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

1 Answer
Jun 11, 2016

#2# or #6#

Explanation:

First note that #1000 = 2^3*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 * 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 >= 100#