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Thread: 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:
    $\displaystyle f(997) = f\left( {f(1002)} \right) = f(999) = f\left( {f(1004)} \right) = f(1001) = 998.$

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

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

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

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

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

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