The 6-digit number 2A683B is divisible by 3. Find the least possible value of (A+B).
2
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of the known digits of 2A683B = 2 + 6 + 8 + 3 = 19, so the digit sum is 19 + (A + B).
For divisibility by 3, 19 + (A + B) must be a multiple of 3. The next multiple of 3 after 19 is 21, giving A + B = 2.
A + B = 2 is achievable (for example A = 1, B = 1 gives 216831), so it is the least possible value.
Hence, the answer is 2.
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Which of the following is/are correct?
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select the correct answer using the code given below:
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