Game of 4 4's

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androl
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Post by androl »

golfinha wrote:
androl wrote:120 = 4!*(4+4/4)

we already had 120 :?
right, but sometimes I find shorter solutions :wink:
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Post by lazza »

127 = (4^4 - sqrt4) / sqrt4
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Post by lazza »

128 = 4 x 4 x (4 + 4)
129 = (4^4 + sqrt4) / sqrt4
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Post by micro »

130 = 4^4/sqrt(4) + sqrt(4)
130 = (4^4+4)/sqrt(4)
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Post by micro »

131 = ?

132 = 44 * ld(4+4)
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Post by lazza »

131 and 133 = ? (can be done with 5 4s)

134 = 44 / .4 + 4!
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Post by androl »

also read this
http://www.math.hmc.edu/funfacts/ffiles/10006.8.shtml
they say it's possible to get all numbers until 112 with only
the four arithmetic operations (+,x,-,/), concatenation (44 is ok and uses up two 4's), decimal points (using 4.4 is ok), powers (using 4^4 is ok), square roots, factorials (using 4! is ok), and overbars for indicating repeating digits (e.g., writing .4 with an overbar would be a way of expressing 4/9)
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JP Simões
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Post by JP Simões »

lazza wrote:134 = 44 / .4 + 4!
44 / 0,(4) = 99
99 + 24 = 123
De férias por período indeterminado...
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Post by micro »

androl wrote:
and overbars for indicating repeating digits (e.g., writing .4 with an overbar would be a way of expressing 4/9)
Then I've got the long missing 107!

107 = (4! + 4!) / .4 overbar - ld(sqrt(4))
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Post by androl »

Zé da Silva wrote:44 / 0,(4) = 99
no, 44 / 0,4overbar = 99
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Post by golfinha »

well, the next one I know is
140=4!*ld(4+4)!-4
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Post by golfinha »

141=?
142=4!*ld(4+4)!-SQRT4
143=4!*ld(4+4)!-ld(SQRT4)
144=(4+SQRT4)!-4!*4!=(4+SQRT4)!*,4/SQRT4
145=4!*ld(4+4)!+ld(SQRT4)
146=4!*ld(4+4)!+SQRT4
147=?
148=4!*ld(4+4)!+4
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Post by JP Simões »

35=!4*4-(4/4)
37=!4*4+(4/4)
107=4*4!+!4+√4
136=(4!+!4)4+4
Last edited by JP Simões on Thu Aug 03, 2006 12:04 pm, edited 6 times in total.
De férias por período indeterminado...
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Post by JP Simões »

EDIT: Never mind...
De férias por período indeterminado...
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