A number is divisible by 3 if the sum of all the digits in the number is divisible by 3
eg. 25632 divides evenly by 3
because 2+5+6+3+2 = 18 which divides by 3
but 25631 has digits that add up to 17
so 25631 does not evenly divide by 3
so let the missing digit in your number be x
then 17X021 has a sum of 1+7+x+0+2+1 = x+11
it is easy to see that x+11 evenly divides by 3 when
x = 1, 4, or 7
What is the sum of all possible digits that could fill in the blank
in 47_021 so that the resulting five digit number is divisible by 3?
1 answer