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A function $f: \Bbb{N^+} \rightarrow \Bbb{N^+}$ , defined on the set of positive integers $\Bbb{N^+}$, satisfies the following properties:

                    $f(n)=f(n/2)$   if $n$ is even

                    $f(n)=f(n+5)$  if $n$ is odd

Let $R=\{ i \mid \exists{j} : f(j)=i \}$ be the set of distinct values that $f$ takes. The maximum possible size of $R$ is ___________.
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Let us assume: $f(1) = x.$

Then, $f(2) = f(2/2) = f(1) = x$
$ f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x.$

Similarly, $f(4) = x$
$f(5) = f(5+5) = f(10/2) = f(5) = y.$

So, it will have two values. All multiples of $5$ will have value $y$ and others will have value $x.$
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Answer is 2
Its Saying we have 2 domains
N+ → N+

  1. So F(1) = F(6) = F(3) = F(8) = F(4) = F(2) = F(1)....It Repeats...  Now F(7) = F(12) = F(6)...Again repeats both above are same...Since F(6) matches in both so same both belongs to same value.We are not getting F(5) above
  2. Now F(5) = F(10) = F(5)..Repeats ...We can see we have different value for multiples of 5 and other natural numbers.
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Answer 2..

for multiples of 5.. f(5)=f(10)...
and one for rest of the numbers in N.
16 votes
16 votes

$\text{let we have f(1) = x. Then, f(2) = f(2/2) = f(1) = x}$

$\text{f(3) = f(3+5) = f(8) = f(8/2) = f(4/2) = f(2/1) = f(1) = x }$
$\text{f(5) = f(5+5) = f(10/2) = f(5) = y. }$

$\text{All  $N^+$ except multiples of 5 are mapped to x and multiples}$ 

$\text{of 5 are mapped to y so ,$\mathbf{Answer\space is\space 2}$}$

 

 

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