## % Change

From the 2013 South Africa Olympiad:

A is a 2-digit number and B is a 3-digit number such that A increased by B% equals B reduced by A%. Find all possible pairs (A, B).

The information presented yields the equation

$A(1+\frac{B}{100}) = B(1-\frac{A}{100})$.

This leads to

$A(100+B) = B(100 - A)$,

so that

$100A+2AB = 100B$,

and hence

$A = \frac{50B}{50+B}$.

Since A is an integer, it follows that 50+B divides 50B. Well, 50+B also divides 50(50+B) = 2500+50B, so in fact 50+B must divide 2500. Remembering that B is a 3-digit number yields just three possible values of B: 200, 450 and 575. Substituting back into the equation above yields corresponding values for A of 40, 45 and 46 respectively. As a final check, we do indeed have

40 increased by 200% = 200 reduced by 40% = 120,

45 increased by 450% = 450 reduced by 45% = 247.5,

46 increased by 575% = 575 reduced by 46% = 310.5.

Therefore the set of possible pairs (A, B) is (40, 200), (45, 450), (46, 575).

A beautiful problem, borne out of a natural question that anyone fluent in percentages might ask themselves.

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