=10!$ meaning that ${\rm p}$ is a divisor of $10!$. We also note that if one of the ${\rm p, \; d }$ is less that $\sqrt{10!}$ the other is forced to be greater that $\sqrt{10!}$. (Equality does not hold in this case since $10!$ is not a perfect square)

Taking the addend of the sum corresponding to ${\rm d}$ along with its pair that corresponds to ${\rm p}$  we have:

$$\frac{1}{{\rm d}+\sqrt{10!}}+ \frac{1}{{\rm p}+\sqrt{10!}}= \frac{1}{{\rm d}+\sqrt{10!}}+ \frac{1}{\frac{10!}{{\rm d}}+\sqrt{10!}}= \frac{1}{\sqrt{10!}}$$

But we have $135$ such pairs hence the initial sum is summed up to:

$$\mathbf{ \sum_{d \mid 10!} \frac{1}{d+\sqrt{10!}}}= \frac{3\sqrt{7}}{112}$$

and the exercise is complete.