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# Interesting Geo prob

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A rectangular prism has a volume of $$720cm^3$$ and a surface area of $$666cm^3$$. Its edges are integers. What are the side lengths of the prism?

I solved this problem the "dumb" way, by trying out every case, but I wonder if there is a "better", more algebraic way to solve it.

Thanks for any help!!

Apr 17, 2022

#1
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$$𝑥𝑦𝑧=720 \\ 𝑥𝑦+𝑥𝑧+𝑦𝑧=333$$,

Im not sure what to do from here, we could make $$𝑥≥𝑦≥𝑧$$, but I don't see how that helps

Apr 17, 2022
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I did that, and I just brute forced it from here...

Surely there's a better way...

BuilderBoi  Apr 17, 2022
#4
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Did It come with a answer key?

qjin27  Apr 17, 2022
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No, but I found the answer already by guessing and checking, $$x = 16$$$$y = 15$$ and $$z = 3$$

BuilderBoi  Apr 17, 2022
#7
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I think because z=8 is the case that z is the largest, and z=1 is when z is at the smallest, so z has to be in between that

qjin27  Apr 17, 2022
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It is a good start

Melody  Apr 18, 2022
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I found online that you can do

$$720≥z^3 \\ 1≤𝑧≤8 \\$$,

Then you try out 8 cases

Apr 17, 2022
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That's interesting, I wonder why it works....

BuilderBoi  Apr 17, 2022
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Here is how I approached it

$$xyz=720\\ xy+xz+yz=333\\ \sqrt[3]{333}\approx 8.9$$

this means that at least one of the dimensions is 8 or less, and at least one is 9 or more

Factor  720 and you get

$$720=2^4*3^2*5$$

the factors 8 and smaller are  1,2,3,4,5,6,8

Try x=8

If x=8 then  yz=720/8 = 90

8y+8z+90=333   so     8y+8z = 333-90 = 243

But 243 is not divisable by 8 so 8 is not one ofthe dimensions.

Try x=6     I ran into the same problem

Try x=5      same problem

Try x=4    same problem

Try x=3

If x=3 then  yz= 720/3 = 240

3y+3z+240=333

3y+3z=93  so y+z=31

31=15+16, that sounds promising

If the sides, x,y and z  are  3, 15, and 16  then

(3*15)+(3*16)+(15*16) = 45+48+240 = 333  Bingo

the sides are   3,15 and 16

Apr 18, 2022
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