If m! is a multiple of k, then m! + 8k will be too, because we're just adding two multiples of k.
From Statement 1, k is some positive integer less than m , and since m! is the product of all integers from 1 to m, k must be in that product somewhere. So m! is a multiple of k, and Statement 1 is sufficient.
Statement 2 also ensures that k < m, so is also sufficient, and the answer is D.
If \(m\) and \(k\) are positive integers, is \(m!+8k\) a
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