Modular Arithmetic Approach

There are several observations/proofs that can be made about PPDIs and PDIs that weren't directly applicable to my programs and don't do much to improve the search for them, yet were interesting for their own sake. One of the first approaches that I tried to take when searching for a bound on the number of PDIs was from a modular arithmetic angle as follows:

Let $M$ be a potential order-k PDI in base 10 with digits $m_{n}m_{n-1}\cdots m_{0}$. If $M$ is in fact a PDI, then

$$M = \sum_{i = 1}^{n}10^{i}m_{i} + m_{0} = \sum_{i = 0}^{n}m_{i}^{k}.$$

Therefore,

$$m_{0} \equiv \sum_{i = 0}^{n}m_{i}^{k} \bmod10 .$$

Using this fact I was able to create Table 2, which shows the necessary conditions for a number to be a potential PDI in base 10 if it consists exclusively of a single digit repeated $d$ times. I created this table be calculating the order of each digit mod 10, resulting in a series of equations of the form $nd^{k} \equiv d \bmod10$, and then solving each resulting equation for $n$. The first column of the table indicates the digit that was being calcated, the second column indicates each possible value of $k$ (exponents) to be considered, and the third column indicates the number of digits,$d$,that must in the number for it to be a possible PDI. If a value for $d$ doesn't appear in the third column, then it is impossible for a number with that many digits to be a PDI if it consists entirely of digits in the first column. For instance, it is impossible for a number consisting entirely of 2's to be a base-10 PDI if the number of digits in it is divisible by 5.

Table 2: Exponent/Number of Digit combinations for potential base-10 PDIs consisting of a single digit

Digit	Exponent	Possible number of Digits
1	$k$	1
2	$1 + 4k$	$1 + 5k$
	$2 + 4k$	$3 + 5k$
	$3 + 4k$	$4 + 5k$
	$5k$	$2 + 5k$
3	$1 + 4k$	$1 + 10k$
	$2 + 4k$	$7 + 10k$
	$3 + 4k$	$9 + 10k$
	$4k$	$3 + 10k$
4	$2k -1$	$1 + 5k$
	$2k$	$4 + 5k$
5	$k$	$2k -1$
6	$k$	$6 + 5k$
7	$1 + 4k$	$1 + 10k$
	$2 + 4k$	$3 + 10k$
	$3 + 4k$	$9 + 10k$
	$4 + 4k$	$7 + 10k$
8	$1 + 4k$	$1 + 5k$
	$2 + 4k$	$2 + 5k$
	$3 + 4k$	$4 + 5k$
	$4k$	$3 + 5k$
9	$2k + 1$	$1 + 10k$
	$2k$	$9 + 10k$

Obviously, this approach was quite tedious, and ruling out PDI's consisting entirely of one digit doesn't provide much significant gain. To see if it really was worth pursuing this avenue further, I wrote a series of short programs that tested all numbers from 100,000 to 999,999, summing their digits raised to the $6^{th}$ power. First I calculated the number of values that were equivalent to the original number mod 10, then the number that were equivalent mod 100, and finally the number that were equivalent mod 1000. The results are shown in table 3.

Table 3: number of potential 6 digit base 10 PDIs based on modular arithmetic

mod value	Number of potential PDIs
10	88,800
100	8,516
1000	962

As would be expected, the more significant digits we choose to keep, the more numbers are excluded from being potential PDIs. However, as explained previously, by considering each number as a sequence of digits rather than a value, we have already reduced the number of potential PDIs for any number of digits to

$$\left( \begin{array}{c} B d \end{array}\right)$$

potential numbers, where $B$ is the base and $d$ is the number of digits. For 6-digit base 10 numbers, this value is 210, which is significantly better than even the best value in the table.