Four Fours/Table of 4s: Difference between revisions
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==Two 4s== | ==Two 4s== | ||
Note: we aren't using fourth roots anywhere, e.g., <math>\sqrt{2} = \sqrt[4]{4}</math>. Adding this would greatly expand the possibilities. | |||
<math> | <math> | ||
Revision as of 21:54, 9 March 2019
Back to Four Fours
One 4
Believe it or not, the rules allow you to do quite a bit with a single 4. The rules say you may combine 4s with any mathematical symbol except numbers. Thus, in addition to 4 alone, we also have:
$ 1 = i^{4} $
$ 2 = \sqrt{4} $
$ 24 = 4! $
The following fractions are also useful:
$ \dfrac{1}{4} $
$ \dfrac{1}{2} = \dfrac{1}{\sqrt{4}} $
$ \dfrac{1}{24} = \dfrac{1}{4!} $
Could possibly add constants (harmonic number), possibly special functions.
Once you allow variables like x into the mix, it's lights out.
One 4 with Variables
$ 5 = \dfrac{ \ln{ \left( \dfrac{ \ln{ \left( \sqrt{ \sqrt{ \sqrt{ \sqrt{ \sqrt{ x } } } } } \right) } }{ \ln{x} } \right) } }{ \ln{\sqrt{4}} } $
Two 4s
Note: we aren't using fourth roots anywhere, e.g., $ \sqrt{2} = \sqrt[4]{4} $. Adding this would greatly expand the possibilities.
$ 1 = \dfrac{4}{4} $
$ 1 = \log_{4}(4) $
$ 2 = 4 - \sqrt{4} $
$ 2 = \dfrac{4}{\sqrt{4}} $
$ 3 = \sqrt{4} + i^{4} $
$ 3 = 4 - i^{4} $
$ 4 = \sqrt{4} \times \sqrt{4} $
$ 4 = \sqrt{4} + \sqrt{4} $
$ 4 = \dfrac{4}{i^4} $
$ 5 = 4 + i^{4} $
$ 5 = \sqrt{ 4! + i^4 } $
$ 6 = 4 + \sqrt{4} $
$ 6 = (4 - i^4)! $
$ 8 = 4+4 $
$ 8 = 4 \sqrt{4} $
$ 12 = \dfrac{4!}{\sqrt{4}} $
$ 16 = 4 \times 4 $
$ 16 = 4^{\sqrt{4}} = (\sqrt{4})^4 $
$ 16 = \sqrt{4^4} $
$ 20 = 4! - 4 $
$ 22 = 4! - \sqrt{4} $
$ 23 = 4! - i^4 $
$ 24 = \sqrt{ (4!)^{\sqrt{4}} } $
$ 25 = 4! + i^4 $
$ 26 = 4! + \sqrt{4} $
$ 28 = 4! + 4 $
$ 44 $
$ 48 = 4! \times \sqrt{4} $
$ 96 = 4! \times 4 $
$ 120 = (4 + i^4)! $
$ 256 = 4^4 $
$ 576 = (4!)^{\sqrt{4}} $
$ 720 = (4 + \sqrt{4})! $
$ 331,776 = (4!)^4 $
$ 16,777,216 = \sqrt{4}^{4!} $
$ 281,474,976,710,656 = 4^{4!} $
Three 4s
These lists blow up pretty fast... as you can see, focusing on using a smaller number of 4s can force you to be creative. This makes it possible to combine 4 4's beyond the integers from 1 to 20, and keep on going...
$ 2^n = (\sqrt{4})^n $
where $ n $ is any number expressible with two 4's.
$ 2 = \dfrac{4+4}{4} $
$ 2 = \sqrt{4} \times \left( \dfrac{4}{4} \right) $
$ 3 = \dfrac{ \ln{(4+4)} }{ \ln{\sqrt{4}} } $
$ 4 = 4 \times \left( \dfrac{4}{4} \right) $
$ 4 = 4 + 4 - 4 $
$ 5 = 4 + \dfrac{4}{4} $
$ 7 = 4 + 4 - i^{4} $
$ 7 = 4 + \sqrt{4} + i^{4} $
$ 8 = 4 \times \left( \dfrac{4}{\sqrt{4}} \right) $
$ 9 = 4 + 4 + i^{4} $
$ 10 = 4 + 4 + \sqrt{4} $
$ 11 = \dfrac{ 4! - \sqrt{4} }{ \sqrt{4} } $
$ 12 = 4+4+4 $
$ 18 = 4 \times 4 + \sqrt{4} $
$ 20 = 4 \times 4 + 4 $
$ 26 = 4! + 4 - \sqrt{4} $
$ 28 = 4! + \sqrt{4} + \sqrt{4} $
$ 30 = 4! + 4 + \sqrt{4} $
$ 30 = \dfrac{(4+\sqrt{4})!}{4!} $
$ 30 = \dfrac{(4 + i^4)!}{4} $
$ 32 = 4(4+4) $
$ 32 = 4^{\sqrt{4}} \sqrt{4} $
$ 36 = (4+\sqrt{4})^{\sqrt{4}} $
$ 46 = 4! + 4! - \sqrt{4} $
$ 47 = (4!)(\sqrt{4}) - i^4 $
$ 49 = (4!)(\sqrt{4}) + i^4 $
$ 50 = 4! + 4! + \sqrt{4} $
$ 60 = \dfrac{(4+i^4)!}{\sqrt{4}} $
$ 64 = \dfrac{4^4}{4} $
$ 64 = (\sqrt{4})^{4 + \sqrt{4}} $
$ 74 = (4A)_{4 \cdot 4} $
(That is, 74 is 4A in hexidecimal, or base 16 = 4 * 4)
$ 75 = (4B)_{4 \cdot 4} $
$ 76 = (4C)_{4 \cdot 4} $
$ 77 = (4D)_{4 \cdot 4} $
$ 78 = (4E)_{4 \cdot 4} $
$ 79 = (4F)_{4 \cdot 4} $
$ 80 = 4(4! - 4) $
$ 116 = ( (4 + i^4)! - 4 $
$ 118 = ( (4 + i^4)! - \sqrt{4} $
$ 119 = ( (4 + i^4)! - i^4) $
$ 120 = \dfrac{(4+i^4)!}{i^4} $
$ 121 = ( (4 + i^4)! + i^4) $
$ 122 = ( (4 + i^4)! + \sqrt{4}) $
$ 124 = ( (4 + i^4)! + 4 $
$ 180 = \dfrac{(4+\sqrt{4})!}{4} $
$ 252 = 4^4 - 4 $
$ 254 = 4^4 - \sqrt{4} $
$ 258 = 4^4 + \sqrt{4} $
$ 260 = 4^4 + 4 $
$ 1,024 = 4^4 \times 4 $
$ 4,096 = (4+4)^{4} $
$ 65,536 = (4 \times 4)^{4} $