My Math Forum Obtain 64 with two 4s

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 August 26th, 2017, 10:57 AM #11 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 Okay, c ya in 4^4 / 4 minutes then Thanks from agentredlum, raul21 and Joppy
 August 26th, 2017, 09:13 PM #12 Senior Member   Joined: Jun 2017 From: India Posts: 234 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 Thanks from raul21
August 27th, 2017, 02:16 PM   #13
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Joined: Apr 2014
From: zagreb, croatia

Posts: 234
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Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology
Quote:
 Originally Posted by raul21 I have no patience. $\displaystyle \sqrt{\sqrt{\sqrt{4^{4!}}}}$
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 04:24 PM.

 August 27th, 2017, 04:59 PM #14 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 1,954 Thanks: 132 Math Focus: Trigonometry $\displaystyle \sqrt{4^{\Gamma(4)}}=64$

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