lean_workbook_plus_67057

Let $n=\overline{abcde}$ . We have $\overline{abcde}=100\overline{abc}+\overline{de}$ , therefore we want $\overline{abc}+\overline{de}$ $=100a+10b+c$ $+10d+e$ $\equiv 0\pmod {11}$ $\Leftrightarrow a-b+c-d+e$ $\equiv 0\pmod {11}$ . Thus, $q+r\equiv 0\pmod {11}$ implies $a-b+c-d+e$ $\equiv 0\pmod {11}$ . Conversely, $a-b+c-d+e$ $\equiv 0\pmod {11}$ implies $\overline{abc}+\overline{de}$ $\equiv 0\pmod {11}$ , i.e. $n$ satisfies the condition. Thus, $n$ satisfies the conditions if and only if $n\equiv 0\pmod {11}$ , and the answer is therefore just the number of $5$ -digit multiples of $11$ , which is just $\lfloor \frac{100000}{11}\rfloor -\lceil \frac{10000}{11} \rceil +1$ $=8181$ .