Hint:

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$1-\frac{1}{2}+...+\frac{1}{1319}$=$(1+\frac{1}{2}+...+\frac{1}{1319})-2(\frac{1}{2}+...+\frac{1}{1318})$=$(1+\frac{1}{2}+...+\frac{1}{1319})$-$(1+\frac{1}{2}+...+\frac{1}{659})$=$\frac{1}{660}+...+\frac{1}{1319}$=$(\frac{1}{660}+\frac{1}{1319})+...+(\frac{1}{989}+\frac{1}{990})$=$\frac{1979}{660*1319}+...+\frac{1979}{989*990}$. Since $1979$ is prime, $p$ is divisible by $1979$.

given, $ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...-\frac{1}{1318}+\frac{1}{1319} $ We notice that the negative terms have even denominators. Now we write each fraction of the form $ \frac{1}{2k} $ as , $ \frac{1}{2k} - \frac{1}{k} $ So our given quantity can be written as, $ \frac{p}{q}=(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1318}+\frac{1}{1319}) - 2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1318})$ $ = (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1318}+\frac{1}{1319}) - (\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{659}) $ $ = (\frac{1}{660}+\frac{1}{661}+\frac{1}{662}+...+\frac{1}{1318}+\frac{1}{1319}) $ Now, we notice $ \frac{1}{660 + k} + \frac {1}{1319-k} = \frac {1979}{(660+k)(1319-k)} $ Therefore, we can write , $ \frac{p}{q} = ( \frac{1}{660} + \frac{1}{1319}) + \cdots + (\frac{1}{989} + \frac{1}{990}) $ $ = \frac {1979}{660.1319} + \frac {1979}{661.1318} + \cdots + \frac{1979}{989.990} $ $ = 1979\frac{p'}{q'} $, where $ \frac{p'}{q'} $ is the sum of the fractions $\frac {1}{660.1319} , \frac {1}{661.1318} , \cdots , \frac{1}{989.990} $ . Therefore, we have $ pq' = 1979p'q $ ......................................... $ (1) $ Now, $ q' $ is the product of all the integers from $ 660 $ to $ 1939 $, each relatively prime to $ 1979 $ ( which is itself a prime), so, $ 1979 \nmid q' $. Hence from the above relation $ (1) $ we have $ 1979 | p $, as required. link

We first write \begin{align*} \frac{p}{q} &=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots-\frac{1}{1318}+\frac{1}{1319}\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-2\cdot\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{1318}\right)\\ &=1+\frac{1}{2}+\cdots+\frac{1}{1319}-\left(1+\frac{1}{2}+\cdots+\frac{1}{659}\right)\\ &=\frac{1}{660}+\frac{1}{661}+\cdots+\frac{1}{1319} \end{align*}Now, observe that \begin{align*} \frac{1}{660}+\frac{1}{1319}=\frac{660+1319}{660\cdot 1319}=\frac{1979}{660\cdot 1319} \end{align*}and similarly $\frac{1}{661}+\frac{1}{1318}=\frac{1979}{661\cdot 1318}$ and $\frac{1}{662}+\frac{1}{1317}=\frac{1979}{662\cdot 1317}$, and so on. We see that the original equation becomes \begin{align*} \frac{p}{q} =\frac{1979}{660\cdot 1319}+\frac{1979}{661\cdot 1318}+\cdots+\frac{1979}{989\cdot 990}=1979\cdot\frac{r}{s} \end{align*}where $s=660\cdot 661\cdots 1319$ and $r=\frac{s}{660\cdot 1319}+\frac{s}{661\cdot 1318}+\cdots+\frac{s}{989\cdot 990}$ are two integers. Finally consider $p=1979\cdot\frac{qr}{s}$, and observe that $s\nmid 1979$ because $1979$ is a prime, it follows that $\frac{qr}{s}\in\mathbb{Z}$. Hence we deduce that $p$ is divisible with $1979$.

$\frac{p}{q}=\sum_{i=1}^{1319}\frac{1}{i} -2\sum_{i=1}^{1318}\frac{1}{i}$ $\implies \frac{p}{q} =\sum_{i=660}^{1319}\frac{1}{i}=\sum_{i=660}^{989}(\frac{1}{i} + \frac{1}{1979-i}) =1979*(\sum_{i=1}^{989} \frac{1}{i(1979-i)})$ Hence $1979|p$

Who was the author of this problem interesting problem?

This problem is simply done if we apply catalan identity

$\sum_{i=1}^{1319}\frac{(-1)^{i+1}}{i} \equiv 0 (mod \ 1979)$ $\iff \sum_{i=1}^{1319}\frac1i - 2 \cdot \sum_{i=1}^{659}\frac1{2i} \equiv 0 (mod \ 1979)$ $\iff \sum_{i=1}^{1319}\frac1i - 2 \cdot \sum_{i=1}^{659}\frac1{i} \equiv 0 (mod \ 1979)$ $\iff \sum_{i=660}^{1319}\frac1i \equiv 0 (mod \ 1979)$ But this is clearly true (Just use Gaussian pairing; $(660 + t) + (1319 - t) \equiv 0 (mod \ 1979)$

Notice that $$\sum_{i=1}^{1319}{\frac{(-1)^{i+1}}{i}}=\sum_{i=1}^{1319}{\frac{1}{i}}-2\sum_{i=1}^{659}{\frac{i}{2i}}=\sum_{i=660}^{1319}{\frac{1}{i}}=\sum_{i=0}^{329}{\frac{1979}{(660+i)(1319-i)}}.$$$\square$

orl wrote: If $p$ and $q$ are natural numbers so that \[ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, \]prove that $p$ is divisible with $1979$. Sum $=\sum_{k=1}^{1319} \frac{1}{k}-2(\sum_{k=1}^{659} \frac{1}{2k})=\sum_{k=660}^{1319} \frac{1}{k}= \sum_{k=660}^{989} (\frac{1}{k} + \frac{1}{1979-k})=1979 \sum_{k=660}^{989} \frac{1}{k(1979-k)}= 1979 \cdot \frac{p_s}{q}.$ The denominator $k(1979-k)$ is prime to $1979$, since this number is a prime. So the gcd of the denominator is not a multiple of $1979$, it must be that the numerator is a multiple of $1979$, as desired.

let $a = 1 -\frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{1318} + \frac{1}{1319}$ $a+2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1318}) = 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{1318} + \frac{1}{1319}$ $a+1+\frac{1}{2}+\frac{1}{3}+... +\frac{1}{659}= 1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots + \frac{1}{1318} + \frac{1}{1319}$ $a= \frac{1}{660} + \frac{1}{661} + \frac{1}{662} + \cdots + \frac{1}{1318} + \frac{1}{1319}$ When you pair the first and last terms, the numerator is $1979$ and the denominator does not divide $1979$ because it is the product of $2$ values and $1979$ is prime. This works for all the tems that are opposite to each other. So the numerator divides $1979$

It suffices to show that $1-\tfrac12+\tfrac13-\tfrac14+\cdots-\tfrac1{1318}+\tfrac1{1319} \equiv 0 \pmod{1979}$. We see that $$1-\frac12+\frac13-\frac14+\cdots-\frac1{1318}+\frac1{1319}=\left(1+\frac13+\frac15+\cdots+\frac1{1319}\right)-\left(\frac12+\frac14+\frac16+\cdots+\frac1{1318}\right)=\left(1+\frac12+\frac13+\cdots+\frac1{1319}\right)-2\left(\frac12+\frac14+\frac16+\cdots+\frac1{1318}\right)=\left(1+\frac12+\frac13+\cdots+\frac1{1319}\right)-\left(1+\frac12+\frac13+\cdots+\frac1{659}\right)=\frac1{660}+\frac1{661}+\frac1{662}+\cdots+\frac1{1319}=1979\left(\frac1{660 \cdot 1319}+\frac1{661*1318}+\frac1{662*1317}+\cdots+\frac1{989 \cdot 990}\right) \equiv 0 \pmod{1979}.$$Since $1979$ is prime, the numerators and denominators of each fraction are relatively prime. Finally, we see that $1979 \mid p$, as desired. $\blacksquare$

orl wrote: If $p$ and $q$ are natural numbers so that \[ \frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}, \]prove that $p$ is divisible with $1979$. This nowadays would be an AMC problem like 15 (maybe even easier). $RHS=\sum_{n=1}^{1319}\frac{1}{n}-2\sum_{i=2j}^{j=(1->659)}\frac{1}{i}=\sum_{k=660}^{1319}\frac{1}{k}.$ We pair up 1/k with 1/(1979-k), which makes the numerator always a multiple of 1979. Then the RHS is a sum of fractions with numerators multiples of 1979, so clearly the entire sum will have a numerator divisible by 1979.

$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319}$ $\Rightarrow \frac{p}{q}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \ldots +\frac{1}{1318}+\frac{1}{1319} -2(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{1318})$ $\Rightarrow \frac{p}{q}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \ldots +\frac{1}{1318}+\frac{1}{1319} -(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{659})$ $\Rightarrow \frac{p}{q}=\frac{1}{660}+\frac{1}{661}+\frac{1}{662}+ \ldots +\frac{1}{1318}+\frac{1}{1319}$ $\Rightarrow \frac{p}{q}=(\frac{1}{660}+\frac{1}{1319})+(\frac{1}{661}+\frac{1}{1318})+(\frac{1}{989}+\frac{1}{990})$ $\Rightarrow \frac{p}{q}= \frac{1979}{x_1}+\frac{1979}{x_2}+\frac{1979}{x_3}+\ldots+\frac{1979}{x_{330}}$, where $x_i's \in \mathbb{Z}$ $\Rightarrow \frac{p}{q}= \frac{1979p_0}{q_0}$ As $1979$ is a prime and all the primes in the denominator $<1979$, we conclude that $gcd(1979,q_0)=1$ $1979 \vert p$, and we're done!

100th post Observe that this is equivalent to proving that $$\frac{p}{q}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319} \equiv 0 \pmod{1979}.$$Now, $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots -\frac{1}{1318}+\frac{1}{1319} = \frac{p}{q}=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \ldots +\frac{1}{1318}+\frac{1}{1319}\right) - 2 \left(\frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{1318} \right)$$$$ = \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \ldots +\frac{1}{1318}+\frac{1}{1319}\right) - \left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+ \ldots +\frac{1}{1318}+\frac{1}{659}\right)$$$$ = \left(\frac{1}{660}+\frac{1}{661}+\frac{1}{662}+ \ldots +\frac{1}{1318}+\frac{1}{1319}\right).$$Now, $$\frac{1}{k} + \frac{1}{1979 - k} \equiv \frac{1979}{k\dot (1979-k)} \equiv 0 \pmod{1979} \text{(as 1979 is prime)} $$and by pairing opposite terms, we are done. $\square$

NT but it's actually algebra. \[\frac{p}{q}=\sum_{i=1}^{1319} \frac{1}{i}-2 \left( \sum_{i=1}^{659} \frac{1}{2i} \right)=\sum_{i=1}^{1319} \frac{1}{i}-\sum_{i=1}^{659} \frac{1}{i}=\sum_{i=660}^{1319} \frac{1}{i}=\sum_{i=660}^{989} \frac{1}{i}+\frac{1}{1979-i}=\sum_{i=660}^{989} \frac{1979}{i(1979-i)} \implies 1979 \mid p\]