Integer function equation

come on guys is it tough ??

wats surprising??
f(x)=√3x
so f(11)=11√3

@archana: not so fast young lady, the range of f too lies in natural numbers

oh rite!!

sir is dat 33.

i didnt get that. my answer is 20. mebbe rohan can confirm

sir mine is rong,i made a huge mistake as u know.

One request: If guys have seen it already, pls dont post without hiding. That way it will help those who are trying it out afresh.

yes i am getting 20 ....

heres my process ..

substitute f(n) in place of n (can be done as f:N→N)

we have

f(f(f(n)))=3f(n)

=>

f(3n)=3f(n)

further

now substitute n=1 in original ..

f(f(1))=3 ..

now f(1) of course cant be greater than 3 other wise we would have f(1)=3+k but f(3+k)=3 [f has decreased where as it is mentioned that f is strctly increasing]

so we must have f(1)≤3

now if f(1)=3 we have f(3) also 3 thus not srictly increasing ...
if we have f(1)=1 then we get again f(1)=3 but putting n=1 .. thus contradiction ...

Thus f(1)=2 ;

further pluggint this value in original equation

f(2)=3
Now the main step is over .. just a few calculations here and there ..

using the second derived equation f(3n)=3f(n)

we have f(3)=3*2=6
f(6)=3f(2)=9
now f(6)-f(3)=3 and 6-3 also =3
further it is mentioned that f is strictly increasing

so we must have f(4)=7 and f(5)=8

thus we must have f(12)=3f(4)=21
further f(9)=3f(3)=18

thus applying the same thing in the interval of f(9) to f(12) we have f(10)=19 and f(11)=20

And here's mine.....sorry if it's the same as rohan's
f(x)<x is not possible f(x)=x is not possible either, so the only situation we have is
x<f(x)<3x
f(1)=2
So now computing the values we get
f(9)=18 & f(18)=27
There are 8 naturals between 9 7 18, and between 18 and 27, also using the fact that f(x) is increasing and maps naturals to naturals,
\boxed{f(11)=20}

1 edit in line number 7,
"There are 8 naturals between 9 & (in place of 7) 18, and between 18 and 27, also using the fact that f(x) is increasing and maps naturals to naturals"