August 26th, 2017, 11:57 AM  #11 
Math Team Joined: Apr 2010 Posts: 2,778 Thanks: 361 
Okay, c ya in 4^4 / 4 minutes then 
August 26th, 2017, 10:13 PM  #12 
Senior Member Joined: Jun 2017 From: India Posts: 232 Thanks: 6 
Here is one more, If A=1, B=2, C=3........Z=26 4 + 4 = FOUR + 4 = ( F + O + U + R ) + 4 = ( 6 + 15 + 21 + 18 ) + 4 = 60 + 4 = 64 
August 27th, 2017, 03:16 PM  #13 
Senior Member Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology  Sorry guys, I'm doing this to explain it to people on another forum. I can't edit. So, we have $\displaystyle \sqrt{\sqrt{\sqrt{4^{4!}}}} = \sqrt{\sqrt{\sqrt{4^{24}}}} = \sqrt{\sqrt{\sqrt{4^{12 \times 2}}}} = \sqrt{\sqrt{\sqrt{\left(4^{12}\right)^{2}}}} = \sqrt{\sqrt{4^{12}}} = \sqrt{\sqrt{4^{6 \times 2}}} = \sqrt{\sqrt{\left(4^6\right)^{2}}} = \sqrt{4^{6}} = \sqrt{4^{3 \times 2}} = \sqrt{\left(4^3\right)^{2}} = 4^{3} = 4 \times 4 \times 4 = 16 \times 4 = 64.$ Last edited by skipjack; August 27th, 2017 at 05:24 PM. 
August 27th, 2017, 05:59 PM  #14 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,428 Thanks: 118 
$\displaystyle \sqrt{4^{\Gamma(4)}}=64$


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