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A rectanglar cuboid with dimensions 6 x12 x 15 is cut into

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Director
Joined: 07 Jun 2004
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A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 14:01
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A rectanglar cuboid with dimensions 6 x12 x 15 is cut into exact number of equal cubes . What will be the least number of cubes.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 17:09
2*4*5= 40 cubes.

size of each cbe = 3 X 3
Manager
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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Updated on: 09 Jan 2005, 08:22
1
Should be 6.

Total surface area = 684.

6s^2*n = 684, where n is the least number of cubes
s^2*n = 114

prime factorization of 114 = 6 * 19

(sqrt(19))^2 * 6 = 114

n = 6.

Originally posted by Dan on 08 Jan 2005, 21:43.
Last edited by Dan on 09 Jan 2005, 08:22, edited 1 time in total.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 21:48
rxs0005 wrote:
A rectanglar cuboid with dimensions 6 x12 x 15 is cut into exact number of equal cubes . What will be the least number of cubes.

cut the rectangular cubide solid with 3 x 3 x 3 size. if we cut the solid with these dimensions, there will be 40 cubes. this is the least number of equal cubes. more than this we can make 1x1x1 dimensions equal 1080 cubes as well.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 22:00
Side of each cube = n
Total number of cubes = k

6*12*15 = n^3 * k

k = 1080/n^3

Now all we need to do is have a max possible value of n that will result in a integer k.

n = 2, k = non integer

n = 3, k = 40

n = 4, k = non integer

n = 5, k = non integer

n = 6 , k = non integer

n = 7 , k = non integer

n = 8, k = non integer

n = 9 , k = non integer

n > 9 , k < 1

Moreover the choices if there are any would drive the answer

regards
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 22:29
maybe I am wrong, but we're interested in this question in the total surface area of the cuboid rather than in its volume since we want to 'cut' it rather than 'fill' it.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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08 Jan 2005, 22:39
I get 5 cubes as the least number of cubes that you can get from a 6x12x15 rectangular cuboid:

Did the problem similar to neelesh:

k = 1080/n^3 = (6x12x15)/n^3 = (2*3*2*2*3*3*5)/n^3 = (6^3*5)/n^3

when n = 6 k = 5...could be wrong though...whats the OA?
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 01:25
Hi. I think it should be 40 cubes.

Since, we are looking at the least no. of same sized cubes, we are now looking for the max. volume of a cube that can be cut equally.

Now looking at the dimension of cuboid : 6*12* 15 , the max. side that can be cut will be gcd of 6,12 and 15 which is 3. This implies that the max. side of the cube will be 3 with a max. volume of 27.

Total volume of the cuboid = 1080

So the max. no. of cubes of the same side will be 1080/27 = 40
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 05:16
I think the answer is 5 cubes with a side of 6.
( will explain if my answer is correct)
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 08:45
5 cubes for me.

6x12x15 = nx^3 where n is number of cubes and x cube dimension.

2^3*3^3*5 = nx^3

For n to be least x3 has to be max....we get that for n = 5 and

x^3 = 2^3*3^3

x===> 6
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 10:31
Since we want to find out the least number of cubes then we should get the largest cube that can be made so highest common factor of the dimensions of 6 12 15 thats 3 so each cube will be volume 3^3 = 27

Total cubes = 6 * 12 * 15 / 27 = 40

rxs0005
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 10:40
Why is it not x(side) = 6, which will give you a lesser no of cubes i.e. 5 as I explained earlier ?
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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09 Jan 2005, 22:48
neelesh wrote:
Side of each cube = n
Total number of cubes = k

6*12*15 = n^3 * k

k = 1080/n^3

Now all we need to do is have a max possible value of n that will result in a integer k.

n = 2, k = non integer

n = 3, k = 40

n = 4, k = non integer

n = 5, k = non integer

n = 6 , k = non integer

n = 7 , k = non integer

n = 8, k = non integer

n = 9 , k = non integer

n > 9 , k < 1

Moreover the choices if there are any would drive the answer

regards

When n = 6, k=5. So the least number of cube is 5.

rxs0005, is OA = 40?
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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10 Jan 2005, 06:44
i get it now. been understanding less number of cubes.. its the other way around.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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10 Jan 2005, 09:08
The answer is not 5 cubes with each side = 6. Is that because for the side of 15, we cannot cut it to exact number? ie 15/6 is not an integer?

For the cubes with side = 3, each side can be divided into an integer. ie 6, 12, 15 /3 = 2, 4, 5

If so I would agree with the answer 40.
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Re: A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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10 Jan 2005, 09:22
I still do not understand why 5 is not the answer. They are asking for the least number of cubes. If you divide 1080/6^3 you get 5, that is less thatn 40 and isn't that the question, the least number of cubes you can get? I think the OA is wrong!
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A rectanglar cuboid with dimensions 6 x12 x 15 is cut into  [#permalink]

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15 Jan 2005, 21:44
The cube has to be cut exactly. I made a mistake in stating that 5 is the answer.

If the cube is of dimensions 6^3, you would get only 4 cubes and not 5 . The fifth one will be of dimensions 3*6*12. This is leftover, a cuboid, and so 4 as answer cannot satisfy the question's condition.
A rectanglar cuboid with dimensions 6 x12 x 15 is cut into   [#permalink] 15 Jan 2005, 21:44