﻿ The Digit Sums of Powers of Positive Integers
San José State University

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Thayer Watkins
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 The Digit Sums of Powers of Positive Integers

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

#### DS(N) = n mod 9

with 0=9 mod 9.

A more remarkable property of digit sums is that for N and M, the decimal representations of two numbers

#### DS(N*M) = DS(DS(N)*DS(M))

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., Nk. From the above property of digit sums

#### DS( Nk) = DS( (DS(N))k)

Thus to examine the digit sums of the k-th 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.

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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 Sk be the set of possible digit sums for the k-th powers of the digits. Then S2 = {1, 4, 7, 9}. The values of Sk for k=2 through 12 are:

The Sets of Possible Digit Sums
for the Powers of the Decimal Digits
PowerSet 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 S6 must be the intersection of the sets S2 and S3. In general

#### Spq = Sp∩Sq

.

Thus S10= {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.