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Reposting puzzle since no one has solved it yet: You are allowed to use any number of mathematical operators or functions (e.g. +, -, *, /, sin, cos, tan, factorial, sqrt, powers, etc) EXCEPT floor and ceiling functions. The only constants that can appear in the solution are two digits - both of which must be '2'. Use these rules to make the number 5.

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Can we use decimal point?
yes - looks like you may have found one of the solutions (I know 2 ways of doing this) :-)
Well, don't you know I am the smart one? :P :D

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Other answers:

of course I do :-)
Anyways, I won't spoil the problem for others, but here is something related
Wow - I guess you really do learn something new everyday - I was not aware of this - thx
You are welcome :)
btw - there is a more /pure/ solution which does not involve using a decimal point
I suppose we can't use variable substitution too, isn't ?
what do you mean @FoolForMath?
do the powers need to be 2 or anything?
remember the rules @Zed - you can ONLY have two constants in the resulting expressions and BOTH must be the digit '2'
ahh i see, read the question wrong
Hm something like this: x=2; x^x + x^(x-x) = 5 :D wait that's with only one 2 :D
no sorry @FoolForMath - that is not allowed.
CLUE: the /purer/ solution involves right-angled triangles
Hm then how about this one \( \csc^2(\cot^{-1}(2)) \) , Checking:
thats the one! - well done @FoolForMath can the others find the other solution?
Thanks asnaseer :) Should I deleted it ?
Well done FoolForMath
that is how I wrote it to avoid the "-1"
no need to delete your answer - lets see if anyone can find the other way of doing this
CLUE for other way: uses one decimal point, one squareroot and one power
The other one is a bit hard in my opinion, I have seen this problem before sooo ;) anyways, the latex version of my solution \( \csc^2(\arccot(2)) \)
I wrote arctan above which was wrong.
btw hey your earlier post is not correct 5 \( \neq \)cosec^2(arctan(2))
wait, arccot is defined in latex no ?
See: \[ \arccot \arctan \]
okay i think this is right \[\sqrt{0.2^{-2}}=5\]
yup - you got it @Zed - well done!
yep! you got that right :)
Yay! That was tricky :)
the only thing I don't like about this way is that sqrt(x) is really x^-1/2
yeah i agree
so it /feels/ like cheating - but nevertheless - well done @Zed
Hero wants hard, lets give him a bit harder :D I just found two other ways of getting the same result, could anybody wanna try ?
Guys if you are trying, then please note it is not a good one and the second is ugly lol, and there is not much of useful maths we can learn from those two.
no - I was just waiting for @Hero to return to take up the challenge :-)
lol :D asnaseer, wanna try something from my sleeve ? ;)
gulp! - I know your reputation @FoolForMath - but lets for it - you gonna post to the left?
*lets GO for it
yeah sure :D
ummm this is kind o really easy.. at first I thought it was hard but i realized i was thinking too much.... 2^2 +2^3 = (12)^2 = 144 /2 = 72 /2 =36 /2 =18/2 =9 √9 = 3+2 = 5
no need for big guns this was basic problem solving skills
you didn't read the rules properly - the ONLY constants allowed are two digits - both of which must be the digit 2. e.g. 2^2+2^3 uses three 2's and a 3 - breaking the rules.
eh it was one of three or four answers ive got more. threre is also something interesting that happens between radians, but let me guess the implied conversion of a radian as 180 counts as a number other than 2 ? I still dont believe this requires higher level math I will stab at this later for now I have a date w/ Mathematica 8 and a vat of coffee. Thanks for clarifying !
:-D - enjoy the coffee!

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