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Math Help - f(84)

  1. #1
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    f(84)

    let f function from z to z(integers)

    and

    f(n)=n - 3 at n > 999

    f(n)= f(f(n+5) at n < 1000




    find
    f(84)
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  2. #2
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    Here is a start:
     f(997) = f\left( {f(1002)} \right) = f(999) = f\left( {f(1004)} \right) = f(1001) = 998.

    f(998) = f\left( {f(1003)} \right) = f(1000) = 997.

    One more:
    f(995) = f\left( {f(1000)} \right) = f(997) = 998.
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  3. #3
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    Note that,
    f(n) for n\leq 999 is determined by the parity of n.
    Meaning f(n)=998 if n is odd. And f(n)=997 if n is even.

    We can show this by Strong Induction.
    First,
    f(999)=f(f(1004))=f(1001)=998
    f(998)=f(f(1003))=f(1000)=997
    f(997)=f(f(1002))=f(999)=998
    f(996)=f(f(1001))=f(998)=997

    Now, if n\leq 995 and odd we have,
    f(n)=f(f(n+5))
    But by induction and the fact n+5 is even
    f(n)=f(997)=998

    And, if n\leq 995 and even we have,
    f(n)=f(f(n+5))
    But by induction and the fact n+5 is odd
    f(n)=f(998)=997

    The above gives us an explicit formula for f(n).
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