1come on guys is it tough ??
I look at askiitians.com only when I am VERY VERY BORED.
But one post took me quite by surprise, when the student asked,
if f:N→N is a strictly increasing function satisfying f(f(n)) = 3n for all n, find f(11).
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12 Answers
341@archana: not so fast young lady, the range of f too lies in natural numbers
341One request: If guys have seen it already, pls dont post without hiding. That way it will help those who are trying it out afresh.
1yes 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
11And 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}
111 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"
341i started out as soumik did, but his method is much more economical. I basically trudged through finding the value of f for every number between 1 and 11 both inclusive :D.
Good work!