Estonian Math Competitions

(a) Let the given two-digit number be $\overline{ab}$. The only number that can be obtained by changing the order of digits is $\overline{ba}$. The difference of these numbers is $(10a+b)-(10b+a)=9(a-b)$. To obtain the largest difference, $a$ must be as large as possible and $b$ as small as possible. Since $b=0$ is impossible, we must have $a=9$ and $b=1$ giving $a-b=8$. Hence the desired largest difference is $72$.

(b) Let the given three-digit number be $\overline{abc}$. Interchanging the last two digits can change it by less than $100$. Interchanging the first two digits can change the number by at most $720$ by part (a) of the problem. It remains to study cases where changing the order of digits results in $\overline{cab}$, $\overline{bca}$ or $\overline{cba}$.

The difference of numbers $\overline{abc}$ and $\overline{cab}$ is $(100a+10b+c)-(100c+10a+b)=9(10a+b-11c)$. To obtain the largest difference, $a$ and $b$ must be as large as possible and $c$ as small as possible. Since $c$ is the first digit of the number, $c=0$ is impossible, whence the largest difference is obtained if $a=b=9$ and $c=1$. This difference is $991-199=792$.

The difference of numbers $\overline{abc}$ and $\overline{bca}$ is $(100a+10b+c)-(100b+10c+a)=9(11a-10b-c)$. To obtain the largest difference, $a$ must be as large as possible and both $b$ and $c$ as small as possible, i.e., $a=9$, $b=1$ and $c=0$. This difference is $910-109=801$.

* The difference of numbers $\overline{abc}$ and $\overline{cba}$ is $(100a+10b+c)-(100c+10b+a)=99(a-c)$. To obtain the largest difference, $a$ must be as large as possible and $c$ as small as possible. Since $c\ne 0$, the largest difference is obtained if $a=9$ and $c=1$. Then the difference of the three-digit numbers is $99\cdot 8=792$.

Consequently, the desired largest difference is $801$.