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## A Nice Solution of Putnam 1981 B5

putnam

Here’s a nice solution of Putnam 1981 B5 that I haven’t seen anywhere else (so far). The main idea is to sum bitwise,' rather than termwise.'

Let $S_k$ denote the set of positive integers with the $k$th bit set, counting from the right starting at $k=0$. Then we have

$\sum_{n=1}^\infty \frac{B(n)}{n^2 + n} = \sum_{k=0}^\infty \sum_{n \in S_k} \frac{1}{n^2 + n}.$

A bit of thinking shows that $n \in S_k$ if and only if $\lfloor 2^{-k} n\rfloor = 2m+1$ is odd. Thus the sum becomes

## A Geometric Solution of Putnam 2003 B5

putnam
geometry

Update 2020-09-08: This solution has been added to Kiran Kedlaya’s solution page.

I've found an elegant geometry-only proof of Putnam 2003 B5, which, to the best of my knowledge, hasn't yet been discovered.

We begin with a diagram of the problem:

## Primes $p$ with $\tan{p} > p$

number-theory

Inspired by Matt Parker’s recent video, I decided to search for primes $p$ with $\tan{p} > p$, the first of which is the 46-digit

$p = 1169809367327212570704813632106852886389036911.$

$p = 116980936732 \dots 886389036911.$

How do you go about finding more?

Well, we want $\tan{p}$ to be big — very big. From high-school trigonometry, we know that this occurs when $p$ is just a tiny bit less than a half-integer multiple of $\pi$. In other words, we want

fourier-analysis
webgl
monte-carlo
optimization