Competition problem — Algebra — Maths Olympiad Prep

Library / /13 of 24

Algebra Difficulty 5.8 AIME, harder Prove it Croatia

If $a$ and $b$ are positive integers, then $\overline\{a.b\}$ is a decimal number obtained by writing the number $a$ , then the decimal point and then the number $b$ . For example, if $a = 20$ and $b = 17$ , then $\overline\{a.b\} = 20.17$ and $\overline\{b.a\} = 17.2$ .
Determine all pairs $(a, b)$ of positive integers such that $\overline\{a.b\} \cdot \overline\{b.a\} = 13$ .

Solution

If $(a, b)$ is a solution, then so is $(b, a)$ – and vice versa. Therefore, we may assume that $a \ge b$ .
If $a \ge 10$ , then necessarily $b = 1$ , and the only possibilities for $a$ are $10$ , $11$ and $12$ .
We can verify that in those cases the product is not $13$ .
Therefore, numbers $a$ and $b$ are both smaller than $10$ , i.e. they are digits. The condition is then equivalent to
$\overline\{ab\} \cdot \overline\{ba\} = 1300.$
Since the last digit of the product is $0$ , assuming $a \ge b$ the only option is $a = 5$ , $b = 2$ . We can check that $5.2 \cdot 2.5 = 13$ , i.e. those numbers satisfy the condition of the problem.
Therefore, the solutions are $(a, b) = (2, 5)$ and $(a, b) = (5, 2)$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.