We have the number 153, which has the following special property:
$$153 = 1^3 + 5^3 + 3^3$$
How can we find more numbers like this mathematically (so without making guesses (or even educated guesses) but purely by mathematics)?
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1If you require cubes, all such numbers are here: http://oeis.org/A046197. – Sarastro Feb 09 '13 at 12:00
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1Project Euler question? Or homework? – Paresh Feb 09 '13 at 12:01
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Addition and multiplication are mathematics --- just try lots of numbers (systematically). – Gerry Myerson Feb 09 '13 at 12:01
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@Paresh My math teacher noticed I am bored all the time during math class so yesterday he decided to challenge me. A friend of his recently gifted him a book with all kinds of math problems and we both tried to solve some of them, we didn't have time for this one and quite frankly I wouldn't know how to do it. – TimBurton Feb 09 '13 at 12:02
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@GerryMyerson So basically we can only make educated guesses and just try? For example, I know almost certainly that 153 is the smallest number with this property (except for 1), so we can rule a lot of numbers out – TimBurton Feb 09 '13 at 12:08
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3@TimBurton I see. You seem to have a very nice and dedicated teacher. Good luck! – Paresh Feb 09 '13 at 12:10
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Sometimes you can write down equations and do clever things and find answers; sometimes you just have to get your hands dirty and run the numbers. Maybe there's a clever way to do this one, and I don't see it. – Gerry Myerson Feb 09 '13 at 12:14
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@TimBurton Are you looking for just cubes no matter how many digits the number has? – Git Gud Feb 09 '13 at 12:14
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@GitGud Yes, just like a number which is the sum of the cubes of its digits – TimBurton Feb 09 '13 at 12:15
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@TimBurton Ok. At first I interpreted is as you wanting to find $(a_na_{n-1}\cdots a_1){\text{base}10}$ such that $a_n^n+a{n-1}^{n}+\cdots + a_1^n=(a_na_{n-1}\cdots a_1)_{\text{base}10}$. – Git Gud Feb 09 '13 at 12:16
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If you google 153 fish you get a lot of religious stuff. Enough said, I suppose. – Will Jagy Feb 09 '13 at 20:31
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By the way, G H Hardy singled out the problem of finding numbers equal to the sum of the cubes of the digits for criticism in his book, A Mathematician's Apology: "These are odd facts, very suitable for puzzle columns and likely to amuse amateurs, but there is nothing in them which appeals to the mathematician." http://amathematiciansapology.pandamian.com/16/ – Gerry Myerson Feb 09 '13 at 22:29
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First note that if a number has $5$ digits then it's at least $10000$ but the sum of the cubes of the digits is at most $5\times9^3=3645$. So $5$-digit (and bigger) numbers can't work.
Now we've reduced it to a finite problem --- we just have to try all the numbers from $1$ to $9999$. Actually, if it's a $4$-digit number it's no bigger than $4\times9^3=2916$, so that cuts it way down. With a bit of thought you can probably rule out lots of these numbers without even trying them.
Gerry Myerson
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We have the trivial single-digit solutions $1=1^3$ and $0=0^3$.
For two digits, we need $10a+b=a^3+b^3$, i.e. $a(10-a^2)=b(b^2-1)$ and thus necessarily $1\le a\le 3<\sqrt{10}$, giving $9$ or $12$ or $3$ on the LHS, whereas the RHS grows like $0, 6, 24, \ldots$ - no solution.
– Hagen von Eitzen Feb 09 '13 at 12:53 -
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Evidently 370 and 371 also work. That writer suggests these as fixed points of the "sum the cubes of the decimal digits" function and says most starting numbers eventually arrive at one of the three. Not sure why you couldn't get bigger cycles. Here is one, (55, 250, 133). – Will Jagy Feb 09 '13 at 20:08
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@Will, the cycle structure is given at http://en.wikipedia.org/wiki/Happy_number under the heading, Cubing the digits rather than squaring. There are four fixed points (other than $0$, $1$), two $3$-cycles, and two $2$-cycles. – Gerry Myerson Feb 09 '13 at 22:26
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