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The digit sum of a number's representation in any base is the repeated sum of digits under a single digit is arrived at. For example, consider the number 543 in base 10. The sum of its digits is 12 and the sum of the digits of 12 is 3, so the digit sum of 543 is 3. Let D(N) be the digit sum of the representation of a number N. The digit sum arithmetic of addition, multiplication and division is dealt with elsewhere. Here the arithmetic of exponentiation is dealt with.
Since exponentiation by integral powers is simply repeated multiplication
but further simplification is possible.
Let δ be a decimal digit. Consider the possibility that there might exist an integer ε>0 such that
If such an ε exists then
where b%ε is the remainder of b upon division by ε.
Since 26=64 and D(64)=1, ε(2)=6. Sixty four is also 43 so ε(4)=3. Likewise 64 is 82 and hence ε(8)=2. For 5 it is found that 56=15625 and D(15625)=1 so &epilon;(5)=6. For 7 it is found that 73=343 and D(343)=1 so ε(7)=3.
For an illustration consider 1414. The digit sum of 14 is 5 and ε(5)=6. The remainder for 14 upon division by 6 is 2. Thus
No such ε exists for 3, 6 and 9. However for a equal to 3, 6 or 9 D(ab) is equal to a for b=1 and equal to 9 for any b>1 because any power of these numbers above one is a multiple of 9.
There is some extension of the results for negative powers for a equal to 2, 4, 51/ and 8 since D(1/2)=5, D(4)=7, D(1/5)=2, D(1/8)=8. There is also an extension of the concept of digit sum to repeating decimals in which the digit sum of 1/7 can be construed to be 4.
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