Prismatic puzzle

This week’s Riddler Classic is simple to state, but tricky to figure out:

What whole number dimensions can rectangular prisms have so that their volume (in cubic units) is the same as their surface area (in square units)?

Here is my solution:
[Show Solution]

5 thoughts on “Prismatic puzzle”

  1. You can use ordering to show that c is at least 3, this leaves the most space for b, a. If c is 3 then b is at least 7, again maximizing a. If c is 3 and b is 7 then a is 42.

    Fun fact is that in higher dimensions you can keep going along this route to get that if the max side for dimension d is m_d then m_{d+1}=m_d*(m_d+1)

    Unfortunately this becomes really big really fast

  2. As someone who enjoys solving The Riddler as well, I always look forward to the way you approach these problems, it’s very cool! I just brute forced it in R like so: but I was left thinking about how to approach this more rigorously, so thanks for that.

    This doesn’t use ordering, but tells me that: a>=3 and b>(2 a)/(a – 2) and c = (2 a b)/(a (b – 2) – 2 b). Does that provide any additional insight?

  3. Divide your first equation by 2abc and it reduces to finding three unit fractions whose sum is 1/2. This feels much simpler, although the casework will be the same. The same idea works in higher dimensions: the answer is the number of ways to represent 1/2 as a sum of n unit fractions. Equivalently, it’s the number of ways to represent 1 as a sum of n+1 unit fractions, where one of the fractions must be 1/2. This implies an upper bound on the dimensions in terms of the Sylvester sequence.

    It’s very surprising that the answer is not known for 7. According to, it was calculated in 2004 that there are 159330691 ways to express 1 as the sum of 8 unit fractions, and we just need to know how many of those contain the fraction 1/2. In general, the answer to this problem will be at most A002966(n+1) (set one of the n+1 fractions equal to 1/2), but at least A002966(n) (double all the denominators).

  4. Hi, These 10 solutions are the same as those found by tesselation/tiling equations for 3 polygons of sides n. 4 such combinations tile the plane (e.g. 6,6,6, three hexagons, 4,8,8 two octagons and a square) while 6 are “forbidden”, including my favorite, the 42-gon! Can someone cleverer than I am explain why this appears to be the same math problem? Great job with this proof!

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