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Four 4s ongoing challange!


RyanJ

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I don't know whether this is a contradiction of the rules, but I don't care :P

 

If [imath]\sigma(n)[/imath] is the sum of the divisors of n, then:

 

[imath]\sigma\left(4! + (4-\frac{4}{4})! - \frac{4}{4}\right) = 30[/imath]

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Don't consider this an actual answer, but it's the best I can do (I got stuck with 31, after all). Someone more advanced than I in math probably has an actual way of doing it.

 

4!+(4!/4)+(≈(sqrt(sqrt4)))=31

≈ meaning the rounded value--I dunno mathematical shorthand, but that was the best I could do lol. Darn the limit of fours, but the approximate value of 1.414 is 1 :P.

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you either need practice with math or with latex. (unless i helluv just read that wrong... whcih im looking at now : P)

 

(4! +4 *4)/4= 10

 

throw in some parentheses and your good to go:

((4! +4) *4)/4= 30

 

 

not sure what went wrong with this one' date=' but

(4!+4)/4 +4=11 not 31[/quote']

 

 

you missed my other factorials, its not (4! + 4*4)/4, its (4! + 4*4!)/4

4! = 24, so this is equal to 24/4 + 24 = 30.

 

same with the other one, missed another ! there :D

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I've got 33...

 

[math]4!+\frac{4-.4}{.4} = 33[/math]

 

34...35 and 36 are easy, the next challenge is 37 and I'm too tired...

 

[math]4! + \frac{4!}{4} +4 = 34[/math]

 

[math]\frac{4.4}{.4} + 4! = 35[/math]

 

[math]4!+4+4+4 = 36[/math]

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We are still stuck at 30... Sure there's an easy answer;

 

[math]4! + 4 + 4 - \sqrt{4} = 30[/math]

 

But [math]\sqrt{4}[/math] is [math]4^{-2}[/math] and it's like using a 2.

 

yeah' date=' but if [math']\sqrt{4}[/math] isnt allowed, then [math]4![/math] shouldnt be either because its like saying [math]4 * 3 * 2 * 1[/math]

 

personally, i feel that people using 44, 4.4, .4, etc are bending the rules more than using [math]\sqrt{4}[/math]

 

 

 

that being said,

 

[math]4! + (4! + \sqrt{4})/ (\sqrt{4}) = 37[/math]

 

(thanks to whelck for that one)

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Edit: Oops. Miscalculated.

 

That said, [math]\sqrt{4}[/math] probably shouldn't be allowed since it's [math]^2\sqrt{4}[/math] which includes another number, and [math].4[/math] should properly include a 0, as well. Still, they might as well be allowed now, I suppose.

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