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Thread: Formula to always get 28

  1. #1
    Sep 2009

    Formula to always get 28

    I need to create a formula where the user would start with x and do at least 4 steps (multiplication, division or subtraction... no *0 allowed) to that number and always get 28. For example
    x = any number
    x+5 = n
    2n+10 = y
    2y+ 6 = z
    final step
    z+3 = 28

    obviously that is nor correct.

    Thank you!
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  2. #2
    Apr 2009
    You want the constant function $\displaystyle f(x)=28$ that doesn't depend on x. Now you can't multiply by 0, but subtracting x from itself x-x=0 always.
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  3. #3
    Member pflo's Avatar
    Apr 2009
    Albuquerque, NM

    Create Algebra Puzzles - Mathemagic

    The trick to creating your own algebra puzzles is just keep track of what you are doing to the original number, then undo it in a different order.

    For example:
    Pick a number (n)
    Add three so that a=n+3
    Multply by 4 so that b=4a (and $\displaystyle b=4(n+3)=4n+12$)
    Add 120 so that c=b+120 (and $\displaystyle c=4n+12+120$... how many 4's is that.?. $\displaystyle c=4(n+3+30)=4(n+27)$)

    So now what? Lets divide by 2 to throw the unsuspecting person off, then we'll have a bunch of 2's instead of a bunch of 4's. Then we can subtract a few 2's before losing the original number and getting to our goal...

    Divide by 2 so that d=c/2 (and $\displaystyle d=2(n+27)$)
    Subtract 14 so that e=d-14 (and $\displaystyle e=2(n+27)-2(7)=2(n+20)=2n+40$)
    Subtract twice the original number so that f=e-2n (and $\displaystyle f=40$)
    Subtract 12 so that g=f-12 (and $\displaystyle g=28$)

    Again, just keep track of what you've got in terms of 'n' over on the side while you are making this up. And think of things in terms of "I've got this many 4's" or "this many 2's" so you can divide later on.
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