Fun with Repeating Decimals

So you may have noticed that when you divide an integer by 11, if it is not a multiple of 11, you get an alternating repeating decimal that looks like “.ABABABABABABAB…”. You may have also noticed that A and B always add up to 9. Let’s discuss why that is.

So let’s divide n by 11, where n is an integer less than 11. You only need to consider integers less than 11, because any improper fraction can be expressed as an integer plus a fraction where the numerator is less than the denominator.
For example:  \frac{48}{11} = 4 \frac{4}{11}

Now if you convert that fraction into a decimal you get an alternating repeating decimal, \frac{4}{11} = 0.363636363..., where 3 and 6 add to 9.
This decimal can also be expressed in another form by turning the decimal places into a sum:
\frac{4}{11} = 3 \cdot \frac{1}{10} + 6 \cdot \frac{1}{100} + 3 \cdot \frac{1}{1000} + 6 \frac{1}{10000} + ...

We can then generalize this for any integer n less than 11.
\frac{n}{11} = A \cdot \frac{1}{10} + B \cdot \frac{1}{100} + A \cdot \frac{1}{1000} + B \frac{1}{10000} +....
Where A and B are integers.
So now we want to prove that A+B = 9.

First let’s put the above equation into a summation form, since the sum is infinitely long, but can be expressed simply because A is multiplied by odd powers of 10, and B is multiplied by even powers of 10. This is done to simplify the expression so there is one instance of A and one instance of B.
\frac{n}{11} = A \cdot \sum_{j=0}^{\infty} 10^{-(2j+1)} + B \cdot \sum_{j=0}^{\infty} 10^{-(2j+2)}

Now we want to simplify this a bit more to get rid of the infinite sum.
First let’s pull out some factors of \frac{1}{10} to make the sums in each term equivalent.
\frac{n}{11} = A \frac{1}{10} \sum_{j=0}^{\infty} 10^{-2j} + B \frac{1}{100} \sum_{j=0}^{\infty} 10^{-2j}
= \left( A \frac{1}{10} + B \frac{1}{100} \right) \sum_{j=0}^{\infty} 10^{-2j}
Now the sum can be reduced to a number: \sum_{j=0}^{\infty} 10^{-2j} = 1.01010101... = \frac{100}{99}

So now we have:
\frac{n}{11} = \left( A \cdot \frac{1}{10} + B \cdot \frac{1}{100} \right) \frac{100}{99}
which can be reduced to a nice expression:
9n = 10A + B
From this we want to prove that A+B=9, where n, A, B are integers.

To prove this, let’s plug in a more general equation A+B = \mu and solve for \mu. The result is:
n = A + \frac{\mu}{9}
Now since n and A are integers, \frac{\mu}{9} must also be an integer, meaning \mu must be a multiple of 9. Since A and B can only hold one digit, the only options are 9 and 18. But \mu =18 only occurs when n=11, which corresponds to \frac{11}{11}=1.
So for the case n<11, it must be that A+B=9.


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