If 321y72 is a multiple of 6, where y is a digit, what is the least value of y?

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To determine the least value of y such that 321y72 is divisible by 6, we need to check the divisibility rules for 6. A number is divisible by 6 if and only if it is divisible by both 2 and 3.

1. Divisibility by 2: The last digit must be even. Here, the last digit is 2, which is even, so the number is divisible by 2 regardless of y.

2. Divisibility by 3: The sum of the digits must be divisible by 3.

Sum of digits = 3 + 2 + 1 + y + 7 + 2 = 15 + y

We want 15 + y to be divisible by 3.

Since 15 is divisible by 3, y must be such that y mod 3 = 0.

Possible digits for y that satisfy this are 0, 3, 6, 9.

Among these, the least value is 0.

Therefore, the least value of y is 0.

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