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Find f(84)

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,587
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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kaliprasad

Well-known member
Mar 31, 2013
1,279
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 $>=$
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,587
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!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,734
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. :)
 

kaliprasad

Well-known member
Mar 31, 2013
1,279
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 $f(84) = 997$