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Here is the question: A rectangular prism has a volume of $720$ cm$^3$ and a surface area of $666$ cm$^2$. If the lengths of all its edges are integers, what is the length of the longest edge?

This is from a previously timed competition. Quick answers will be the most helpful.

You can set up equations and use Vieta's formulas to get $x^3+bx^2+333x-720$. How do I solve the problem after this?

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    $\begingroup$ “This is from a timed competition. Fast answers are best”. Would that no be cheating then? $\endgroup$
    – Jack LeGrüß
    Oct 9, 2020 at 19:23
  • $\begingroup$ @JackLeGrüß This is from 2011, why would I cheat $\endgroup$
    – user807252
    Oct 9, 2020 at 19:25
  • $\begingroup$ @JackLeGrüß There are many timed competitions, such as AMC. Have you ever even participated in one? $\endgroup$
    – user807252
    Oct 9, 2020 at 19:25
  • $\begingroup$ @AndrewChin Please post answer in answer fields, and not in comments. $\endgroup$
    – amWhy
    Oct 9, 2020 at 19:26
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    $\begingroup$ @chem1kal: I was enquiring from you,..., not telling you. (No, I have not participated before). $\endgroup$
    – Jack LeGrüß
    Oct 9, 2020 at 19:26

2 Answers 2

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We have $$xyz=720$$ and $$xy+xz+yz=333,$$ where $x$, $y$ and $z$ are naturals.

Now, let $x\geq y\geq z$.

Thus, $$720\geq z^3,$$ which gives $$1\leq z\leq8$$ and $z=3$ is valid, which gives $x=16$ and $y=15.$

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Factorize $720$ as a product of three factors, then pairwise, their products sum to $333$. Once you see this, you know that two of the factors must be odd.

From trial and error, I get the longest side to be $16$, with the other two sides being $3$ and $15$. Verifying:

$$3\times15\times16=720,\qquad 2\times(3\times15+15\times16+16\times3)=666.$$

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