Good solutions to this were sent in by Chen Jinquan, age 14, The Chinese High School, Singapore, Anders Muller and Sammy Heatley, age 14, Bentley Park College, Rebecca Watson, age 13, Henry Box School and Andrei Lazanu, age 14, School 205, Bucharest, Romania. Rolf Matthews, age 16 sent in a very nice generalisation. Well done all of you!

“I observed the pattern, then I proved it. Writing the decimal expansions we get recurring decimals and I use the brackets to denote that the set of digits inside the brackets is repeated over and over indefinitely.

1/9 = 0.11111… = 0.(1)
1/99 = 0.010101… = 0.(01)
1/999 = 0.001001001 … = 0.(001)
1/9999 = 0.000100010001… = 0.(0001) and so on.

The general pattern is:

I multiply the number by 10ⁿ

Now, I subtract the first expression from the second one:

So, I have proved that I am right.”

Here is Rolf Mathews’ generalization.

”A generalization can be stated as follows: Given a number (a₁)(a₂)(a₃)(a₄)….(an) where each (ai) corresponds to the ith digit of the number, we can create an infinitely repeating decimal of the form: 0.a₁a₂a₃a₄….a _na₁a₂a₃a₄….a _n … by writing it as a fraction with repeated nines as the denominator, that is

a₁a₂a₃a₄….a_n /(a sequence of n 9's).
A proof of this is given for the sequence .234523452345....

Let k=.234523452345....
Then 10000k=2345.23452345....
and 9999k=2345.
Hence k=2345/9999 as desired.

This proof can clearly be modified to prove the generalization for any sequence of (ai)s.”