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The Digit Sum Arithmetic of Exponentiation

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

D(a^{b}) = (D(a))^{b} for a and b nonnegative integers

but further simplification is possible.

Let δ be a decimal digit. Consider the possibility that there might exist an integer
ε>0 such that

D(δ^{ε}) = 1

If such an ε exists then

D(δ^{b}) = δ^{b%ε}

where b%ε is the remainder of b upon division by ε.

Since 2^{6}=64 and D(64)=1, ε(2)=6. Sixty four is also 4^{3} so ε(4)=3. Likewise 64 is
8^{2} and hence ε(8)=2. For 5 it is found that 5^{6}=15625 and D(15625)=1 so &epilon;(5)=6. For
7 it is found that 7^{3}=343 and D(343)=1 so ε(7)=3.

For an illustration consider 14^{14}. The digit sum of 14 is 5 and ε(5)=6. The remainder for 14 upon division
by 6 is 2. Thus

D(14^{14}) = D(5^{14}) = D(5^{2}) = D(25) = 7

No such ε exists for 3, 6 and 9. However for a equal to 3, 6 or 9 D(a^{b}) 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.