undoubtedly (I suck at maths) that`s why I showed my working out

Normally, multiplications are made before additions and subtractions. So your equation would result in 24 - 4 - 2 = 18, with only a little correction;

 

[math]\sqrt{4}(4!-\sqrt{4})-\sqrt{4} = 42[/math]

 

[math]44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43[/math]

 

[math]4! + 4*4 +4 = 44[/math]

 

[math]44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45[/math]

 

...and copying YT2095;

 

[math]\sqrt{4}(4!-\sqrt{4})+\sqrt{4} = 46[/math]

 

[math]4!+4!-\frac{4}{4} = 47[/math]

 

... With [math]\sqrt{4}[/math] it's quite easy

4!+4! =48

 

notice how I only do the easy ones!

... With [math]\sqrt{4}[/math'] it's quite easy

 

With root 4 its way to easy.

You need to use four 4, even if our definition of "4" is quite... creative.

 

[math]4!+4!+4-4 = 48[/math]

I thought it said in the original post, four 4`s or Less, so I didn`t bother to cancel the remaining ones

This may be an inappropriate place to post this question. I will ask anyway. How do you get the small number images to appear when you write an equation?

I`ve no idea, I think it`s something to do with Latex commands.

 

!4+!4 = 48

then 48 + 4/4 =49

 

same again for 48 then add (sqrrt4) = 50

 

 

Grrr... I cannot do 51 without using five 4`s, this one`s got me beat!

This may be an inappropriate place to post this question. I will ask anyway. How do you get the small number images to appear when you write an equation?

 

Use the code from; http://amath.colorado.edu/documentation/LaTeX/Symbols.pdf

 

And place your LaTeX code between [ math] and [ /math] without the space after the [

 

So, rewriting YT2095's solutions for 49 and 50;

 

[math]!4+!4 + \frac{4}{4} = 49[/math]

[ math]!4+!4 + \frac{4}{4} = 49[ /math]

 

[math]!4+!4 + \frac{4}{\sqrt{4}} = 50[/math]

[ math]!4+!4 + \frac{4}{\sqrt{4}} = 50[ /math]

[math]\frac{4!-\sqrt{4}}{.4} - 4 = 51[/math]

I know 51 has been done, but as this has been relayed and was considered some time ago, we thought it should be added anyway...

 

From YT2095: I can't get out of my shed untill later but I think the answer is 4!+4!=48+int pi =51

 

and from IRC....

13:55 <+RICHARDBATTY> he called it in at 12:21 uk time

Numbers apart from 4 aren't allowed.

Numbers apart from 4 aren't allowed.

 

The mathematical constants are, the 48 there is not

 

Cheers,

 

Ryan Jones

The 48 results from 4!+4!, but in any case:

3. You may use any mathematical operations and symbols (not including symbols for other numbers, like Pi, e, etc.)[/b'] you wish.

If we can use numbers like pi and not have them count towards the 4 4s total, then we could just use [math]\frac{\pi}{\pi}[/math] to get every number.

The 48 results from 4!+4!, but in any case:If we can use numbers like pi and not have them count towards the 4 4s total, then we could just use [math]\frac{\pi}{\pi}[/math'] to get every number.

 

Good point

 

I thought I put it as you were allowed, after what you said its probably better that its not then

 

Cheers,

 

Ryan Jones

[math]44-\frac{\sqrt{4}+\sqrt{4}}{4} = 43[/math]

...

[math]44+\frac{\sqrt{4}+\sqrt{4}}{4} = 45[/math]

 

Too many 4s' date='

 

[math']43 = 44-\tfrac{4}{4}[/math]

 

[math]45 = 44+\tfrac{4}{4}[/math]

 

And for anybody complaining that 44 is not valid, you can always replace every instance of it with (4!-4).

I think we're on 52...

 

[math]4!+4!+\sqrt{4}+\sqrt{4}=52[/math]

Good point

 

I thought I put it as you were allowed' date=' after what you said its probably better that its not then

[/quote']

 

so is my post (by proxy) #61 allowed then, or not?

so is my post (by proxy) #64 allowed then, or not?

 

When it starts to get hard, then you may use mathematical symbols but they will count as a four, that should resolve the problem of overuse of the constants

 

Cheers,

 

Ryan Jones

I refute it thus! Your new rule sux.

I refute it thus! Your new rule sux.

 

Fine, read my last post again. Last compramise. You can use them in any way but no more than 4 numbers, constants or what ever in total

 

 

Cheers,

 

Ryan Jones

[math]\frac{4!-\sqrt{4}}{.4} - \sqrt{4} = 53[/math]

 

[math]4! + 4! + \sqrt{4} + 4 = 54[/math]

 

I don't think using constants in any form should be allowed... It defeats the purpose of the game. Why not just leave it as 4 4s, other numbers excluded? Constants will make things far too easy.

[math]\frac{4!-(4-\sqrt{4})}{.4}=55[/math]

Ha I get the easy number:

[math]

4!+4!+4+4=56

[/math]

And the next:

[math]

\frac{4!-\sqrt{4}}{.4}+\sqrt{4}=57

[/math]