In the six-digit integer 3A6,792, what is the largest digit A so that the six-digit integer will be divisible by 3?
+312 The requirement for divisibility by 3 is that all the digits have to add up to a multiple of 3.
3 + 6 + 7 + 9 + 2 = 27
This is already divisible by 3 so you can say that the largest multiple of 3 from 1 - 10 is the value of A.
3 * 1 = 3
3 * 2 = 6
3 * 3 = 9
A = 9; the sum of the digits is 36 which is 12 * 3
