Prove analytically that for any nonnegative integer k there exists a power of 3, terminating with 0..001, i.e. a sequence of k zeroes followed by digit 1,
e.g. k=0==> 3^0=1, 3^4 =81;
k=1==> 3^20=3486784401,
k=2==> 3^100= 515377520732011331036461129765621272702107522001.
Bonus question: Show that replacing the number 3 by any odd number, not terminating by 5, leaves your proof valid.

(In reply to Adding zeros (full solution) by Jer)

Jer,

I  like your reasoning and appreciate your solution.

However, prior to publishing my solution- let me pose  this question:

Why pigeons will like it?

 

Ady

Posted by Ady TZIDON on 2010-03-28 21:22:12