Show that all the divisors of any number of the form 19...9 (with an odd number of nines) end in 1 or 9. For example, the numbers 19, 1999, 199999, and 19999999 are prime (so clearly the property holds), and the (positive) divisors of 1999999999 are 1, 31, 64516129 and 1999999999 itself.
Show further that this property continues to hold if we insert an equal number of zeroes before the nines. For example, the numbers 109, 1000999, 10000099999, 100000009999999, and 1000000000999999999 are prime, and the (positive) divisors of 10000000000099999999999 are 1, 19, 62655709, 1190458471, 8400125030569, 159602375580811, 526315789478947368421, and 10000000000099999999999 itself.
By way of initial exploration, we can ignore even numbers and 5 and all its multiples, since they will never be divisors of 2*10^n1. The same is true of 3 and 13, and their multiples.
In terms of multiplicative orders, it seems that it is possible for a number, N, ending in 3 or 7 to be a divisor of a number of the form 2*10^n1, but only if n is even, since the form of the power will be 2*10^((N1)+2m) where 2m is the even number representing the first case on which 2*10^n modN=1. However, numbers ending in 1 or 9 can perform as divisors in both the odd and even cases; so are the only ones available as divisors of a number of the form 19...9, with an odd number of nines (i.e. where the power of 10 is also odd).
Addendum: See further my subsequent post. In fact 'regular' primes ending in 1 or 9 for which there is a solution 2*10^n1 =k*P for some value of n, that is, those having both 2 and 10 as primitive roots, will ONLY be found in the cases with an odd number of 9's; while 'irregular' i.e. noncyclic P ending in 1 or 9, may appear in either odd or even cases.
I still haven't considered the second part of the puzzle.
Edited on December 16, 2012, 7:45 am

Posted by broll
on 20121214 06:43:41 