MHB Finding $f(84)$ with the Defined Function $f$

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The function \( f \) is defined for integers, with two cases: \( f(n) = n - 3 \) for \( n \geq 1000 \) and \( f(n) = f(f(n + 5)) \) for \( n < 1000 \). The discussion centers on finding \( f(84) \), with participants noting a typo in the function's definition regarding the inequality. The correct interpretation is crucial for solving the problem accurately. Acknowledgment of errors and the importance of checking work before posting are highlighted as part of the conversation. The thread emphasizes the collaborative nature of problem-solving in mathematical discussions.
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The function $f$ is defined on the set of integers and satisfies

\[ f(n)=\begin{cases}
n-3, & \text{if} \,\,n\geq 1000 \\
f(f(n+5)), & \text{if}\,\, n< 1000
\end{cases}
\]

Find $f(84)$.
 
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anemone said:
The function $f$ is defined on the set of integers and satisfies

\[ f(n)=\begin{cases}
n-3, & \text{if} \,\,n\leq 1000 \\
f(f(n+5)), & \text{if}\,\, n< 1000
\end{cases}
\]

Find $f(84)$.

I think 1st condition should be $>=$
 
Ah, thanks to kaliprasad for pointing it out! Sorry all! I made a blunder, I think for those who know me well, you can tell this wasn't the first time I made a typo in my posting, hehehe...but, I apologize. This shouldn't happen in the first place, I should have checked both before and after posting too!
 
anemone said:
Ah, thanks to kaliprasad for pointing it out! Sorry all! I made a blunder, I think for those who know me well, you can tell this wasn't the first time I made a typo in my posting, hehehe...but, I apologize. This shouldn't happen in the first place, I should have checked both before and after posting too!

No one is perfect, and only those who do things put themselves at the risk of making a mistake. When I see that someone contributing to a site has possibly made a typo or other error in a post, I tend to send them a PM so they can correct it with minimal embarrassment. :)
 
As for the value of f(n) for n less than 1000 can be computed if we now for n+5 so let us compute f(n) for n = 1004 to 995 downwards
$f(1004) = 1004- 3 = 1001$
$f(1003) = 1003 - 3 = 1000$
$f(1002) = 1002-3 = 999$
$f(1001) = 1001 - 3 = 998$
$f(1000) = 1000-3 = 997$
$f(999) = f(f(1004)) = f(1001) = 998$
$f(998) = f(f(1003)) = f(1000) = 997$
$f(997) = f(f(1002)) = f(999) = 998$
$f(996) = f(f(1001)) = f(998) = 997$
$f(995) = f(f(1000)) = f(997) = 998$

Now as 84 + 915 = 999 or 84 + 183 * 5 = 999 So let us evaluate f(999-5k) for some k

we have $f(994) = f(f(999)) = f(998) = 997$
$f(989) = f(f(994)) = f(997) = 998$
$f(984) = f(f(989)) = f(998) = 997$
from the above we see that $f(999-kn)$ is $998$ if k is even and is $997$ if k is odd

as $999- 84 = 183 * 5$ so we get $f(84) = 997$
 
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