Dodgy decimals

From the 2014 American Invitational Maths Examination:

The repeating decimals 0.ababab… and 0.abcabc… satisfy

0.ababab… + 0.abcabc… = 33/37.

Find the digits a, b and c.

Converting recurring decimals into fractions is one of the more intriguing topics taught at GCSE level in UK schools. The following facts are useful:

0.010101… = 1/99

0.001001001… = 1/999.

It is a worthwhile exercise trying to prove these if you weren’t already aware of them.

Armed with this information, we turn the given equation into

\dfrac{10a+b}{99} + \dfrac{100a+10b+c}{999} = \dfrac{33}{37}.

Multiplying both sides by 11 x 999 yields

111(10a+b) = 11(100a+10b+c) = 9801,

which simplifies to

2210a + 221b = 9801 - 11c.

Now the left hand side is divisible by 221, so the right hand side must also be divisible by 221. Since c is a digit between 0 and 9, the right hand side must be between 9702 and 9801. The only multiple of 221 in this range is 44 x 221 = 9724, which gives us c = 7.

So

2210a + 221b = 9724.

Dividing both sides by 221 gives us

10a + b = 44.

Finally, since a and b are again digits between 0 and 9, the only possibility is that a = b = 4.

We have our final answer: \boxed{(a,b,c) = (4,4,7)}.

The thing that stuck out for me when tackling this problem was just how much algebra and number theory was packed into it! This makes it a very good exercise for anyone wanting to test their problem solving skills.

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