I_am_the_Omega
Jan 08, 2005
 Doesn't 9304 convert to 00111001 00110011 00110000 00110100? ... 
chimpe81
Jan 09, 2005
 To Easy 
DakarMorad
Jan 09, 2005
 Omega: 9304 is 8192 + 1024 + 64 + 16 + 8.
Orange: Well, your a first. ;)
Sorry that this teaser was so difficult. It's my first. 
saucyangel
Jan 10, 2005
 ok, i NEVER would have figured that out! (well, maybe after i sat there and thought about it for an hour or three...) good one! 
cloud_strife
Jan 18, 2005
 err... what is a binary?? 
Atropus
Jan 18, 2005
 Odd.. I guessed it had to do with binary.. but it was really just too obscure.
For your next one perhaps you chould add a hint ^_^ 
GodsGrace2005
Jan 23, 2005
 I don\'t get it and what is binary? 
DakarMorad
Jan 24, 2005
 Atropus: I\'ll keep that in mind.
Binary is a system of counting that uses only 1s and 0s instead of 19 as digits.
1 is one,
10 is two,
11 is three,
100 is four,
etc. 
CPlusPlusMan
Jan 30, 2005
 I wouldn't necessarily call a binary conversion a formula, but great teaser anyways! When I saw it wasn't a function, it had me really thrown off. I'd never of even guessed of binary! 
Gandalf
Feb 14, 2005
 it was hard but when my sister got it i felt so embarresed evn though shes older then me 
sftbaltwty
Feb 17, 2005
 haha..i always knew there was reason i stopped taking math and stuck to english........ 
waffle
Feb 27, 2005
 How were we ever supposed to arive at that answer? 
ben2
Apr 07, 2005
 great one 
(user deleted)
Apr 23, 2005
 i got 24345
a 6th order polynomial will pass through all those points.
eq looks something like this:
f(x) = 6.43551381261098E12*x^4 + 3.63684663115652E08*x^3  0.0000674643206344868*x^2 + 0.0452940004077013*x + 6.42249974693123
but yeah adding the digits in a bianary representation will give you something else 
(user deleted)
Apr 23, 2005
 sorry that's the 4th order regression eq 
sweetime
May 16, 2005
 i know what binary is, but have never used it in my whole life.
how does 10010001011000 = 19? 
darthforman
May 22, 2005
 
solidtanker
Jun 10, 2005
 These kinds of puzzles are not my favorite because anyone can come up with an arbitrary system to convert one number into another. There are infinite ways to do so. 
rashad
Jun 11, 2005
 I feel so jealous because some of you understood it and I didn't get a single atom of it!!! 
schatzy228
Aug 27, 2005
 great teaser,,those who didnt like it just dont get the concept of "teaser",,,,but its all good 
soccercow10
Aug 29, 2005
 HuH!?!?!?
that was a fun teaser to try and find out !!
even though i didnt
all i have to say is creative.....creative indeed 
i_am_hated
Sep 28, 2005

!!! 
usaswim
Oct 28, 2005
 
mrbrainyboy
Nov 18, 2005

The hardest teaser in the whole site...
wow 
lovefrenzy
Nov 30, 2005
 what 
qqqq
Dec 20, 2005
 My head hurts. 
teen_wiz
Feb 09, 2006
 Ow. My brain is killing me. 
coolblue
May 21, 2006
 So many zeroes, and who the heck heard of the binary system? 
sftball_rocks13
Jun 14, 2006
 huh....... 
soccercow10
Nov 20, 2006
 can someone please explain this to me? lol sorry too hard 
(user deleted)
Feb 02, 2007
 To all who dont get it:
If you dont know binary you're screwed before you even started. Look up "binary" in wikipedia if you dont even know what it is.
A summary: Computers use binary to represent data, since computers work with circuits that have two states, off (0) and on (1). A computer can represent numbers, strings, or you're favourite MP3 as a string of 0's and 1's. Now that thats out of the way...
With his conversion system, a table of values goes like this:
0,1,10,11,100,101,110,111,1000,1001,1010
Number = Binary = Value*
0 = 0 = 1
1 = 1 = 2
2 = 10 = 3
3 = 11 = 4
4 = 100 = 4
5 = 101 = 5
6 = 110 = 5
7 = 111 = 6
8 = 1000 = 5
9 = 1001 = 6
10 = 1010 = 6
11 = 1011 = 7
12 = 1100 = 6
13 = 1101 = 7
14 = 1110 = 7
15 = 1111 = 8
*given that you add 1 for every zero and 2 for every one.
and so on. I actually never new how to convert binary, but I now do just by looking at the conversions.
Good very hard teaser 
sftball_rocks13
Feb 27, 2007
 Um... My brain hurts but this was pretty good, I learned binary in school this year, but I would have NEVER gotten that good teaser! 
MrDoug
Mar 18, 2007
 I don't like this one because it doesn't have a clear (single) correct answer. There are lots of formulas that give the given numbers. For example, one can construct (as already stated) a 4thdegree polynomial which takes on all the valued specified (or infinitely many polynomials of degree 5 or higher), and any of these qualify as a "formula."
It might help to give some clue as to what you had in mind, such as "The formula I have in mind only applies to integers, and it always gives an integer value." This at least rules out continuous mathematics and identifies it as a discrete problem, which is apparently what you intended. 
brainglewashed
Jun 14, 2007
 DANG I SAID 1,000,000,000 
Pojuer
Jun 28, 2007
 too hard 
Brainyday
Nov 11, 2007
 I am confused. 
UlsterCharlotte
Jan 14, 2008
 I agree with MrDoug. This is WAY too obscure. You realize right away that there are multiple answers. Not good at all. Who proofreads/screens these things anyway? 
annvie9
Mar 30, 2008
 I would only get this answer if I sat there for a whole day. But if I did, I would staring at the ceiling doing nothing anyways. 
(user deleted)
May 07, 2008
 As far as I remember, you can't have 14 digits in a binary number  it has to be in sets of 4 (i.e. the number 5 in binary would be 0101, not '101'). So,
9304 becomes 0010 0100 0101 1000, which has 11 zeros and 5 ones, which is 21. 
Natrix
May 14, 2008
 If you want people to understand this add a hint that says "This number willbe converted into binary." 
EvilMonkeySpy3
Dec 02, 2008
 darrhhhhhararrrrrrr...... i'm only in seventh grade.... i had absolutely no idea..... XP 
piratechicken92
Dec 04, 2008
 that was waaaaay to hard for me2 
javaguru
Dec 10, 2008
 Lame. As mentioned before, arbitrarily obscure without a unique or obviously correct answer.
And to greenrazi: You're are probably thinking of hexadecimal (base 16), where each digit can have one of 16 values. A binary representation of a hexadecimal number would have a granularity of 4 bits. 
(user deleted)
Apr 10, 2009
 Too make things easy to understand i just got the need to post a comment... Here it is(look at CanadaAotS comments)...
14 = 1110
lets convert 1 to 2 and 0 to 1
14 = 2+2+2+1 = 7
15 = 1111
15 = 2+2+2+2 = 8 
rashad
May 07, 2009
 Wonderful,yet ...impossible. 
(user deleted)
Jul 21, 2009
 The answer is not A FORMULA. 
xandrani
Jan 21, 2010
 There is more than one solution. The binary answer is more succinct and sweeter therefore it is the 'official' answer, but this also works:
a = 0.00000000003090981409468774
b = 0.000000087318835992468036
c = 0.000023696152096488413
d = 0.028997981886427873
e = 6.6192125424396062
f(x) = a(x^4) + b(x^3) + c(x^2) + dx + e
So answer would be:
f(9304) = 163621 
xandrani
Jan 23, 2010
 Note that the above function should strictly have read:
f(x) = floor(a(x^4) + b(x^3) + c(x^2) + dx + e)
Where floor rounds down to the nearest integer.
Note that this can also be written as:
f(x) = ⎣a(x^4) + b(x^3) + c(x^2) + dx + e⎦
See:
http://mathworld.wolfram.com/CeilingFunction.html 
xandrani
Jan 23, 2010
 I have noticed a few comments stating that the answer is not a function... however aside from the function above I posted (which is one solution), I now also post another function which fits the other solution. Almost anything can be made in to a function.
g(x) = 1 + g(x  2^⎣logx/log2⎦)
Where x ≠ 0 and g(0) = 0
f(x) = 1 + ⎣logx/log2⎦ + g(x)
Let's try and solve for x = 9304:
f(9304) = 1 + 13 + g(9304)
g(9304) = 1 + g(9304  2^13) = 1 + g(1112)
g(1112) = 1 + g(1112  2^10) = 1 + g(8
g(8 = 1 + g(88  2^6) = 1 + g(24)
g(24) = 1 + g(24  2^4) = 1 + g(
g( = 1 + g(8  2^3) = 1 + g(0) = 1
So iterating out we get:
9(24) = 1 + 1 = 2
9(8 = 1 + 2 = 3
9(1112) = 1 + 3 = 4
g(9304) = 1 + 4 = 5
So therefore:
f(9304) = 1 + 13 + 5 = 19 
xandrani
Jan 23, 2010
 The smiley faces with glasses should be '8 )'. 
xandrani
Jan 23, 2010
 I have noticed a few comments stating that the answer is not a function... however aside from the function above I posted (which is one solution), I now also post another function which fits the other solution. Almost anything can be made in to a function.
g(x) = 1 + g(x  2^⎣logx/log2⎦)
Where x ≠ 0 and g(0) = 0
f(x) = 1 + ⎣logx/log2⎦ + g(x)
Let's try and solve for x = 9304:
f(9304) = 1 + 13 + g(9304)
g(9304) = 1 + g(9304  2^13) = 1 + g(1112)
g(1112) = 1 + g(1112  2^10) = 1 + g(8 )
g(8 ) = 1 + g(88  2^6) = 1 + g(24)
g(24) = 1 + g(24  2^4) = 1 + g(8 )
g(8 ) = 1 + g(8  2^3) = 1 + g(0) = 1
So iterating out we get:
9(24) = 1 + 1 = 2
9(8 ) = 1 + 2 = 3
9(1112) = 1 + 3 = 4
g(9304) = 1 + 4 = 5
So therefore:
f(9304) = 1 + 13 + 5 = 19 
xandrani
Jan 23, 2010
 Damn smilies! I post yet again:
I have noticed a few comments stating that the answer is not a function... however aside from the function above I posted (which is one solution), I now also post another function which fits the other solution. Almost anything can be made in to a function.
g(x) = 1 + g(x  2^⎣logx/log2⎦)
Where x ≠ 0 and g(0) = 0
f(x) = 1 + ⎣logx/log2⎦ + g(x)
Let's try and solve for x = 9304:
f(9304) = 1 + 13 + g(9304)
g(9304) = 1 + g(9304  2^13) = 1 + g(1112)
g(1112) = 1 + g(1112  2^10) = 1 + g(88 )
g(88 ) = 1 + g(88  2^6) = 1 + g(24)
g(24) = 1 + g(24  2^4) = 1 + g(8 )
g(8 ) = 1 + g(8  2^3) = 1 + g(0) = 1
So iterating out we get:
9(24) = 1 + 1 = 2
9(88 ) = 1 + 2 = 3
9(1112) = 1 + 3 = 4
g(9304) = 1 + 4 = 5
So therefore:
f(9304) = 1 + 13 + 5 = 19 
xandrani
Jan 23, 2010
 There are indeed many solutions to this one, here's another just for fun:
n = floor(x / 230)
f(x) = 14  7(x mod 2) + (1  (x mod 2))((n^2  n) mod 4)
f(9304) = 14 
(user deleted)
Apr 13, 2010
 Another solution (and perhaps the simplest so far)
f(x) = 16  ([(x % 14) * 6] % 23)
f(9304) = 14 