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.

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

 

[anemone]

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

 

[CellCoree]

I would approach this problem by first understanding the given equation and identifying any patterns or relationships between the numbers involved. The expression on the left side of the equation can be rewritten as a sum of fractions with a common denominator of $9!$:
$$\dfrac{1}{9!1!}+\dfrac{1}{7!3!}+\dfrac{1}{5!5!}+\dfrac{1}{3!7!}+\dfrac{1}{1!9!} = \dfrac{9}{9!}+\dfrac{35}{9!}+\dfrac{21}{9!}+\dfrac{35}{9!}+\dfrac{9}{9!} = \dfrac{109}{9!}$$
We can also express the right side of the equation in a similar way, by using the formula for $y! = y\times(y-1)\times(y-2)\times\dots\times3\times2\times1$:
$$\dfrac{2^x}{y!} = \dfrac{2^x}{y\times(y-1)\times(y-2)\times\dots\times3\times2\times1}$$
From this, we can see that the denominator of the right side will always be a multiple of $9!$, since it includes all the numbers from $1$ to $y$. This means that $y$ must be a positive integer that is a multiple of $9$, since the denominator of the left side is also $9!$.

Now, let's look at the numerator of the right side, $2^x$. We know that $2^x$ is the product of $x$ twos, so it must be a multiple of $2$. However, since the denominator of the left side is $9!$, which includes all the numbers from $1$ to $9$, we can see that $2^x$ must also be a multiple of $9$. This means that $x$ must be greater than or equal to $2$, since $2^1 = 2$ is not a multiple of $9$.

To summarize, we have determined that $y$ must be a positive integer that is a multiple of $9$, and $x$ must be a positive integer greater than or equal to $2