If k is a multiple of 24 but not a multiple of 16, which of the following cannot be an integer?
(a) k/8 (b) k/9 (c) k/32 (d) k/36 (e) k/81
Can anyone help me with the prime factorization method to solve this problem…??
24 = 2^3 \times3^1
Constraint :16 = 2^4 → meaning that we cannot have more than three 2 in our number(k).
Only options which has more than three 2’s is option C. \frac{k}{32}, where 32 = 2^5
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thank you very much… i appreciate it