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?
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
To find 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, we need to consider the divisibility rule for 3.
The rule states that a number is divisible by 3 if the sum of its digits is divisible by 3.
In this case, we need to find the sum of the digits that could replace the blank space to make the resulting number divisible by 3.
To determine the possible digits that can fill the blank, we need to examine the digits 0 to 9 and see which ones make the number divisible by 3.
Starting with the number 47021, we can calculate the sum of its digits:
4 + 7 + 0 + 2 + 1 = 14
To reach a multiple of 3, we need to find digits such that when added to 14, the resulting sum is divisible by 3.
The possible digits that make the sum divisible by 3 are 1, 4, and 7.
Therefore, the sum of all the possible digits that could fill in the blank in 47_021 so that the resulting five-digit number is divisible by 3 is 1 + 4 + 7 = 12.