Try this problem!!!!

Suppose f is a function such that f:Z+Z+ f:\mathbb{Z} ^+\to \mathbb{Z} ^+ satisfying

f(1)=1,f(2n)=f(n)f(1)=1, f(2n)=f(n) and f(2n+1)=f(2n)+1f(2n+1)=f(2n)+1,

for all positive integers nn.

Find the maximum of f(n)f(n) when 1n2013 1 \leq n \leq 2013.

Note by Mharfe Micaroz
6 years, 9 months ago

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11 votes

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f(n)f(n) is the number of 11s in the binary expansion of nn. Since 211=20482^{11} = 2048, the smallest integer nn with f(n)=11f(n) = 11 is 20472047. Since f(1983)=10f(1983) = 10, the maximum of f(n)f(n) over 1n20131 \le n \le 2013 is 1010.

Mark Hennings - 6 years, 9 months ago

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This function doubtlessly suffices the condition, & the solution is beautiful. But is this definition of f(n)f(n) unique? Why?

A Brilliant Member - 6 years, 9 months ago

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Each number can be constructed, according to its binary expansion, by applying these rules.

We start with f(1), or "1" in binary. The f(2n) equation will tell us what f provides when we append 0 to our input, that is, f(10), and f(2n+1) tells us what happens when we append a 1.

e.g. f(110101) is computed by f(1) --> f(11) --> f(110) --> f(1101) --> f(11010) --> f(110101).

(all numbers in binary)

Alex Meiburg - 6 years, 9 months ago

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Prove it by (strong) induction on nn.

Mark Hennings - 6 years, 9 months ago

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Please help me solve my problem above!

Mark Mottian - 6 years, 9 months ago

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Beat me to it... although it has been some time since it was posted.

But, yes, although it would have sufficed simply to use 21012^{10}-1 (since there cannot be any more 11s, as this number would then be 21112^{11}-1).

The argument for the sum of the digits of the binary expansion can be seen by simple construction (this is just a quick, non-rigorous sketch) of a two-case inductive proof:

Allow the number nn to be of the form n=ak2k1+ak12k2++a1n=a_k2^{k-1}+a_{k-1}2^{k-2}+\cdots+a_1, where ai{0,1}a_i\in\{0,1\} and ak=1a_k=1.

Case 1

Then, if this number is odd, we must have a1=1a_1=1, hence, we have f(2m+1)=f(2m)+1f(2m+1)=f(2m)+1 for some mZm\in\mathbb{Z}. So, now our number becomes: 2m+1=ak2k1+ak12k2++a221+1 2m+1=a_k2^{k-1}+a_{k-1}2^{k-2}+\cdots+a_22^{1}+1 m=ak2k2+ak12k3++a2 m=a_k2^{k-2}+a_{k-1}2^{k-3}+\cdots+a_2 This adds unity to our count, removes the last digit, and continues the process until the only value left of mm is one.

Case 2

Say that nn is even, then we must have a1=0a_1=0. Clearly, then, we have: f(2m)=f(m)f(2m)=f(m), which, after dividing, gives us a new number with the last digit removed and adds nothing to the count (as the last digit is 0 in base 2).

The same process applies.

Hence, we are done (note that this proof can easily be extended for uniqueness of definition as Paramjit S.'s request).

Hopefully this makes the case a little more clear for those in doubt.

Guillermo Angeris - 6 years, 9 months ago

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a very good thing indeed.good work Mark H.

Rajarshi Chatterjee - 6 years, 9 months ago

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Hi there! Seeing that I am having trouble with starting a discussion, I decided to just post it as a comment instead.

While preparing for the South African National Olympiad, the following problem featured in my training material:

"An 11 sided polygon has its vertices on a circle. How many triangles can be formed using three vertices of the polygon but sharing no side with a side of the polygon?"

Can somebody please provide a solution?


Mark Mottian - 6 years, 9 months ago

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Neerad Nandan - 6 years, 9 months ago

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Sorry, it's not 1009 factorial. :) Lol Careful with the !

A Brilliant Member - 6 years, 9 months ago

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Is it 11?

Daniel Leong - 6 years, 9 months ago

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