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The digit sum of an integer is the sum of the digits of an integer, with the summation repeated until result is a single digit. For example the digit sum of 87 is 6 because 8+7=15 and 1+5=6. Let the digit sum of the representation N of a number n to base 10 be denoted as DS(N). A remarkable fact is that the digit sum of a number is equal to the remainder upon its division by 9 if 0 and 9 are taken to be equivalent. That is to say if the remainder upon division by 9 is zero then the digit sum is 9. In symbols
with 0=9 mod 9.
A more remarkable property of digit sums is that for N and M, the decimal representations of two numbers
For example, consider 5 times 25. The digit sum of 5 is 5 and of 25 is 7. The product of 5 and 7 is 35 and the digit sum of 35 is 8. The product of 5 and 25 is 125 and the digit sum of 125 is 8.
Now consider integral powers of numbers; i.e., N^{k}. From the above property of digit sums
Thus to examine the digit sums of the kth powers of numbers we need only examine the digit sumes of the digits. The sequence of the digit sums of the squares of numbers is {1, 4, 9, 7, 7, 9, 4, 1, 9}. If a number is to be a perfect square then its digit sum must be 1, 4, 7 or 9. If its digit sum is not one of these numbers then it definitely is not a perfect square. However if its digit sum is one of these numbers it may be but is not necessarily a perfect square..
Shown below are the digit sums of a selection of powers. The table does not show the digit sums for multiples of 3 because for any power 2 or above a multiple of 3 is a multiple of 9 and hence its digit sum is 9. The table also does show the case for the digit 1 because all powers of 1 are 1.
The Digit Sums of Powers of the Decimal Digits  

Digits  
Power  2  4  5  7  8 
2  4  7  7  4  1 
3  8  1  8  1  1 
4  7  4  4  7  8 
5  5  7  2  4  8 
6  1  1  1  1  8 
7  2  4  5  7  1 
8  4  7  7  4  1 
9  8  1  8  1  1 
10  7  4  4  7  8 
11  5  7  2  4  8 
12  1  1  1  1  8 
Let S_{k} be the set of possible digit sums for the kth powers of the digits. Then S_{2} = {1, 4, 7, 9}. The values of S_{k} for k=2 through 12 are:
The Sets of Possible Digit Sums for the Powers of the Decimal Digits 


Power  Set of Possible Digit Sums 
2  {1, 4, 7, 9} 
3  {1, 8, 9} 
4  {1, 4, 7, 8, 9} 
5  {1, 2, 4, 5, 7, 8, 9} 
6  {1, 9} 
7  {1, 2. 4, 5, 7, 8, 9} 
8  {1, 4, 7, 9} 
9  {1, 8, 9} 
10  {1, 4, 7, 9} 
11  {1, 2, 4, 5 7, 9} 
12  {1, 9} 
It is interesting to note that any number which is a sixth power is also a cube and a square. Thus S_{6} must be the intersection of the sets S_{2} and S_{3}. In general
Thus S_{10}= {1, 4, 7, 9}∩{1, 2, 4, 5, 7, 8, 9} = {1, 4, 7, 9}.
It is notable that any number with a digit sum of 3 or 6 cannot be a power of two or greater of any number. This is easy to see for numbers like 12, 15, 33 etc. but it not obvious though true for 1221.
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