A Harmonic Conundrum

This one came from user_1312 on reddit with a heading “This is a bit tricky… Enjoy!”. What else can we do but solve it?

Let $m$ and $n$ be positive numbers such that $\frac{m}{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{101}$. Prove that $m-n$ is a multiple of 103.

My first approach was to invoke Wolstenholme’s theorem: the numerator of the $p-1$th harmonic number is a multiple of $p^2$ if $p$ is prime, so $H_{102}={103}^2\frac{a}{b}$ with $b$ and $103$ coprime.

(What’s a harmonic number? It’s the sum of reciprocals. $H_4$, for example, is $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}$; the number we’re looking at is $H_{101}$.

What we have is $H_{102}-1-\frac{1}{102}$, which we could write as $H_{102}-\frac{103}{102}$. Both of those terms are multiples of 103… but (looking more closely, some time on) I don’t see that this proves that $m-n$ is a multiple of 103.

Fortunately, I have another way

Let’s think about $\frac{m-n}{n}=\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{101}$.

If we group those, Gauss-style into 50 pairs: $(\frac{1}{2}+\frac{1}{101})+(\frac{1}{3}+\frac{1}{100})+\dots$, we get $\frac{103}{2\times 101}+\frac{103}{3\times 100}+\dots$.

We can write $\frac{m-n}{n}=103\left(\frac{p}{q}\right)$ for some integers $p$ and $q$. However, $103$ is prime, and doesn’t show up in any of the denominators, so $q$ and $103$ are coprime.

That means $m-n$ must be a multiple of 103, because it can’t cancel with anything on the bottom.

What a lovely puzzle!

* Edited 2019-11-11 to fix LaTeX.