Find x and y if $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\cdots+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

If $x,\,y$ are positive integers such that $\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$.

Find $x,\,y$.

This math problem has exactly the same form of the IMO contest problem.

Method 1:

You might argue that this is not at all a hard problem, because one can approach it by straightforward counting, but that depends on if we recognize the need to multiple both sides of the equation by $20!$.

You wonder, why that specific number, $22!$? If you're observant, you will see that the two figures on each denominator on the LHS all add up to 22, therefore, if we multiply each fraction by $22!$, each fraction will turn into an integer instead. That's why we want to multiple both sides of the equation by $22!$.

We can group the pairs of $\dfrac{1}{1!21!},\,\dfrac{1}{21!1!}$, $\dfrac{1}{3!19!},\,\dfrac{1}{19!3!}$ and so on as well to simplify the LHS of the given equation.

$\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}+\dfrac{1}{11!11!}+\dfrac{1}{13!9!}+\dfrac{1}{15!7!}+\dfrac{1}{17!5!}+\dfrac{1}{19!3!}+\dfrac{1}{21!1!}=\dfrac{2^x}{y!}$

$2\left(\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}\right)+\dfrac{1}{11!11!}=\dfrac{2^x}{y!}$

$2(22!)\left(\dfrac{1}{1!21!}+\dfrac{1}{3!19!}+\dfrac{1}{5!17!}+\dfrac{1}{7!15!}+\dfrac{1}{9!13!}\right)+22!\left(\dfrac{1}{11!11!}\right)=22!\left(\dfrac{2^x}{y!}\right)$

$2\left(\dfrac{22!}{1!21!}+\dfrac{22!}{3!19!}+\dfrac{22!}{5!17!}+\dfrac{22!}{7!15!}+\dfrac{22!}{9!13!}\right)+\dfrac{22!}{11!11!}=\dfrac{2^x22!}{y!}$

This is really the combinatorial formula $n$ choose $r$, where we have:

[MATH] 2\left({22 \choose 1}+{22 \choose 3}+{22 \choose 5}+{22 \choose 7}+{22 \choose 9}\right)+{22 \choose 11}=\dfrac{2^x22!}{y!}[/MATH]

[MATH] 2(695,860)+705,432=\dfrac{2^x22!}{y!}[/MATH]

[MATH] 2,097,152=\dfrac{2^x22!}{y!}[/MATH]

[MATH] 2^{21}=\dfrac{2^x22!}{y!}[/MATH]

Since both $x$ and $y$ are positive integers, we can conclude by now that:

$x=21$; $y=22$

And that's all there is to it.

But, math educators can make optimal use of this problem so to show their students that there is another way to approach this problem as well.

Attention to detail and the ability to look at a problem from all angles are two important traits students must have that are waiting to be cultivated by the teachers and to flourish.

Having said that, this problem can be approached by making smarter choice to avoid using calculator to figure out the values for $x$ and $y$.

Hey math educators out there, the ball is in your court now whether you want to generate easily satisfied students or students who always want to improve and go out of their comfort zone in solving hard challenge math problems.

I will write in my next blog about how to approach this problem without the help of calculator, see ya!