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

.

Multiplying both sides by 11 x 999 yields

,

which simplifies to

.

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

.

Dividing both sides by 221 gives us

.

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: .

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.

### Like this:

Like Loading...

*Related*