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Category: | Math |

Submitted By: | DakarMorad |

Fun: | (1.84) |

Difficulty: | (3.35) |

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?

Show Answer |

Posted by I_am_the_Omega | 01/08/05 |

Doesn't 9304 convert to 00111001 00110011 00110000 00110100? ... | |

Posted by DakarMorad | 01/09/05 |

Omega: 9304 is 8192 + 1024 + 64 + 16 + 8. Orange: Well, your a first. ;) Sorry that this teaser was so difficult. It's my first. | |

Posted by saucyangel | 01/10/05 |

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! :P | |

Posted by cloud_strife | 01/18/05 |

err... what is a binary?? | |

Posted by Atropus | 01/18/05 |

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 ^_^ | |

Posted by God-sGrace2005 | 01/23/05 |

:-? :-? :-? I don\'t get it and what is binary? | |

Posted by DakarMorad | 01/24/05 |

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. | |

Posted by CPlusPlusMan | 01/30/05 |

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! | |

Posted by Gandalf | 02/14/05 |

it was hard but when my sister got it i felt so embarresed evn though shes older then me | |

Posted by sftbaltwty | 02/17/05 |

haha..i always knew there was reason i stopped taking math and stuck to english........ :wink: | |

Posted by waffle | 02/27/05 |

How were we ever supposed to arive at that answer? :-? | |

Posted by ben2 | 04/07/05 |

great one :D | |

Posted by sweetime | 05/16/05 |

i know what binary is, but have never used it in my whole life. how does 10010001011000 = 19? | |

Posted by darthforman | 05/22/05 |

:-? :-? :-? :oops: :cry: :x :oops: | |

Posted by solidtanker | 06/10/05 |

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. | |

Posted by rashad | 06/11/05 |

I feel so jealous because some of you understood it and I didn't get a single atom of it!!! :o :x | |

Posted by schatzy228 | 08/27/05 |

great teaser,,those who didnt like it just dont get the concept of "teaser",,,,but its all good 8) | |

Posted by soccercow10 | 08/29/05 |

HuH!?!?!? :oops: that was a fun teaser to try and find out !! even though i didnt :oops: all i have to say is creative.....creative indeed | |

Posted by i_am_hated | 09/28/05 |

:o !!! | |

Posted by usaswim | 10/28/05 |

:o :o :o :o :o :o :o :o :o :o | |

Posted by mrbrainyboy | 11/18/05 |

:o :o :o The hardest teaser in the whole site... wow :o 8) | |

Posted by lovefrenzy | 11/30/05 |

what :-? | |

Posted by qqqq | 12/20/05 |

My head hurts. :o | |

Posted by teen_wiz | 02/09/06 |

Ow. :o My brain is killing me. :lol: | |

Posted by coolblue | 05/21/06 |

So many zeroes, and who the heck heard of the binary system? :P :P | |

Posted by sftball_rocks13 | 06/14/06 |

huh....... | |

Posted by soccercow10 | 11/20/06 |

can someone please explain this to me? lol sorry too hard | |

Posted by sftball_rocks13 | 02/27/07 |

Um... :D My brain hurts :o but this was pretty good, I learned binary in school this year, but I would have NEVER gotten that :D good teaser! | |

Posted by MrDoug | 03/18/07 |

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. | |

Posted by brainglewashed | 06/14/07 |

DANG I SAID 1,000,000,000 :oops: :oops: :-? :-? :-? :D :D | |

Posted by Pojuer | 06/28/07 |

too hard :o | |

Posted by Brainyday | 11/11/07 |

I am confused. :-? :-? :-? :-? :-? | |

Posted by UlsterCharlotte | 01/14/08 |

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? | |

Posted by annvie9 | 03/30/08 |

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. :D | |

Posted by Natrix | 05/14/08 |

If you want people to understand this add a hint that says "This number willbe converted into binary." | |

Posted by EvilMonkeySpy3 | 12/02/08 |

darrhhhhhararrrrrrr...... :o i'm only in seventh grade.... :( i had absolutely no idea..... XP | |

Posted by piratechicken92 | 12/04/08 |

that was waaaaay to hard for me2 :oops: | |

Posted by javaguru | 12/10/08 |

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. | |

Posted by rashad | 05/07/09 |

Wonderful,yet ...impossible. | |

Posted by xandrani | 01/21/10 |

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 | |

Posted by xandrani | 01/23/10 |

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) = âŽ£a(x^4) + b(x^3) + c(x^2) + dx + eâŽ¦ See: http://mathworld.wolfram.com/CeilingFunction.html | |

Posted by xandrani | 01/23/10 |

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^âŽ£logx/log2âŽ¦) Where x â‰ 0 and g(0) = 0 f(x) = 1 + âŽ£logx/log2âŽ¦ + 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 | |

Posted by xandrani | 01/23/10 |

The smiley faces with glasses should be '8 )'. | |

Posted by xandrani | 01/23/10 |

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^âŽ£logx/log2âŽ¦) Where x â‰ 0 and g(0) = 0 f(x) = 1 + âŽ£logx/log2âŽ¦ + 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 | |

Posted by xandrani | 01/23/10 |

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^âŽ£logx/log2âŽ¦) Where x â‰ 0 and g(0) = 0 f(x) = 1 + âŽ£logx/log2âŽ¦ + 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 | |

Posted by xandrani | 01/23/10 |

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 |

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