The South African Mathematical Olympiad Third Round

If $a>b$, then $a$ has at least five digits, so $a\ge 12345$, and $b$ has at most four digits, so $b\le 9876$. In this case, we have $$\frac{a}{b}\ge \frac{12345}{9876}>1,$$ so the value of $a/b$ that is closest to $1$ is $$\frac{12345}{9876}=1+\frac{2469}{9876}$$ in this case.

On the other hand, if $a<b$ ($a=b$ is impossible, since $a$ and $b$ cannot have the same number of digits), then $a$ has at most four digits and $b$ at least five, so $a\le 9876$ and $b\ge 12345$, which means that $$\frac{a}{b}\le \frac{9876}{12345}<1.$$ Hence in this case the value that is closest to $1$ is $$\frac{9876}{12345}=1-\frac{2469}{12345}.$$ Since the denominator $12345$ is greater than the denominator $9876$, we see that the value of $9876/12345$ is closer to $1$ than that of $12345/9876$ (in fact, we have $9876/12345=4/5$ and $12345/9876=5/4$, so the distances are $1/5$ and $1/4$ respectively). Thus the values of $a$ and $b$ for which $a/b$ is closest to $1$ are $a=9876$ and $b=12345$.