The 6-digit number A678B3 is divisible by 3. Find the least possible value of (A+B).
3
A number is divisible by 3 exactly when the sum of its digits is divisible by 3, so the digit sum of A678B3 must be a multiple of 3.
Sum of the known digits = 6 + 7 + 8 + 3 = 24, so the full digit sum is (A + B) + 24.
Since 24 is already a multiple of 3, (A + B) itself must be a multiple of 3, i.e. A + B can be 0, 3, 6, 9, … . But A is the leading digit of a 6-digit number, so A ≥ 1 and A + B = 0 is impossible.
Therefore the least possible value of A + B is 3 (for example A = 1, B = 2, or A = 3, B = 0).
Check: with A = 1 and B = 2 the number is 167823, whose digit sum is 1 + 6 + 7 + 8 + 2 + 3 = 27 = 3 × 9, so it is divisible by 3; sums of 1 or 2 would give 25 or 26, which are not multiples of 3.
Hence, 3.
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Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum.
Which of the following is/are correct?
1. S is always divisible by 74.
2. S is always divisible by 9.
select the correct answer using the code given below:
How many 3-digit natural numbers (without repetition of digits) are there such that each digit is odd and the number is divisible by 5?
