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$$(V+W+X+Y+Z)^3 = V \times 10^4 + W \times 10^3 + X \times 10^2 + Y \times 10 + Z \\ V, W, X, Y, Z \in \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$$

($V$ through $Z$ are digits in a five-digit number)

Solve for $V$, $W$, $X$, $Y$ and $Z$.

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  • $\begingroup$ Isn't it as simple as putting all the alegraic numbers 0?$(0+0+0+0+0)^3=0 \times 0 \times 0 \times 0 \times 0$? OR $(0+0+0+0+0+0)^3=00000$ $\endgroup$ – ministic2001 May 30 '15 at 12:04
  • $\begingroup$ I will tell you guys the right answer on next saturday $\endgroup$ – shayan48 Jun 1 '15 at 5:40
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The complete solution set including repeated digits and leading zeroes is:

00000 = 0^3
00001 = 1^3
00512 = 8^3
04913 = 17^3
05832 = 18^3
17576 = 26^3
19683 = 27^3

Since every solution must be a cube number, the easiest way of generating the solution set is to compute the cube of the numbers from 0 to 45, and for each result check if it is a legal solution.

It is worth noting that if we increase the allowed length of the number, no extra solutions are found. All 6-digit options are trivially tested, and for 7 digits and up it is easy to show that the digit-sum cubed will contain fewer digits than the original number.

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  • $\begingroup$ Could you share the code you used to find these? $\endgroup$ – user1717828 May 31 '15 at 16:22
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    $\begingroup$ @user1717828 JavaScript: for(a=0;a<46;a++){b=a*a*a;c=0;while(b){c+=b%10;b=Math.floor(b/10)};if(c==a){console.log(a+" "+a*a*a)}} $\endgroup$ – aaaaaaaaaaaa May 31 '15 at 18:31
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    $\begingroup$ @user1717828 Or Python 3: print(*map(str,(i**3 for i in range(1,int(10e5**0.5))if sum(map(int,str(i)))**3==i)),sep="\n") $\endgroup$ – user10203 May 31 '15 at 19:19
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Solution(s) :

If the Digits can not repeat :
(1+9+6+8+3)^3=27^3=19683

If the Digits can repeat :
(1+7+5+7+6)^3=26^3=17576

Details :

If the Digits are different, then maximum 5 Digit number can be 9+8+7+6+5 which has a Digit Sum of 35. The minimum 5 Digit number is 10234, whose cube root is a little more than 21. So the Solution must involve the cube of integers between 22 & 35. Check the cubes of these 14 numbers to find (1) which has Digits with the required sum (2) which has no Repeating Digits, to quickly get the first solution, then continue looking further for solutions which have Repeating Digits, to get the second solution.

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