## Solution to HP A4 Printer Paper Mysterious Question

A while ago, I posted the HP A4 Paper Mysterious Question which goes like this:

Problem of the Week

Suppose $f$ is a function from positive integers to positive integers satisfying $f(1)=1$, $f(2n)=f(n)$, and $f(2n+1)=f(2n)+1$, for all positive integers $n$.

Find the maximum of $f(n)$ when $n$ is greater than or equal to 1 and less than or equal to 1994.

So far no one seems to have solved the question on the internet yet!

I have given it a try, and will post the solution below!

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We will solve the problem in a few steps.

## Step 1

First, we will prove that $\boxed{f(2^n-1)=n}$. We will do this by induction. When $n=1$, $f(2^1-1)=f(1)=1$. Suppose $f(2^k-1)=k$. Then, \begin{aligned} f(2^{k+1}-1)&=f(2(2^k)-1)\\ &=f(2(2^k-1)+1)\\ &=f(2(2^k-1))+1\\ &=f(2^k-1)+1\\ &=k+1 \end{aligned}

Thus, we have proved that $f(2^n-1)=n$ for all integers n.

## Step 2

Next, we will prove a little lemma. Let $g(x)=2x+1$. We will prove, again by induction, that $\boxed{g^n (1)=2^{n+1}-1}$. Note that $g^n(x)$ means the composition of the function g with itself n times.

Firstly, for the base case, $g^1(1)=2+1=3=2^2-1$ is true. Suppose $g^k (1)=2^{k+1}-1$ is true. Then, $g^{k+1}(1)=2(2^{k+1}-1)+1=2^{k+2}-1$. Thus, the statement is true.

## Step 3

Next, we will prove that if $y<2^n-1$, then $f(y). We will write $y=2^{\alpha_1}x_1$, where $x_1$ is odd. We have that $x_1<2^{n-\alpha_1}$. \begin{aligned} f(y)&=f(2^{\alpha_1} x_1)\\ &=f(x_1) \end{aligned}

Since $x_1$ is odd, we have $x_1=2k_1+1$, where $k_1<2^{n-\alpha_1-1}$.

Continuing, we have \begin{aligned} f(x_1)&=f(2k_1+1)\\ &=f(2k_1)+1\\ &=f(k_1)+1 \end{aligned}

We will write $k_1=2^{\alpha_2}x_2$, where $x_2$ is odd. We have $x_2<2^{n-\alpha_1-\alpha_2-1}$. \begin{aligned} f(k_1)+1&=f(2^{\alpha_2}x_2)+1\\ &=f(x_2)+1 \end{aligned}

where $x_2=2k_2+1$, and $k_2<2^{n-\alpha_1-\alpha_2-2}$. \begin{aligned} f(x_2)+1&=f(2k_2)+1+1\\ &=f(k_2)+2\\ &=\cdots\\ &=f(k_j)+j \end{aligned}

where $k_j=1$, $1=k_j<2^{n-\alpha_1-\alpha_2-\cdots-\alpha_j-j}$.

Case 1: All the $\alpha_i$ are 0, then $y=2(\cdots 2(k_j)+1=g^j(1)=2^{j+1}-1$. Then, $j+1, i.e. $j.

Thus, $f(y)=f(k_j)+j<1+n-1=n$.

Case 2: Not all the $\alpha_1$ are 0, then, $1=k_j<2^{n-\alpha_1-\alpha_2-\cdots-\alpha_j-j}\leq 2^{n-j-1}$. We have $2^0=1<2^{n-j-1}$, thus, $0, which means that $j. Thus, $f(y)=f(k_j)+j<1+n-1=n$.

## Step 4 (Conclusion)

Using Step 1, we have $f(1023)=f(2^{10}-1)=10$, $f(2047)=f(2^{11}-1)=11$. Using Step 3, we guarantee that if $y<2047$, then $f(y)<11$. Thus, the maximum value of f(n) is 10.

Ans: 10

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Did you know:

Euler’s “lucky” numbers are positive integers n such that m2 − m + n is a prime number for m = 0, …, n − 1.

Leonhard Euler published the polynomial x2 − x + 41 which produces prime numbers for all integer values of x from 0 to 40. Obviously, when x is equal to 41, the value cannot be prime anymore since it is divisible by 41. Only 6 numbers have this property, namely 2, 3, 5, 11, 17 and 41.