Discovering a Mathematical Reduction Method

By Jonathan Lam on 01/13/18

Tagged: math

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Just another moment of realiation that I felt that I couldn't ignore.

I was attempting an old Math Team problem. It goes as follows:

Solve for x: 1 + 2/(3 + 4/(5 + 6/x)) = 7

And it seemed pretty straightforward. Just keep simplifying the fraction, from the bottom up. Tedious, but do-able. After all, this was a Team round question, the most difficult type of question in the match.

7
= 1 + 2/(3 + 4/(5x/x + 6/x))
= 1 + 2/(3 + 4/((5x + 6)/x))
= 1 + 2/(3 + 4x/(5x + 6))
= 1 + 2/(3(5x + 6)/(5x + 6) + 4x/(5x + 6))
= 1 + 2/((19x + 18)/(5x + 6))
= 1 + 2(5x + 6)/(19x + 18)
= (19x + 18)/(19x + 18) + (10x + 12)/(19x + 18)
=> (29x + 30)/(19x + 18) = 7
29x + 30 = 7(19x + 18)
29x + 30 = 133x + 126
-104x = 96
x = -96/104 = -12/13

Hurray! It's correct! But oh so long and prone to error (I didn't get it right on my first three tries due to arithmetic error)…

But I noticed just after I had solved it that 1 + 2/(some value) = 7. Thus (some value) = 2/6 = 1/3. This eliminated a whole step of creating a single fraction on the left with denominator 19x+18. If we call that fraction z, then we can say that z = 6. And I realized that I could continue this pattern, because z had a similar form: 3 + 4/(another value) = 1/3. Using this method, this solution ensues:

Setting variables:

x = x, y = 5 + 6/x, z = 3 + 4/y

Evaluation:

1 + 2/z = 7 => z = 1/3
3 + 4/y = z = 1/3 => y = -3/2
5 + 6/x = y = -8/3 => x = -12/13

Wow. Just by reducing the function like that, it boils down to three fundamental steps. Not even a single binomial fraction, and constituting almost only one-digit numbers.

I guess the reason I'm so amazed by this is that I've yet to solve so many questions on Project Euler, and that I've never yet discovered a reduction method like this on my own. It's always been brute-forcing up the stack, into the many permutations and bad optimizations. Looking at this problem, I guess the reason is that a lot of what you need to reduce a function is hindsight, a bigger picture. I don't think I ever would have thought of this method without doing it the other way first. It only strengthens my philosophy that there's always a quicker, more optimized way to solve a logical puzzle of at least moderate complexity, and this is what fascinates me so much about math and CS.