4!+(4+4)/4 = 26

4! +4 -(4/4)=27

44-(4*4)=28

4!+4+(4/4)=29

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

 

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]

Also, a quick rule that I'm going to impose:

 

You must wait for at least 3 posts after your last one before you can post another. This stops the post whoring to a certain extent

Well... It could be difficult to respect the new rule as it goes harder. And about 30 it's easy IF sqrt(4) is ok (which, from a technical point of view, isn't).

Well, it's never very difficult; you could just have [math]n = \sum_{1}^{n} \frac{4}{4}[/math]

Except that it's only worling for 4/4 + 4/4 = 2.

Yes, but my point was that you could easily just sum a load of 4/4's for any number. That kind of equation isn't particularly interesting

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 .

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.

Oh, sorry, I thought we had passed 30. Hmm, now I've gotta think about 30.

I got 30

 

[math]

\frac{4! + 4*4!}{4}

[/math]

 

 

and 31

 

[math]

\frac{4! + 4}{4} + 4!

[/math]

 

and 32 is easy

 

[math]

 

4*4 + 4*4

 

[/math]

 

ok im gunna stop now. at least til you guys get stuck again

so yeah... im done for tonight... or at least until i have a lot of caffiene in my system.

 

my bad : P

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

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]

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)

44-4!/4=38

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.

4! +4! -4 -4= 40

What happened to 39?

 

Edit: I'm guessing Xyph took it down, so here's one.

(4!/.4)/4+4!=39

41 = (4 x 4 + .4) / .4

 

√4 is definantly cheating in my opinion. Why not call it the four 4 or 2s ongoing challenge?

ok I`ll have a go at 42.

 

(4!)-(sqrt4)x(srqrt4)-(sqrrt4)=42

 

(4!) = 24

(sqrrt4) = 2

 

so 24 - 2 = 22

22 x 2 = 44

44 - 2 =42

ok I`ll have a go at 42.

 

(4!)-(sqrt4)x(srqrt4)-(sqrrt4)=42

 

 

That equals 18 (I may be wrong), I think your order of operation is wrong.

 

Try:

 

4!x(4/4+4/4)-((4!)/4))