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Bunuel
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Bunuel
The product of digits of a positive three-digit integer k is 120. Is k a multiple of 5?

(1) k is a multiple of 3.
(2) If we switch any digits in k, it will decrease.

Breaking Down the Info:
> We first need to break 120 down into three digits.
> We only have two ways to do this: \(120 = 8*15 = 8*3*5\), and observe we may move a 2 over from 8 to get \(120 = 4*6*5\).
> Thus k has 12 possible candidates, and there are 6 possible numbers from each scenario.

Rephrase the question:
> Does k have a unit digit of 0 or 5? Note we cannot have a unit digit of 0 given the product of digits must be 120, therefore we may even go one step further:
> Does k have a unit digit of 5?

Statement 1 Alone:
> The first case \(120 = 8*3*5\) has a sum of digits \(8 + 3 + 5 = 16\), which is not a multiple of 3; so k itself is not a multiple of 3.
> Then by elimination, we confirmed we must have the second case, which indeed has a sum of digits that is a multiple of 3.
> However, this only confirms which scenario we fall in, and we do not know if 5 is the unit digit. Thus this statement is insufficient.

Statement 2 Alone:
> This gives us the formation of the digits, we must have either 853 or 654.
> Both options answer the question "Does k have a unit digit of 5?" with NO.
> Thus this statement is sufficient.

Answer: B
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