If G is any integer greater than 1, show that the final digits of 2^(10^G + G + 1) are the same as all the...

arithmetic/digits/least.significant/power.of.two New, unsolved

New entry, edited from material collected for the archive and not yet in a released version.

Problem

If G is any integer greater than 1, show that the final digits of
2^(10^G + G + 1) are the same as all the digits of 2^(G+1) in base 10.

[ Editor: problem statement reconstructed from the postings; it was a
hint, by Barry Wolk <wolk@ccm.UManitoba.CA>, to a puzzle "What is the
digit?" whose statement is not in the collected material. ]

Show solution

Barry Wolk (wolk@ccm.UManitoba.CA) writes:

> Hint: If G is any integer greater than 1, show that 2^(10^G + G + 1)
> and 2^(G+1) have the same last (G+1) digits in base 10.

Sharath Manjunath (sharath@vtaix.cc.vt.edu) writes:

> I am kind of perplexed here. 2^(G+1) cannot have (G+1) digits at the
> first place. So how can the theorem be proved unless it is restated?

I think that this is just a typo (or to coin a slightly more
appropriate term - "phraso") on BW's part.  The hint should have read
something like: If G is an integer greater than 1, show that the final
digits of 2^(10^G+G+1) are the same as ALL the digits of 2^(G+1) in
base 10.

-- Mike Fee <M.Fee@irl.cri.nz>

No solution was found in the collected postings.