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Challenge: Create 64 with two 4's

soroban

Well-known member
Feb 2, 2012
409

Using any of [tex]\{+,\;-,\;\times,\;\div,\;x^y,\;\sqrt{x},\;x!\}[/tex]

. . create 64 with two 4's.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: Challenge

In base 15:

44
 

eddybob123

Active member
Aug 18, 2013
76
Re: Challenge

(Clapping)
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Re: Challenge

A Perelman-like solution suffices :

$-\lg \, \log_{\sqrt{4}} \, \underbrace{\sqrt{\sqrt{...\sqrt{4}}}}_{\text{65 times}}$

Or even,

$4^\left ({\log \sqrt{\sqrt{\sqrt{e^{4!}}}}} \right )$

But as $\log$ is not desired, a twist around the base of logarithm of the latter should do :

$\sqrt{\sqrt{\sqrt{4^{4!}}}}$
 
Last edited:

soroban

Well-known member
Feb 2, 2012
409
Re: Challenge

Hello, mathbalarka!

That's it! . . . Nice reasoning!

Alternate form: .[tex]4^{\left(\sqrt{\sqrt{\sqrt{4!}}}\right)}[/tex]
 

mathbalarka

Well-known member
MHB Math Helper
Mar 22, 2013
573
Re: Challenge

Okay, thanks! The other form didn't click to me, really nice!

Now, in my post, the first form using logarithm is referred to as "Perelman-like" as another related, but twisted problem was given in Yakov Perelman's Mathematics Is Fun. He refers a challenger in a congress of physicist in Odessa. I couldn't find further reference, so just named this after him. (a year ago when I found out this kind of approach is in general very doable for these problems)