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Puzzle ID: #20276
Submitted By: DakarMorad Corrected By: lessthanjake789
Submitted By: DakarMorad Corrected By: lessthanjake789
If you use a certain formula on 13, you end up with 7.
Under the same formula, 2352 becomes 16, 246 becomes 14, 700 turns into 16, and 1030 becomes 14.
What would 9304 become?
Under the same formula, 2352 becomes 16, 246 becomes 14, 700 turns into 16, and 1030 becomes 14.
What would 9304 become?
Answer
19. Convert the number to binary, add one for every zero, and add two for every 1.9304 becomes 10010001011000, which has 9 zeros and 5 ones.
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Comments
Doesn't 9304 convert to 00111001 00110011 00110000 00110100? ...
To Easy
Omega: 9304 is 8192 + 1024 + 64 + 16 + 8.
Orange: Well, your a first. ;)
Sorry that this teaser was so difficult. It's my first.
Orange: Well, your a first. ;)
Sorry that this teaser was so difficult. It's my first.
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!
err... what is a binary??
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 ^_^
For your next one perhaps you chould add a hint ^_^
I don\'t get it and what is binary?
Atropus: I\'ll keep that in mind.
Binary is a system of counting that uses only 1s and 0s instead of 1-9 as digits.
1 is one,
10 is two,
11 is three,
100 is four,
etc.
Binary is a system of counting that uses only 1s and 0s instead of 1-9 as digits.
1 is one,
10 is two,
11 is three,
100 is four,
etc.
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!
it was hard but when my sister got it i felt so embarresed evn though shes older then me
haha..i always knew there was reason i stopped taking math and stuck to english........
How were we ever supposed to arive at that answer?
great one
Apr 23, 2005
i got -24345
a 6th order polynomial will pass through all those points.
eq looks something like this:
f(x) = -6.43551381261098E-12*x^4 + 3.63684663115652E-08*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
a 6th order polynomial will pass through all those points.
eq looks something like this:
f(x) = -6.43551381261098E-12*x^4 + 3.63684663115652E-08*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
Apr 23, 2005
sorry that's the 4th order regression eq
i know what binary is, but have never used it in my whole life.
how does 10010001011000 = 19?
how does 10010001011000 = 19?
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.
I feel so jealous because some of you understood it and I didn't get a single atom of it!!!
great teaser,,those who didnt like it just dont get the concept of "teaser",,,,but its all good
HuH!?!?!?
that was a fun teaser to try and find out !!
even though i didnt
all i have to say is creative.....creative indeed
that was a fun teaser to try and find out !!
even though i didnt
all i have to say is creative.....creative indeed
!!!
The hardest teaser in the whole site...
wow
what
My head hurts.
Ow. My brain is killing me.
So many zeroes, and who the heck heard of the binary system?
huh.......
can someone please explain this to me? lol sorry too hard
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
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
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!
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 4th-degree 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.
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.
DANG I SAID 1,000,000,000
too hard
I am confused.
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?
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.
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.
9304 becomes 0010 0100 0101 1000, which has 11 zeros and 5 ones, which is 21.
If you want people to understand this add a hint that says "This number willbe converted into binary."
darrhhhhhararrrrrrr...... i'm only in seventh grade.... i had absolutely no idea..... XP
that was waaaaay to hard for me2
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.
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.
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
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
Wonderful,yet ...impossible.
Jul 21, 2009
The answer is not A FORMULA.
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
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
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
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
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
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
The smiley faces with glasses should be '8 )'.
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
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
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
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
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
n = floor(x / 230)
f(x) = 14 - 7(x mod 2) + (1 - (x mod 2))((n^2 - n) mod 4)
f(9304) = 14
Apr 13, 2010
Another solution (and perhaps the simplest so far)
f(x) = 16 - ([(x % 14) * 6] % 23)
f(9304) = 14
f(x) = 16 - ([(x % 14) * 6] % 23)
f(9304) = 14
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