There are just four 3-digit numbers which are the sums of the cubes of their digits. For example:

$370 = 3^3 + 7^3 + 0^3$ and $371 = 3^3 + 7^3 + 1^3$.

Without using a calculator/computer, can you find the other two 3-digit numbers with this property? Are there any more such numbers?

  • $\begingroup$ Spoilers: answer inside. I’m pretty sure the first part of the answer has to be done by exhaustion. The second part is an actually interesting problem, though $\endgroup$
    – El-Guest
    Commented Jun 5, 2020 at 3:46
  • $\begingroup$ See also OEIS: A005188 $\endgroup$ Commented Jun 5, 2020 at 9:55
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    $\begingroup$ By a difficult and exhaustive search, I've found two more numbers that are the sum of the cubes of their digits: $0, 1$. $\endgroup$ Commented Jun 5, 2020 at 17:32

4 Answers 4


Partial Answer

We are finding digits $a,b,c$ such that $100a+10b+c=a^3+b^3+c^3$. Taking $\pmod 9$, we have $$\big(a^3-a\big)+\big(b^3-b\big)+\big(c^3-c\big)\equiv0\pmod9$$
These are the values of remainder of $a^3-a$ divided by $9$:

a(mod 9)|a^3-a(mod 9)
0       |0
1       |0
2       |6
3       |6
4       |6
5       |3
6       |3
7       |3
8       |0


The $3$ digit numbers that satisfy the condition are either all digits from either groups $(0,1,8,9), (2,3,4), (5,6,7)$ or one digit per group.

  • $\begingroup$ Now it reduces to 42 possibilities, which can be easily bruteforced by hand. $\endgroup$
    – trolley813
    Commented Jun 5, 2020 at 14:10

I happen to know them. Does that count as a valid answer? When I was young, we 'discovered' that repeatedly applying the procedure $abc \to a^3 + b^3 + c^3$ always ended up at one of four numbers; 370, 371,

$153 = 1^3 + 5^3 + 3^3$ or $407 = 4^3 + 0^3 + 7^3$.

For me, it's hard to forget, just like this anecdote about Hardy visiting Ramanujan:

I remember once going to see him when he was ill at Putney. I had ridden in taxi cab number 1729 and remarked that the number seemed to me rather a dull one, and that I hoped it was not an unfavourable omen. "No," he replied, "it is a very interesting number; it is the smallest number expressible as the sum of two cubes in two different ways."

($1729 = 1^3 + 12^3 = 9^3 + 10^3$)

  • 1
    $\begingroup$ Haha well that works too, if you already know the answer :) $\endgroup$ Commented Jun 5, 2020 at 6:53
  • $\begingroup$ That is one of my favorite maths stories, along with Gauss summing 1+2+...+100 in a few seconds in class. I am not entirely sure that the events actually took place, but they certainly make great stories. $\endgroup$ Commented Jun 5, 2020 at 7:03
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    $\begingroup$ Somebody needs to find that taxi cab (or at least the number sign from the top of it) and install it in a mathematics-themed museum somewhere. $\endgroup$ Commented Jun 5, 2020 at 13:45

To show there is no four digit solution, the maximum sum of the cubes of the digits of a four digit number is $4\cdot 9^3=2912$ For a number less than this, the maximum sum of the cubes of the digits is $1+3\cdot 9^3=2188$. The thousands digit must be $1$. To get the sum of cubes up to $1000$ we need a $9$, two $8$s, one $8$ plus two $7$s, or three $7$s. We can check that $1,7,7,7$ and $1,7,7,8$ fail. With two $8$s we have $1^3+2\cdot 8^3=1025$ and all the possibilities fail. Then $1^3+9^3=730$ We need another digit to be at least $4$ to get up to $1000$. This is in the range of hand check as well and nothing works.

  • $\begingroup$ The thousands digit must be 1. Why? $\endgroup$ Commented Jun 5, 2020 at 13:30
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    $\begingroup$ Because the sum of the cubes is less than 2188 at that point and no number with a thousands digit of $2$ matches its sum of cubes because the lower digits cannot contribute enough. $\endgroup$ Commented Jun 5, 2020 at 13:37

There are

  • two 1-digit solutions: $0,1$
  • no 2-digit solutions: $5$ and above have 3-digit cubes. A digit of $4$ would require the number to have another digit of $6$ or above. The 12 possibilities with digits $\le 3$ are easily eliminated.
  • four 3-digit solutions, as indicated in the question.
  • no 4-digit solutions, as Ross Millikan has proved.
  • no higher-digit solutions, as for $n > 4, n \times 9^3$ has fewer than $n$ digits.

So there are six numbers total that are the sum of the cubes of their digits.


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