If a 7-digit number 87321A6 is completely divisible by 6, what is the least value of A?
0
A number is divisible by 6 only if it is divisible by both 2 and 3.
The number 87321A6 already ends in 6, an even digit, so it is always divisible by 2 regardless of the value of A.
For divisibility by 3, the sum of all digits must be divisible by 3: 8 + 7 + 3 + 2 + 1 + A + 6 = 27 + A, so 27 + A must be a multiple of 3.
Since 27 is already divisible by 3, the smallest digit A that keeps 27 + A divisible by 3 is A = 0, because 27 + 0 = 27, which is divisible by 3.
Hence, the least value of A for which 87321A6 is divisible by 6 is 0.
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