*by Tommaso Dorigo*

Two days ago, before returning from Israel, my fiancee Kalliopi and I had a very nice dinner in a kosher restaurant near Rehovot in the company of Eilam Gross, Zohar Komargodski, and Zohar’s wife Olga.

The name of Eilam should be familiar to LHC enthusiasts, as he was the Higgs convener of the ATLAS collaboration when the particle was discovered. As for Zohar, he is a brilliant theorist working in applications of quantum field theory. He is young but already won several awards, among them the prestigious New Horizons in Physics prize.

At some point we were discussing the ability to do calculations by heart, and I mentioned that I could do square roots by heart at the age of five (I think I said four in the conversation – funny how we always increase a little the surprise power of our statements in dinner conversations!); then Zohar bounced off the rest of us a nice little problem: find three cubes that add up to 2017 (that is to say: find three numbers such that, when each of them is elevated to the third power, the results add up to 2017; for instance 1,2,3 is not a solution as 1^3+2^3+3^3 is 36). I quickly decided that the riddle was hard enough to make it a bad idea trying to solve it on the spot, so the conversation went on; but once back home I started to think hard at it, all the while pretending I was listening to my fiancee’s bedtime chat.

After some serious brain overheating, I came up with a solution, and shot it out followed by profanities (thus betraying the lack of attention to my fiancees’ talk): I thought there was a trick that had been concealed in the problem statement, which made the solution much harder to get to. I then proceeded to message Zohar with the solution; he replied saying that the “trick” was not needed, and the solution he intended was different.

So what is your solution to this little problem? Of course it takes just ten lines of code to write a program that solves it, but that would not be much fun… I advise to try it by heart first, and then maybe with paper and pencil if you’re not too good at summing and multiplying numbers. Have fun and let me know the proceedings in the comments thread – then I will disclose more fun facts about the problem and related facts!

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7 January 2017 at 14:31

Hi Tommaso, I guess Zohar implied 11, 7, 7 as the numbers behind the cubes for the 2017 sum; interesting that all of them are primes. [no command lines were harmed in this, just some of my next post’s time]

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7 January 2017 at 15:18

Correct! But the solution I found was different… can you guess it? It involves thinking outside the box a bit…

T.

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7 January 2017 at 15:29

Well, thinking outside the ‘cube’, there is one 0, one 1, one 27 in this number, but I am not sure you mean that..

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7 January 2017 at 16:35

I meant going negative – there are seven more cubic regions of 3D space spanned by xyz integers…

T.

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7 January 2017 at 17:18

So you mean something like: 3375 (15) – 1331 (11) – 27 (3), right?

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7 January 2017 at 23:02

Since you did not specify that they need to be integers, here is a lazy solution: 2017^(1/3), 0, 0

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