Consider all the (non leading zero) positive ten-digit pandigitals arranged in descending order..
Determine the millionth number on the list.
*** Computer program/spreadsheet assisted methodologies are welcome, but extra credit will be given for an analytical solution.
There is an idea of Base Factorial (BF), where the positional values step up through the factorials instead of powers of a base integer. Each digit goes from 0 to n, where n is the position with a positional value of n!.
So a four digit BF number would have positional values of 24, 6, 2, and 1; and the digits allowed for each position are 0-4, 0-3, 0-2, 0-1 respectively. Then as an example 99 = 4*24 + 0*6 + 1*2 + 1*1, which makes 99 = 4011_BF.
So now to the problem. I'll index the list of pandigitals starting from 0. Then the 1,000,000th number corresponds to 999,999th index. Now 999,999 = 266,251,211_BF. As follows from 999,999 = 2*9! + 6*8! + 6*7! + 2*6! + 5*5! + 1*4! + 2*3! + 1*2! + 1*1!.
Next I'll increment the BF digits by 1 and express as ordinals: 3rd,7th,7th,3rd,6th,2nd,3rd,2nd,2nd.
Now for each ordinal I will in turn remove that position from the starting list of descending digits {9,8,7,6,5,4,3,2,1,0}, without replacement.
3rd of {9,8,7,6,5,4,3,2,1,0} is 7
7th of {9,8,6,5,4,3,1,0} is 2
7th of {9,8,6,5,4,3,1,0} is 1
3rd of {9,8,6,5,4,3,0} is 6
6th of {9,8,5,4,3} is 0
2nd of {9,8,5,4,3} is 8
3rd of {9,5,4,3} is 4
2nd of {9,5,3} is 5
2nd of {9,3} is 3
The final digit remaining {9} is the last digit.
So now just read the chosen digits in order to get 7126084539 as the millionth number on the list of positive ten-digit pandigitals arranged in descending order.
Base Factorial (Solution)
