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Find the smallest perfect cube that is divisible by both 8 and 27.

A72
B216
C1728
D5832
Answer & Solution
Correct answer: B. 216
For the answer to be a perfect cube AND divisible by 8 = 2³ AND by 27 = 3³, the smallest such number is 2³ × 3³ = (2 × 3)³ = 6³ = 216. Both 8 and 27 divide 216 (216/8 = 27, 216/27 = 8). And 216 = 6³ is itself a perfect cube.
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