Reverse it!
For positive integers n,
f(1)=1
f(3)=3
f(2n)=f(n)
f(4n+1)=2f(2n+1)-f(n)
f(4n+3)=3f(2n+1)-2f(n)
How many possible values of n (which is less than or equal to 1988) are there so that f(n)=n?
The answer is 92.
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1 solution
We have an easy way to find f(n). First you convert n to binary form, and reverse it, then change it back to decimal form, and thats f(n).
As example, lets say we want to find f(23).
1)convert 23 to binary form.
2 3 1 0 = 1 0 1 1 1 2
2)reverse 10111 so it become 11101.
3)change 11101 to decimal form.
1 1 1 0 1 2 = 2 9 1 0
So f(23)=29.