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# A box measuring 54 inches long by 36 inches wide by 12

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Joined: 12 Oct 2012
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A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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Updated on: 19 Nov 2012, 07:53
4
10
00:00

Difficulty:

55% (hard)

Question Stats:

67% (02:19) correct 33% (02:39) wrong based on 192 sessions

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A box measuring 54 inches long by 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 number of cubes that can accomplish this objective?

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

you need the highest edge of the cube to ensure smallest number of cubes fit in the box. )

Originally posted by Mbawarrior01 on 19 Nov 2012, 07:33.
Last edited by Mbawarrior01 on 19 Nov 2012, 07:53, edited 3 times in total.
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19 Nov 2012, 07:41
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1
A box measuring 54 inches long by 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 number of cubes that can accomplish this objective?
17
18
54
108
864
you need the highest edge of the cube to ensure smallest number of cubes fit in the box. )

least number of cubes will be required when the cubes that could fit in are biggest.
6 is the biggest number that could divide all three, 54, 36 and 12.
Thus side of cube must be 6, and total number of cubes = 54/6 * 36/6*12/6 = 108

Ans D it is.
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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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29 Aug 2013, 14:42
Perfect use of HCF, I solved wrong and got 18.
Later realized that 6 is the highest common factor of these numbers.
.... silly conceptual mistake, hope I will not do it on G day.
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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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30 Aug 2013, 17:23
use volumes

volume of box = 54 (2*3^3) * 36 (2^2*3^2) * 12 (2^2*3) = 2^5*3^6

the greatest cub that divide 2^5*3*6 is 2^3*3^3 and the remainder is thus the biggest number of cubes is 2^2*3^3 = 27*4 = 108
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A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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19 Dec 2013, 23:54
1
Find HCF of 54, 36 & 12 = 6

$$\frac{54 * 36 * 12}{6 * 6 * 6}$$
= 108

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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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01 Nov 2015, 12:05
try to see which kind of cubes can fit it the box, so that no empty space is there:
let's see the common factors of 12, 36, and 54
6, 3, 1
12 can't be, since 54 is not divisible by 12.
9 can't be since 12 is not divisible by 9.

so, to minimize the number of cubes, we need to have cubes with the maximum possible edge - 6.
12/6 = 2
36/6 = 6
54/6 = 9
2*6*9 = 108. D.
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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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23 Apr 2018, 17:37
Mbawarrior01 wrote:
A box measuring 54 inches long by 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 number of cubes that can accomplish this objective?

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

The smallest number of identical cubes that can fit into the box without any space left unfilled is one with an edge that is the greatest common factor (GCF) of the three dimensions of the box. Since the dimensions of the box are 54, 36 and 12, their GCF is 6. We should fit 6-inch cubes inside the box, and the number of cubes we can fit is:

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

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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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02 Jun 2018, 08:11
JeffTargetTestPrep wrote:
Mbawarrior01 wrote:
A box measuring 54 inches long by 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 number of cubes that can accomplish this objective?

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

The smallest number of identical cubes that can fit into the box without any space left unfilled is one with an edge that is the greatest common factor (GCF) of the three dimensions of the box. Since the dimensions of the box are 54, 36 and 12, their GCF is 6. We should fit 6-inch cubes inside the box, and the number of cubes we can fit is:

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

Hi JeffTargetTestPrep
Do we need to assume that the edge of the cube will be INTEGER?
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Re: A box measuring 54 inches long by 36 inches wide by 12  [#permalink]

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19 Sep 2019, 18:20
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Re: A box measuring 54 inches long by 36 inches wide by 12   [#permalink] 19 Sep 2019, 18:20
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