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. 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.
  3. 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...
  4. 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)
  5. GJG

    GJG Well-Known Member

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

    Cheers, everyone.
  6. 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.
  7. flugtastic

    flugtastic Member

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

    GJG Well-Known Member

  9. 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.
  10. GJG

    GJG Well-Known Member

    I surrender.

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