The teacher looked at Ted’s arithmetic sheet. The answer was correct, but there was something suspicious about the working: Ted had simply cancelled the 6s, top and bottom.

‘You can’t do that,’ growled the teacher. ‘You were just lucky you got the right answer. Look, suppose we take 49/98, and cancel the 9s, then we get ... . No! Ignore that example!

Let us take 16/64, and cancel the 6s. Now we get ... . Really, I don’t know where this new maths is leading us.’

There is in fact only one other ratio of two-digit numbers lying between 0 and 1 which simplifies correctly by such illegal cancellation. What is it?

Hint 1

Remember that a < c, and that a, b and c are all integers.

Hint 2

We need to investigate
^ab/_bc = ^a_/c, remembering that ab stands for 10a + b.

Solution

We wish to solve the equation

10a + b = a .
10b + c    c

Multiplying out gives 10ac + bc = 10ab + ac, or 9ac + bc = 10ab.

We need to find solutions for this equation for which a, b and c are single digits. This suggests we ‘solve’ for one of a, b, or c. Since we know that a < c, we try solving for b in terms of a and c. We obtain

b = 9ac / (10a – c ).

We now have to substitute values for a and c which make b an integer. We do this systematically, (discounting the trivial cases a = b = c ). For a = 3, 5, 6, 7, 8, 9 there are no non-trivial solutions. This leaves:

a = 1: c = 4, b = 6 (16/64 = 1/4);  
c = 5, b = 9 (19/95 = 1/5).
a = 2: c = 5, b = 6 (26/65 = 2/5).
a = 4: c = 8, b = 9 (49/98 = 4/8).

Hence the only remaining fraction that gives the correct result with this illegal cancellation is 19/95.

Extensions

1. Investigate some other possible illegal cancellations.
For example, are the following ever valid?

^ab/_ac = ^b/_c, ^ab/_cb = ^a/_c, ^ab/_ca = ^b/_c<sup>.

2. We could extend this investigation to quotients of 3 digits. One might expect this to be a harder problem!</sup>

Hint 1

Hint 2

Solution

Extensions