Magic ... ?

Discussion in 'Off-Topic Chat' started by GJG, Jan 16, 2008.

  1. GJG

    GJG Well-Known Member

    Have a look at this ...

    I've seen something very similar to this before, and I remember there was a very simple explanation as to the trick behind it, but I can't for the life of me remember what it was.

    Any ideas?
  2. theMouthPiece Related Searches

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  3. Bayerd

    Bayerd Active Member

    It's based around mulitples of nine. Look at the multiples of nine on the grid, they each have the same symbol.
  4. flugtastic

    flugtastic Member

    hehe i no how this 1 works!!!
    you pick any number eg 67. add the 2 numbers together = 13.minus that from 67 = 54.... its a multiple of 9 (9,18,27,36 etc.) you look at the final chart you will notice that all the multiples of 9 have the same number...
  5. MoominDave

    MoominDave Well-Known Member

    And it's a multiple of 9 because -

    Write your original number (for example, take 35) as digits a and b. Then
    35 = 10*a + b
    The number you subtract is a+b (3+5=8 in our example), so the final number you deal with is
    (10*a + b) - (a + b)
    which, if we remove the brackets, becomes
    10*a + b - a - b
    10*a - a
    which is obviously always divisible by 9
    (our example gives 35 - 8 = 27 = 3*9)
  6. GJG

    GJG Well-Known Member

    Mmmm ... yes, I thought it was simple ... ;)

    Cheers, everyone.
  7. MoominDave

    MoominDave Well-Known Member

    It works for larger numbers too - take a number with digits (any number of digits missing in the middle).
    Then, this number is
    z + 10*y + 100* x + 1000*w + ........... + 10^(n-2)*b + 10^(n-1)*a

    Add up the digits - you get a + b + c + d + .... + y + z

    Subtract one from the other, you get
    9*y + 99*x + 999*w + ........ + [(n-2) 9s]*b + [(n-1) 9s]*a
    in which every term divides by 9.

    The sum of its digits subtracted from a (positive integer) number of any size is divisible by 9 - not just 2-digit numbers, as the puzzle specifies.
  8. flugtastic

    flugtastic Member

    think il go for a lie down now ....
  9. GJG

    GJG Well-Known Member

  10. MoominDave

    MoominDave Well-Known Member

    This follows the same principle - digit-sums and divisible-by-9-ness

    Write down a 4-digit number -
    Jumble up the digits -
    efgh, where e, f, g and h are a, b, c and d in some order
    Assume that abcd > efgh (okay to do this, the only difference in the answer is a minus sign). Then
    abcd - efgh
    = 1000*(a-e) + 100*(b-f) + 10*(c-g) + (d-h)
    Now we can say that 999*(a-e) + 99*(b-f) + 9*(c-g) is divisible by 9, whatever the digits are.
    So abcd - efgh is divisible by 9 if the difference between these two sums is.
    The difference between these two sums is
    (1000-999)*(a-e) + (100-99)*(b-f) + (10-9)*(c-g) + (d-h)
    a - e + b - f + c - g + d - h
    a + b + c + d - e - f - g - h

    But efgh is just abcd in a different order, so this is the same as
    = 0
    which obviously divides by 9.

    So the difference between the four-digit number and its scrambled version is always divisible by 9. Now we know that a number that is divisible by 9 has a digit sum that is also divisible by 9.
    You give the puzzle three of the digits of the answer - it just finds the next multiple of 9 bigger, and takes the difference to get the remaining digit. This is why it asks you not to pick a zero - it can't tell whether the number is 0 or 9.

    An example:
    My first number: 9713
    My first number, scrambled: 1379
    Difference of the two: 9713 - 1379 = 8334
    I give the puzzle "438". It calculates 4+3+8 = 15; next multiple of 9 bigger than 15 is 18. 18-15 = 3. Tada!

    Like the other puzzle, this one also works for arbitrary numbers of digits.
  11. GJG

    GJG Well-Known Member

    I surrender.
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