FunctionCeption!

Algebra Level 5

The function $f$ is defined on the set of integers and satisfies $f(n)=\begin{cases} n-3&\mbox{if}\ n\ge 1000\\ f(f(n+5))&\mbox{if}\ n<1000\end{cases}$

Find $f(84)$ .

Image Credit: Wikipedia Penrose Stairs

The answer is 997.

2 solutions

Parth Lohomi
Mar 10, 2015

We start by finding values of the function right under $1000$ since they require iteration of the function.

$f(999)=f(f(1004))=f(1001)=998$ $f(998)=f(f(1003))=f(1000)=997$ $f(997)=f(f(1002))=f(999)=998$ $f(996)=f(f(1001))=f(998)=997$ Soon we realize the $f(k)$ for integers $k<1000$ either equal $998$ or $997$ based on it parity. (If short on time, a guess of $998$ or $997$ can be taken now.) If $k$ is even $f(k)=997$ if $k$ is odd $f(k)=998$ .

$84$ has even parity, so $\boxed{f(84)=997}$ .

Nice question..did it same way -

f(84) = f(f(f(f(....f(999)))))))))) (total 184 times f) f(999) = f(1001) = 998 f(998) = f(1000) = 997 f(997) = f(999) = f(1001) = 998

And so on..

Thus for 2 'f's we get 997, so for 184 it will be the same, = 997.

Ayan Jain - 6 years, 3 months ago

good man!! $\ddot\smile$

Parth Lohomi - 6 years, 3 months ago

Can you say how to calculate f(994) the same way without using the pattern......

Hari prasad Varadarajan - 6 years, 3 months ago

f(994) = f(f(999)) = f(f(f1004)) = f(f(1001)) = f(998) = f(f(1003)) = f(1000) = 997.

Ayan Jain - 6 years, 3 months ago

Thanks.........

Hari prasad Varadarajan - 6 years, 3 months ago

Harvindersingh Chavan
May 19, 2015

Good question! I finally solved it. 84 is even parity, so f (84) comes out to be 997

FunctionCeption!

Image Credit: Wikipedia Penrose Stairs

The answer is 997.

2 solutions

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