A surprisingly elegant and easy proof for a Putnam problem !

Image source : https://www.hmc.edu/about/2023/02/26/first-years-impress-in-putnam-competition/

William Lowell Putnam Mathematical Competition is known for its very hard Math Olympiad style problems. It is organized in America and Canada for university students. Many people got 0 for the test. However, what if I said there is the problem that has a very unique , simple yet elegant that even a secondary students can understand and solve.

Yes, it is. That’s the Putnam 2006 paper question B2

source : https://math.hawaii.edu/home/pdf/putnam/2006.pdf

Thinking and planning

This problem give us a set of n element real numbers and we need to prove that exists a subset S of X as such.

Reading here I got two notice

1. We only need to prove there’s exist and the upper bound is 1/n+1
2. We can choose a really close number of m, so we can actually “don’t care” about the left hand side integer part.

Hence, from the first notice , we have the tuition of using pigeonhole principle where we use a overwhelming numbers to ensure the exists of something.

For the second notice , we have a note that we only need to deal with the fractional part because we can choose the integer m to eliminate the integer part.

Introducing pigeonhole principle

The prove is straight forward

Solution to the Putnam problem

From notice 2 , we have the dividing strategies

We want to achieve M_S from this two interval because that’s when we can create the m as such ( due to the absolute value sign) .

Here is the magic, we using pigeonhole principle to prove the existence of two M in the same interval. Thus the next part is straight forward :

All of the prove is from my Math paper I wrote. For more details, reading the paper **here** . The paper has various idea of using Pigeonhole principle hope you all enjoy !