Apologises for the vague title; I couldn't think of anything better to call it.

I'm currently working on the following question:

Consider the equation $\sum_{i=1}^{5} \frac{1}{i} = \frac{X}{5Y}$. Without making explicit calculations on the left hand side, show that $X \equiv Y$ modulo $125$.

I'm unsure how to approach this really. I've noted that $125 = 5^3$ of course, but I don't know how significant that will be. I've tried some simple things like multiplying by $5! \cdot s$ to try and eliminate all denominators, but that doesn't seem to be of any use to me?

Can anyone give a nudge in the right direction for me? Thanks in advance!

  • $\begingroup$ A related notion is that of a harmonic number $H_n$, that is a sum of $\frac{1}{k}$ terms for $k=1$ to $n$. $\endgroup$ – hardmath Jan 30 '16 at 21:29

I think one way to look at this is by subtracting $\frac 1 5$ from both sides of the equation to yield the following:

$$\sum_{i-1}^5\frac 1 i-\frac 1 5=\frac{X}{5Y}-\frac 1 5$$

$$\sum_{i-1}^4\frac 1 i=\frac{X-Y}{5Y}$$

The lowest common denominator of the numbers $1$ through $4$ does not have $5$ as a factor since $5$ is prime. However, the denominator on the right side does have $5$ as a factor. This means that both the numerator and denominator must have been multiplied by some multiple of $5$ from the original reduced fraction. Therefore, $5 | (X-Y)$ since $X-Y$ is the numerator of the resulting fraction from this multiplication.

Now, we want to show $X \equiv Y \pmod{125}$. However, if we show that $125 | (X-Y)$, then $X-Y \equiv 0 \pmod{125}$ and thus $X \equiv Y \pmod{125}$. Therefore, we just need to prove $125 | (X-Y)$.

Now, express $\sum_{i=1}^4\frac 1 i$ as $\frac{p}{4!}$ for some $p \in \mathbb{N}$. If $25 | p$, then the numerator of the reduced fraction is also divisible by $25$ because by reducing a fraction with denominator $4!$, there is no way we can get rid of $25$ as a factor since $4!$ and $25$ are coprime.

Thus, if we show $25 | p$, then since $X-Y$ is a multiple of $5$ times the numerator of the reduced fraction, $125 | (X-Y)$ as $5*25=125$. Thus, by showing $25 | p$, we will have proven the theorem.

I have not solved this problem yet either, so I don't know if I'm leading you in the right direction, but hopefully, this will give you something to go off of. Good luck!

  • $\begingroup$ Thank you for this. I'm still working on this, but your post certainly has given me a direction I wouldn't have thought of exploring. :-) $\endgroup$ – user143137 Jan 31 '16 at 19:41
  • $\begingroup$ @user143137 Glad I could help! $\endgroup$ – Noble Mushtak Jan 31 '16 at 21:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.