Question 717356
If a number is divisible by 3 then the sum of the digits of that number are also divisible by 3. So we want the 4 and the 5 and the "four different digits" to add up to a number divisible by 3.<br>
Probably the easiest way to do this is to pick three of the digits and then see what 4th digit would make the sum of the digits be divisible by 3. Let's say the 3 of the 4 digits are 1, 2 and 3 and let's call the one missing digit "n". So the digits are 4123n5. Now let's add the digits: 4+1+2+3+n+5. This simplifies to 15+n. So what digit (or digits) would make 15+n divisible by 3? I'll leave it up to you to figure it out. Hint: There are 4 digits that would make 15+n divisible by 3. One of them is the same as one of the digits already in the number. So there are three possible 4th digits (when you pick 1, 2 and 3 to start with) which would make the number divisible by 3 and would be different from the others.