On the cover of HP Printer A4 Paper, there is a mysterious Math Problem, which is strangely not solved by anyone on the internet yet! At least, when I gave it a search on Google, no one seems to have solved it…

Click on the images above to zoom in!

I will also type out the question in case the image is not clear:

**Problem of the Week**

Suppose is a function from positive integers to positive integers satisfying , , and , for all positive integers .

Find the maximum of when is greater than or equal to 1 and less than or equal to 1994.

I will give it a try when I have free time this week.

Anyone who is interested to try please leave your answer in the comments below! I will help to verify if it is correct. If it is, you may be the first person on the internet to solve the HP A4 Paper Math Problem! 🙂

You may also forward this problem to your friends to try!

Update: Solution can be found here (Solution to HP A4 Printer Paper Mysterious Question)

Problems of Number Theory in Mathematical Competitions (Mathematical Olympiad Series) is a Math Olympiad book written by 2 Chinese authors. As everyone may know, China is one of the big guns of International Math Olympiad, winning the most medals out of all the countries almost each time, due to the huge talent pool of kids, and also their **specialized Chinese Math training**. So, to discover the **Chinese Math secrets of Number Theory**, this book cannot be missed. (By the way, this book is written in English, in case of any misunderstanding.)

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It counts the number of bits that are set in the binary representation of the number.

For powers of two, f() => 1

f(64) = f(32) = f(16) = f(8) = f(4) = f(2) = f(1) ==> 1

Consider the 3 rules: 1) f(1) = 1 2) f(2n) = f(n) 3) f(2n+1) = f(2n) +1

Rule 2 is another way of saying that any even number 2n has the same bit pattern as the number n, only shifted over one bit position.

For any odd number, the value is one more than the previous even number, since bit zero is set for the odd number, and clear for the even, so rule 3 is upheld.

f(1) = 1 – rule 1

f(2) = 1 – rule 2

f(3) = 2 – rule 3

f(4) = 1 – rule 2

f(5) = 2 – rule 3

f(6) = 2 – rule 2

f(7) = 3 – rule 3

f(8) = 1 – rule 2

The maximum is 10.

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Interesting solution based on bits. Thanks for the comment!

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