The 7-digit number 6A9317B is divisible by 3. Find the least possible value of (A + B).
1
A number is divisible by 3 exactly when the sum of its digits is divisible by 3.
Sum of the digits of 6A9317B = 6 + A + 9 + 3 + 1 + 7 + B = 26 + A + B.
26 leaves remainder 2 when divided by 3, so A + B must leave remainder 1, i.e. A + B can be 1, 4, 7, 10, … Since 6 is the leading digit, A and B may each be 0, so the smallest possible value of A + B is 1.
Check: with A = 0 and B = 1 the number is 6093171, whose digit sum is 6 + 0 + 9 + 3 + 1 + 7 + 1 = 27, which is divisible by 3; A + B = 0 would give digit sum 26, which is not.
Hence, 1.
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