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# Richard cuts a $4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}$ rectangular prism into congruent cu

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Richard cuts a $4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}$ rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?

Aug 23, 2022

#1
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How many will he produce?

Hello Guest!

Richard cuts a $$4\frac{1}{2} ~\text{ft} \times 7\frac{1}{2} ~\text{ft} \times 11\frac{1}{4}~ \text{ft}$$ rectangular prism into congruent cubes without any part of the prism leftover. If he makes the cubes as large as possible, how many will he produce?

Richard divides the length, width and height of the cube into two halves each. In this way he produces with just three cuts 2 times 2 times 2 equals 8 cubes.

!

Aug 23, 2022
edited by asinus  Aug 23, 2022
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Thanks CPhill!

The greatest commom divisor of the three dimensions:

"Das also war des Pudels Kern!"

!

asinus  Aug 23, 2022
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Let us make  the edges of the cube  100  times as large so that  we have a prism  of

450, 750 and 1125  (this prism will be similar to the smaller one )

The greatest commom divisor of the three dimensions   =  75 = the edge length of each cube

On the top  of the prism we can divide one dimension into  450/75  = 6 parts

And, again on the top, we can divide another dimension into  750/75 = 10 parts

So, we have  6 * 10 =  60 cubes in each layer

And we can divide the remaining dimension into  1125/75 = 15 layers  (parts)

So....the max number of  cubes =   6 * 10 * 15 =   900

Aug 23, 2022
edited by CPhill  Aug 23, 2022