*As punishment for misbehaving, Johnny's teacher made him compute*\(\frac{6446266}{9669399} - \frac{6666442}{9999663}\)

*. "That's easy," said Johnny, as he sauntered out of the classroom. "It's the same thing as*\(\frac{2446666}{3669999} - \frac{2446666}{3669999}\)

*, which is zero."*

Reader jeff solved this puzzle in brief. Here's a more leisurely walkthrough, in case interesting.

If you see a fraction like \(\frac{24}{48}\), it's pretty easy to realize that it equals \(\frac{1}{2}\). How about a fraction like \(\frac{1234332}{2468664}\)? This is also \(\frac{1}{2}\). How can you tell? One way is to compare the numerator and the denominator one place value at a time. If you do that, you'll see that each digit in the denominator is twice as large as its corresponding digit in the numerator. Therefore the denominator itself is twice as large as the numerator.

The reasoning behind this "twice as large" example works for other ratios as well. For example, in the fraction \(\frac{123}{369}\), each digit in the denominator is three times as large as the corresponding digit in the numerator. That makes the denominator itself three times as large as the numerator—so the fraction equals \(\frac{1}{3}\).

In the cases above, we "sized" the denominator relative to the numerator, but you could just as well reverse the comparison. Thus, in \(\frac{123}{369}\), we can notice (along the lines of the reader's comment) that each digit in the numerator is one-third as large as the corresponding digit in the denominator. That makes the numerator itself one-third of the denominator—so the fraction equals \(\frac{1}{3}\).

Those two ways of looking at \(\frac{123}{369}\) could be expressed respectively as

\(\frac{123}{369}=\frac{123}{300+60+9} = \frac{123}{3\cdot100+3\cdot20+3\cdot3} = \frac{123}{3\cdot(100+20+3)} = \frac{123}{3\cdot123} = \frac{1}{3}\)

and

\(\frac{123}{369} = \frac{100+20+3}{369} = \frac{\frac{1}{3}300 + \frac{1}{3}{60}+\frac{1}{3}{9}}{369} = \frac{\frac{1}{3}(300+60+9)}{369} = \frac{\frac{1}{3}\cdot369}{369} = \frac{1}{3}\).

One last example: In \(\frac{462}{693}\), each digit in the numerator is two-thirds of the corresponding digit in the denominator. That makes the numerator itself two-thirds of the denominator—so the fraction equals \(\frac{2}{3}\).

When every digit in the numerator is a constant multiple of its corresponding digit in the denominator, that remains true even if the digits are permuted, as long as the digits in the numerator and the digits in the denominator are permuted in the same way. Thus \(\frac{462}{693} = \frac{426}{639} = \frac{246}{369}\), etc. One way to understand Johnny's cheeky response is to observe that he has permuted digits in the two given fractions in such a way as to make them manifestly identical.

Exercise: \(\frac{66336363}{88448484} - \frac{12221}{48884}\) = ?

***

The principles at work in this puzzle might merit a little more discussion. If so, consider this scenario about a hypothetical hotel that has one guest room, one boardroom, and one ballroom. In each of the three rooms there is a party going on.

- In the guest room there are 3 people, 2 of whom are crashing the party. Clearly, in this room \(\frac{2}{3}\) of the people are crashers.
- In the boardroom, there are 60 people, 40 of whom are crashing that party. So in this room, \(\frac{40}{60} = \frac{2}{3}\) of the people are crashers too.
- In the ballroom, there are 900 people, 600 of whom are crashing that party. So in this room as well, \(\frac{600}{900} = \frac{2}{3}\) of the people are crashers.

What fraction of people in the entire hotel are crashers? One way to find out is to add up all of the crashers and then divide by the total number of guests: \(\frac{600 + 40 + 2}{900 + 60 + 3}=\frac{642}{963}\). But another way is to realize that since \(\frac{2}{3}\) of every room is crashers, \(\frac{2}{3}\) of the entire hotel is crashers. The two methods must agree, so \(\frac{642}{963}=\frac{2}{3}\).

***

More abstractly, we can say that if \(\frac{a}{b}=\frac{c}{d}\), then for any \(k\), \(\frac{ka+c}{kb+d} = \frac{c}{d}\). (One can verify this by cross-multiplying. For simplicity I'm taking all of the variables to be positive.) Applying this principle with \(\frac{a}{b}=\frac{4}{6}\), \(\frac{c}{d}=\frac{2}{3}\), and \(k=10\), we have \(\frac{10\cdot4+2}{10\cdot6+3} = \frac{2}{3}\). Applying the principle once more with \(\frac{a}{b}=\frac{600}{900}\), \(\frac{c}{d}=\frac{40 + 2}{60 + 3}\), and \(k=100\), we have \(\frac{100\cdot6+40 + 2}{100\cdot9+60 + 3} = \frac{40 + 2}{60 + 3} = \frac{2}{3}\). And we could keep going to create larger and larger fractions, all equal to \(\frac{2}{3}\). Choosing the \(k\)'s to be powers of 10 creates a series of equivalences that can be read off directly from the digit strings of the multi-digit numbers in the numerator and denominator.

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