6
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 Jun 5 at 3:46
  • $\begingroup$ See also OEIS: A005188 $\endgroup$ – Daniel Mathias Jun 5 at 9:55
  • 6
    $\begingroup$ By a difficult and exhaustive search, I've found two more numbers that are the sum of the cubes of their digits: $0, 1$. $\endgroup$ – Paul Sinclair Jun 5 at 17:32
8
$\begingroup$

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

So

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Now it reduces to 42 possibilities, which can be easily bruteforced by hand. $\endgroup$ – trolley813 Jun 5 at 14:10
7
$\begingroup$

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."

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

| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Haha well that works too, if you already know the answer :) $\endgroup$ – Dmitry Kamenetsky Jun 5 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. $\endgroup$ – Dmitry Kamenetsky Jun 5 at 7:03
  • 2
    $\begingroup$ 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. $\endgroup$ – Darrel Hoffman Jun 5 at 13:45
6
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The thousands digit must be 1. Why? $\endgroup$ – Ross Presser Jun 5 at 13:30
  • 2
    $\begingroup$ 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. $\endgroup$ – Ross Millikan Jun 5 at 13:37
4
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.