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A box measuring 54 inches long by 36 inches wide...

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by AAPL » Sun Jan 14, 2018 7:47 am
A box measuring 54 incches by long 36 inches wide by 12 inches deep is to be filled entirely with identical cubes. No space is to be left unfilled. What is the smallest numbre of cubes that can accomplish this objective?

A. 17
B. 18
C. 54
D. 108
E. 864

The OA is D.

I don't have clear this PS question, I should find the maximum common divisor of the 3 dimensions of the box, right? I appreciate if any expert explain it for me. Thank you so much.
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by [email protected] » Sun Jan 14, 2018 10:38 am
Hi AAPL,

We're told that a box measuring 54 inches by 36 inches by 12 inches is to be filled entirely with identical CUBES and NO space is to be left unfilled. We're asked for the SMALLEST number of cubes that can accomplish this objective.

Since the length, width and height of a cube are the SAME - and we have to fill ALL of the space in the box - we need the length of the cube to be a factor of 54, 36 and 12 (and since we want the LEAST number of cubes possible, we need the greatest common factor of all three numbers). In this case, that would be 6.

6 divides evenly into 54 (9 times), 36 (6 times) and 12 (2 times), so the number of 6x6x6 cubes that will fill this box is (9)(6)(2) = 108 cubes

Final Answer: D

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by Scott@TargetTestPrep » Mon Aug 05, 2019 4:15 pm
AAPL wrote:A box measuring 54 incches by long 36 inches wide by 12 inches deep is to be filled entirely with identical cubes. No space is to be left unfilled. What is the smallest numbre of cubes that can accomplish this objective?

A. 17
B. 18
C. 54
D. 108
E. 864
Since we want to fit the smallest number of cubes in the box, we want each cube to be as large as possible. Since the greatest common factor of 54, 36 and 12 is 6, the largest cube is 6 x 6 x 6. Therefore, the smallest number of cubes we can fit in the box is:

(54 x 36 x 12)/(6 x 6 x 6) = 9 x 6 x 2 = 108

Answer: D

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