# If 345,987,091,543,0?4 is divisible by 8, then what can the missing digit be?

Jun 11, 2016

$2$ or $6$

#### Explanation:

First note that $1000 = {2}^{3} \cdot {5}^{3}$ is divisible by $8 = {2}^{3}$, so we can ignore the initial $345 , 987 , 091 , 543$ part of the number when assessing divisibility by $8$.

We can ignore the leading $0$ of the remaining $3$ digits too.

So we are looking for $2$ digit numbers ending in $4$ and divisible by $8$.

$24 = 8 \cdot 3$ works

The next solution is $24 + 40 = 64$, since $40$ is the least common multiple of $10$ and $8$.

There are no more solutions since $24 - 40 < 0$ and $64 + 40 \ge 100$