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Surely 2 cubed is out because that uses a 3.
And 2^2 + (2/2) has 4 twos.

connor o'connor wrote:2 cubed-(2/2) gives 8-1 =7
2 sqared +(2/2) gives 4+1 =5

to make the 7 you used 3 2's and a 3.
and for the 5 you used 4 2's

or is this allowed? it would make it a lot easier if "cubed" was actually a function

connor o'connor wrote:If a function is input-process-output then hex is a function
I know this is semantics but I could call a function the connor function which would add 3 to any number processed. As long as it did the same thing each time, ie it was constant in its output then as a function it would be valid.
so 2 connor would = 5

I think we're fuzzing the line between standard maths notation and advanced processes here.

Surely the idea of this is to use the former, otherwise I could simply say:

2n+2p+2l=11 ?

Or

function seige($int)
{
$int=11;
return $int;
}
seige(2+2+2);
// ergo seige(2+2+2)=11


That's cheating!

I disagree, the function statement y^2 is yxy, just as the function statememnt of y^3 is yxyxy. The use of the characters (not numbers) in the above is shothand to let you know how many times the functon should be multiplied by itself.
If it helps these are shorthand of y^n

sorry lawrence and seige, we all posted at the same time so missed your last posts.

cubed and square are functions

seige, I know I was being anoying. Your function is perfectly valid. I love it

That's cheating!

we're magicians we always cheat

Hang on...

2a+2b^2+c uses all three 2s and could make any number you like. Just add values for a, b and c.

Tomo wrote:Hang on...

2a+2b^2+c uses all 2s and could make any number you like. Just add values for a, b and c.

That's what I said above:

seige wrote:Surely the idea of this is to use the former, otherwise I could simply say:

2n+2p+2l=11 ?

Which is why I think this should have been restricted to standard math's notation.

Cubed and squared numbers are out, then, surely?

imaginary numbers are perfectly valid too

connor o'connor wrote:imaginary numbers are perfectly valid too

2x2+(i^2) would indeed give us that 3.
but you can't use i, since it's a number and not a function so...

if Magical_Trevor has the answers, perhaps he could tell us which methods we have got right.. I would say squares and cubes would be fine, after all most people only know + - x / ^2 ^3. Roots sin tan etc are out as they will give decimals. We have been talking ! which not a lot of people know about.
I think we may be making this a bit more complicated than it is.

I is not a real number it's just in our imagination. so in my book it's valid.

It is interesting though as to wether I is a number or a function. It cannot exist but IS used as a process in the input-process-output model so it is used as a function even if it is a number.
It exists to be a function even though it is a number. ODD

I also believe what is meant is using mathematical functions, not programming ones (afterall, this way you could get anything from anything).
And since they're only to involve three 2s, it means that ^3 etc are out, except of course for ^2 and ^(2+2) and so on, i.e. by using the 2s - otherwise you could just get, say, (2+2-2)^x where x is whatever value you need to get to your answer. The fact that you can't use any other numbers also means that you can't use "e"s, "i"s, "pi"s and whatever else you can think of, so although you may use sines and cosines, without "pi"s you won't get much out in the form of integers.
That's at least how I interpret the rules.

Oh, and a way to do 10:
(Sum from -2 to +2) of "2"
i.e.

2
SIGMA 2
n = -2

but as there's no "n" in the bit after the SIGMA, we just add up 5 2s, one for -2, one for -1, one for 0, one for 1 and one for 2. And since we're using a definte sum, introducing such a dummy variable n shouldn't be against the rules.

That said, maybe it is possible to find a definite integral between -2 and 2 (definite, hence value outcome without variables), where the bit in the integral contains one 2 and only variables otherwise, to get to certain other integers.
I'm working on it...

Edit: Why does this machine capitalise the "i", meaning square root of -1?

Ok, it is, in fact, possible to get to at least any rational number using definite integrals, I will try to outline one way of doing it:

Note that

2
Integral (x^2) dx = 16/3
-2

Hence, to get, say 7, we need to multiply it by a value 7*3/16=21/16, which we do withouth using numbers by just putting it as
(x+x+x+x+....+x)/(x+x+x+...+x) where there are 21x at the top and 16 at the bottom, so

. ........ 2
7 = Integral [(21x)/(16x) * x^2 dx]
. ........-2

where we write 21x as x+x+x+x+... etc. Our third "2" apart from the 2 limits is the square of the x which doesn't cancel with the others. (The dots around the integral are just to stop the auto-correct from displaying the 2 and -2 right at the front, ignoring all my spaces).
By just replacing the 21 by 3n, we find the right solution for any number n we want. Of course, it works for any rational number, not just integers, though I'm sure there are other ways as well.

Final point, after having gone through all this - of course, it's equally possible by just using sums. As I said above,

2
SIGMA (2) = 10
n = -2

So I just say for any number k one to twelve (or any, really)

2
SIGMA [(k*n)/(10n) * 2] = k
n = -2

where k*n and 10n is just written as (n+n+n+n....)

Well, if you are allowed to use any math symbol then....

2 x 2 x 2 ≈ 10
2 x (2/2) x π ≈ 9
2 x 2 x 2 = 8
2 x 2 x 2 ≈ 7
2 + 2 + 2 = 6

2 + 2 - 2 = 2
etc.