MHB Solving for Positive Integer $x$ and $y$ in $\dfrac{2^x}{y!}$

[anemone]

If $\dfrac{1}{9!1!}+\dfrac{1}{7!3!}+\dfrac{1}{5!5!}+\dfrac{1}{3!7!}+\dfrac{1}{1!9!}=\dfrac{2^x}{y!}$ find $x,\,y$ where they are positive integers.

 

[Nono713]

Spoiler

Multiply through by $9!$ to get:
$$1 + 2^2 \cdot 3 + \frac{2 \cdot 3^2 \cdot 7}{5} + 2^2 \cdot 3 + 1 = \frac{2^x \cdot 9!}{y!}$$
Simplfying a bit, we get:
$$\frac{2 \cdot 5 + 2^3 \cdot 3 \cdot 5 + 2 \cdot 3^2 \cdot 7}{5} = \frac{2^x \cdot 9!}{y!}$$
Now note that the numerator in the LHS happens to equal $256 = 2^8$ hence:
$$\frac{2^8}{5} = \frac{2^x 9!}{y!}$$
And so the solution must satisfy:
$$y! = 2^{x - 8} \cdot 5 \cdot 9!$$
Observe that $y \geq 10$ since there are two factors of 5 in the RHS, and that $y < 11$ since 11 is prime and so does not divide the RHS. Thus it follows that $y = 10$, and we get:
$$10! = 2^{x - 8} \cdot 5 \cdot 9! ~ ~ ~ \implies ~ ~ ~ 2 \cdot 5 = 2^{x - 8} \cdot 5 ~ ~~ \implies ~ ~ ~ x = 9$$
So the solution is $(x, y) = (9, 10)$.

 

[Greg]

Spoiler

It's not too tedious to construct Pascal's triangle for this one.

$$\frac{1}{9!1!}+\frac{1}{7!3!}+\frac{1}{5!5!}+\frac{1}{3!7!}+\frac{1}{1!9!}=\frac{1}{10!}\sum_{n=0}^4\binom{10}{2n+1}=\frac{2^9}{10!}\Rightarrow(x,y)=(9,10)$$

 

[anemone]

Thanks to both of you for participating and providing us those awesome solutions! :)