what number is equal to the sum of its digits of its cube?

Answer 1

0, 1, 8, 17, 18, 26, 27 are all of them.
0^3 = 0 and 0=0
1^3 = 1 and 1=1
8^3 = 512 and 5+1+2=8
17^3 = 4913 and 4+9+1+3 = 17
18^3 = 5832 and 5+8+3+2=18
26^3 = 17576 and 1+7+5+7+6=26
27^3 = 19683 and 1+9+6+8+3 = 27

Answer 2

Clearly, as the previous answers have shown, there is more than one such number. One is likely to ask (if one is a math major at least) whether there are infinitely many such numbers. My intuition tells me no. Let's for the moment restrict ourselves to possitive integers (negatives would require some extra rules). Let's try to put a cap on what the solutions could be. Suppose we have a three digit number. When we cube it, it will become a number with between 7 and 9 digits. However, even if we use the maximum number, 999,999,999 (which we can't even actually achieve), the sum can be at most 81, which cannot be equal to our three-digit number. Thus we have an upper bound of 99.

We can use the same process on lower numbers. Take, say, eighty-something. This cubed has 6 digits, for a maximum of 54, not enough. In fact, the earliest this is possible is 54 (54^3 has 6 digits...it doesn't work, but it gives us an easy upper bound).

This is enough to write a program to check all possibilities in seconds:

1 8 17 26 27

These are all the positive integers that meet the condition.
Another interesting problem might be to extend this to rational numbers, but I won't attempt that here.

Answer 3

1 will satisfy this condition since 1^3 = 1
Then we have 8 , 8 X 8 X 8 = 512 = 8
17 X 17 X 17 = 4913 = 17
27 X 27 X 27 = 19683 = 27

Answer 4

17
17^3=4913
4+9+1+3=17

Answer 5

8

8^3 = 512

Answer 6

666