# Find all the numbers such that “Sum of cube of digits=Number itself”

I want to find all the numbers such that "Sum of cube of digits=Number itself".

For example, I am seeking for numbers like $153$ as $1^3+5^3+3^3=153$

I know by hit and trial that there are 6 such numbers ( $0,1,153,370,371,407$), But I need it in a pure mathematical way.

• $153, 370, 371, 407$. See, e.g., Narcissistic Numbers – lulu Dec 7 '16 at 17:19
• Man I said I know only six numbers are there those are 0,1,153,370,407,,, But I did it with the help of a program .How to do it mathematically??? – Atul Mishra Dec 7 '16 at 17:22
• To be clear: this is just a finite problem. If $f(n)$ denotes the sum of the cubes of the digits of $n$ then obviously $f(n)≤729\times L(n)$ where $L(n)$ is the number of digits in $n$. But $L(n)=\lceil \log_{10}n \rceil$ so... – lulu Dec 7 '16 at 17:22
• Not following you. Those numbers do not appear anywhere in your post. – lulu Dec 7 '16 at 17:22
• Well, I got it now. :) :) :) – I am Back Dec 8 '16 at 15:07

Let me show you what I have done:

Suppose such a number exist (sum cube of digits=number itself), then it will be of the form (as usual): $$10^{n-1}a_0+10^{n-2}a_1+.........10a_{n-2}+a_{n-1}={a_0}^3+{a_1}^3+.......{a_{n-2}}^3+{a_{n-1}}^3$$ Which on further solving becomes, $$a_0(10^{n-1}-a_0{^2})+a_1(10^{n-2}-a_1{^2})+.........+a_{n-2}(10-a_{n-2}{^2})+a_{n-1}(1-a_{n-1}{^2})=0$$ $$a_0(10^{n-1}-a_0{^2})+a_1(10^{n-2}-a_1{^2})+.........+a_{n-2}(10-a_{n-2}{^2})=a_{n-1}(a_{n-1}{^2}-1)$$

Now, the hardest part comes:

The terms of left are all positive and they increase as we keep on going towards left.

We can only compare the terms $$a_{n-1}(a_{n-1}{^2}-1)$$ and $$a_{n-2}(10-a_{n-2}{^2})+a_{n-3}(100-a_{n-3}^2)$$ because the shortest of other terms is too far from comparison.

Now if we make some comparison, we will have $$a_{n-1}(a_{n-1}{^2}-1)=a_{n-2}(10-a_{n-2}{^2})+(a_{n-3})(100-a_{n-3}{^2})$$

Now let's form a table for the separated functions:

$$\begin{array}{c|c} a_{n-2}& a_{n-2}(10-a_{n-2}{^2})\\\hline 0& 0\\ 1& 9\\ 2& 12\\ 3& 3\\ 4& -24\\ 5& -75\\ 6& -156\\ 7& -273\\ 8& -592\\ 9& -639 \end{array}$$

$$\begin{array}{c|c} a_{n-1}& a_{n-1}(a_{n-1}^{2}-1)\\\hline 0& 0\\ 1& 0\\ 2& 6\\ 3& 24\\ 4& 60\\ 5& 120\\ 6& 210\\ 7& 336\\ 8& 504\\ 9& 720 \end{array}$$

$$\begin{array}{c|c} a_{n-3}& a_{n-3}(100-a_{n-3}^{2})\\\hline 0& 0\\ 1& 99\\ 2& 192\\ 3& 273\\ 4& 336\\ 5& 375\\ 6& 384\\ 7& 357\\ 8& 288\\ 9& 171 \end{array}$$

Now just look at the tables and find the numbers ($a_{n-1},a_{n-2},a_{n-3}$) which follow the following condition:

$$a_{n-3}(100-a_{n-3}{^2})+a_{n-2}(10-a_{n-2}{^2})=a_{n-1}(a_{n-1}{^2}-1)$$

A quick look at the table will give you following set of values of ($a_{n-3},a_{n-2},a_{n-1}$.

$1. 000$ or $0$

$2.001$ or $1$

$3.153$

$4.370$

$5.371$

$6.407$

• This is a lot of work! (+1). I am standing by my claim that simply checking the numbers up to $2473$ is the way to go. – lulu Dec 7 '16 at 17:46
• Yes I know @lulu, As you claimed here. But showing own work gives the greatest feeling. – I am Back Dec 7 '16 at 17:47
• I liked your work @lone wolf And now I am able to understand what you were saying lulu ,,,, sorry for the quarrel – Atul Mishra Dec 7 '16 at 17:48
• No argument from me! Nothing better than digging into the details. – lulu Dec 7 '16 at 17:48