3579?45, make it divisible by 3 and 5.
3579?45 already ends in 5, so it's divisible by 5 no matter what number we pick for ?.
A number is divisible by 3 if all of its digits add up to 3.
3+5+7+9+4+5 = 33 (divisible by 3)
Whatever ? is, 33 + ? has to be divisible by 3.
But since 33 is already divisible by 3, that means ? has to be divisible by 3.
3, 6, and 9 are each divisible by 3, so ? can be 3, 6, or 9.