The 7-digit number 5A931B6 is divisible by 3. Find the least possible value of (A + B).
0
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of the known digits = 5 + 9 + 3 + 1 + 6 = 24, so the digit sum is 24 + A + B.
Since 24 is already a multiple of 3, A + B must be a multiple of 3 (0, 3, 6, …). The least value is A + B = 0, with A = 0 and B = 0.
Check: with A = B = 0 the number is 5093106; its digit sum is 24 and 5093106 ÷ 3 = 1697702 exactly, so it is divisible by 3 and it remains a 7-digit number.
Hence, 0.
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select the correct answer using the code given below:
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