Mind Reader
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http://trunks.secondfoundation.org/files/psychic.swf
http://digicc.com/fido/
Actually it is just a mathematical trick, but can you figure it out?
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lol...didnt really try the 2nd one..but 1st 1 i know lar..takes me a few mins to figure it out but i did it
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the first one if u notice it..the image that shows up...it is alwaes the multiple of 9.
however..everytime u reload the pg..the image WILL change...so that dumb pple like the thread-starter would be duped by the so-called mind reader..and wun suspect a thing..
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yesh u r so right...not bad eh yuko-ogura..Originally posted by yuko-ogura:the first one if u notice it..the image that shows up...it is alwaes the multiple of 9.
however..everytime u reload the pg..the image WILL change...so that dumb pple like the thread-starter would be duped by the so-called mind reader..and wun suspect a thing.. -
You haven't even solve the second one and you are calling me dumb?Originally posted by yuko-ogura:the first one if u notice it..the image that shows up...it is alwaes the multiple of 9.
however..everytime u reload the pg..the image WILL change...so that dumb pple like the thread-starter would be duped by the so-called mind reader..and wun suspect a thing..
I already solve both.
Let me give you a clue, both the magic numbers are related. -
related?i see..
u mean 1 is the mother of 2 and 2 is the daughter of 3 while 3 is the grandpa of 4 and so on...?
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lame...Originally posted by yuko-ogura:related?i see..
u mean 1 is the mother of 2 and 2 is the daughter of 3 while 3 is the grandpa of 4 and so on...? -
im beginning to see the relationship between the nos..
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i dont get it.Originally posted by yuko-ogura:the first one if u notice it..the image that shows up...it is alwaes the multiple of 9.
however..everytime u reload the pg..the image WILL change...so that dumb pple like the thread-starter would be duped by the so-called mind reader..and wun suspect a thing.. -
which part?Originally posted by TwinTurbo_Supra:i dont get it.
did u try playing it? -
ok now i get it. apparently there are only a few possible answers in the whole table, and those confirm correct 1 are the same symbols. the rest are just to trick u.Originally posted by yuko-ogura:which part?
did u try playing it? -
dats good..now try the second one and fit wad the thread-starter has been saying..
1 the mother of 2 and 2 the daughter of 3 while 3 being the grandfather of 4 and so on.. -
i got no idea how the 2nd 1 works.Originally posted by yuko-ogura:dats good..now try the second one and fit wad the thread-starter has been saying..
1 the mother of 2 and 2 the daughter of 3 while 3 being the father of 4 and so on.. -
precisely..ask the thread-starter?Originally posted by TwinTurbo_Supra:i got no idea how the 2nd 1 works.
im sure he wun let u down. -
actually, both quizzes involve the magic of the number 9.
in quiz 2, the sum of all the digits in the difference is a multiple of 9, hence it is only a matter of 9 minus (reminder of adding the digits you supplied divided by 9).
missing digit = 9 - ( sum % 9 )
there is a proof on subtracting a number from a number whose digits were jumbled up will yield a number with a multiple of 9 (i.e. 7981-1987=5994, 63-36=27 etc), but the book (Queen of Mathematics) is hidden in my collections. got time then put up. -
No time to consult my book. *yawns* This is my version.
Mod sign (triple-dash hyphen) is indicated as '='.
Let N be the original number.
For some N is a member of the real number set,
N
= 10³·a + 10²·b + 10¹·c + 10·d
= 999a + 99b + 9c + (a + b + c + d)
= (a + b + c + d) mod 9
>>> Since 999k, 99k or 9k are all multiples of 9
>>> This effectively means the remainder when N is divided by 9 is equal to the sum of all its digits. And N can be any real number, regardless of how many digits it has, so this trick need not be limited to 3 or 4 digits but any number of digits.
For some number M which has the same digits (a, b, c, d) as the abovementioned N, in any order, the same remainer can be obtained when its constituent digits are divided by 9, i.e.
M
= 10³·b + 10²·c + 10¹·d + 10·a
= 999b + 99c + 9d + (b + c + d + a)
= (b + c + d + a) mod 9
N - M = [ (a + b + c + d) - (b + c + d + a) ] mod 9 = 0 mod 9
(N - M) leaves a remainder of 0 when divided by 9, i.e. (N - M) is a multiple of 9.
Now, we know for any number which is a multiple of 9, the sum of its constituent digits is also a multiple of 9. Deducing from this, a missing digit can easily be derived by obtaining the sum of the other constituent digits and subtracting from the next highest multiple of 9.
Proof:
For some real number P which is a multiple of 9,
P
= 10³·a + 10²·b + 10¹·c + 10·d
= (a + b + c + d) mod 9
= 0 mod 9 (since P is a multiple of 9)
a + b + c + d = 0 mod 9
>>> (a + b + c + d) is a multiple of 9.
a + b + c + d = 9k, where k is some real integer
Let a be the missing digit.
a = 9k - (b + c + d)
>>> The missing digit can be obtained by subtracting the sum of the other constituent digits from a multiple of 9 such that 1 <= a <= 9
With regards to not choosing 0 as the missing digit, it is due to an overlapping result, i.e. a remainder of 0 or 9 may be obtained for certain values of P, yielding 2 different results. Hence, a certain set of values have to be defined for the missing value. In this case, it is either
1. 0 <= a <= 8, or
2. 1 <= a <= 9. -
woah..reminds me of mathematical induction...Originally posted by ^tamago^:No time to consult my book. *yawns* This is my version.
Mod sign (triple-dash hyphen) is indicated as '='.
Let N be the original number.
For some N is a member of the real number set,
N
= 10³·a + 10²·b + 10¹·c + 10·d
= 999a + 99b + 9c + (a + b + c + d)
= (a + b + c + d) mod 9
>>> Since 999k, 99k or 9k are all multiples of 9
>>> This effectively means the remainder when N is divided by 9 is equal to the sum of all its digits. And N can be any real number, regardless of how many digits it has, so this trick need not be limited to 3 or 4 digits but any number of digits.
For some number M which has the same digits (a, b, c, d) as the abovementioned N, in any order, the same remainer can be obtained when its constituent digits are divided by 9, i.e.
M
= 10³·b + 10²·c + 10¹·d + 10·a
= 999a + 99b + 9c + (b + c + d + a)
= (b + c + d + a) mod 9
N - M = [ (a + b + c + d) - (b + c + d + a) ] mod 9 = 0 mod 9
(N - M) leaves a remainder of 0 when divided by 9, i.e. (N - M) is a multiple of 9.
Now, we know for any number which is a multiple of 9, the sum of its constituent digits is also a multiple of 9. Deducing from this, a missing digit can easily be derived by obtaining the sum of the other constituent digits and subtracting from the next highest multiple of 9.
Proof:
For some real number P which is a multiple of 9,
P
= 10³·a + 10²·b + 10¹·c + 10·d
= (a + b + c + d) mod 9
= 0 mod 9 (since P is a multiple of 9)
a + b + c + d = 0 mod 9
>>> (a + b + c + d) is a multiple of 9.
a + b + c + d = 9k, where k is some real integer
Let a be the missing digit.
a = 9k - (b + c + d)
>>> The missing digit can be obtained by subtracting the sum of the other constituent digits from a multiple of 9 such that 1 <= a <= 9
With regards to not choosing 0 as the missing digit, it is due to an overlapping result, i.e. a remainder of 0 or 9 may be obtained for certain values of P, yielding 2 different results. Hence, a certain set of values have to be defined for the missing value. In this case, it is either
1. 0 <= a <= 8, or
2. 1 <= a <= 9. -
See, there are always a higher mountain.Originally posted by yuko-ogura:woah..reminds me of mathematical induction... -
Originally posted by Hydro:See, there are always a higher mountain.