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Let $n$ be a positive integer such that $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n} = \frac{4}{11}}$ then $\displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} = \frac{m}{p}.$

The question further "Is $m+p$ a prime?"

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    @Kanna: I think the confusion here is that the question *does* "make sense" (it is well-defined and easily understood) when $(m,p)\ne1$, and you mean to say it is not "*sensible*" (the answer "no" is so obvious that even asking would seem strange).2012-03-18

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At first attempt, I was tempted to do $\displaystyle{3+4+\cdots+3n = \frac{3n(3n+1)}{2}-3}$, but there are $4$ expressions of that sort, it is better to find a general form of it like this

$ \begin{align*} k+(k+1)+(k+1)+\cdots+kn &= \frac{1}{2} \left[ kn(kn+1)-k(k-1) \right]\\ &= \frac{1}{2}\left[ k(n+1)(kn-k+1)\right] \tag{1} \end{align*} $

Appying $(1)$ to $\displaystyle{\frac{3+4+\cdots+3n}{5+6+\cdots+5n}} = \frac{3(n+1)(3n-3+1)}{5(n+1)(5n-5+1)} =\frac{3(3n-1)}{5(5n-4)} = \frac{4}{11}$, leads us to $\displaystyle{\frac{3n-2}{5n-4}=\frac{20}{33}}$ and further to an expresion $99n-66=100n-80 \Rightarrow n=14$

$ \displaystyle{\frac{2+3+\cdots+2n}{4+5+\cdots+4n}} =\frac{2(n+1)(2n-2+1)}{4(n+1)(4n-4+1)} = \frac{30\times27}{60\times53} =\frac{27}{106} $

$m+p=27+106=133$. But $133 = 7\times19$. Therefore the answer is "No, $m+p=133$ is not a prime".

(For those wondering what did I just change, for clarity I changed the right side to be $\frac{m}{p}$, the answer still stays)

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    Formally speaking, from the second equation you get $m=27s$ and $p=106s$, hence $m+p=133s$ for some positive integer $s$. This does not change the answer, though: it is not a prime for any $s$ including $s=1$.2012-03-18
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$k+(k+1)+...+kn=\frac{k}{2}(n+1)(kn-k+1)$ (just the half of the sum of the first and the last terms times the number of the terms)$=\frac{k}{2}(kn^2+n-(k-1))$

Now, from the first equation we get $n$: $33(3n^2+n-2)=20(5n^2+n-4)$ $n^2-13n-14=0$ $(n-14)(n+1)=0$ $n>0\Rightarrow n=14$

And from the second equation we get $m$: $m=n\frac{2n^2+n-1}{8n^2+2n-6}=\frac{189}{53}=3.566...$

So, $m+n$ is not prime, because it is not even integer...

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    Wow, while I was typing this, you have already changed $m/n$ to $m/p$ in the second equation... Then, from the second equation $m/p=27/106$, i.e. $m=27s$, $n=106s$, and $m+n=133s=7\times19\times s$, not a prime.2012-03-18
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Hint:

$\sum_{i=1}^ni=\dfrac{n(n+1)}{2}$

After a bit of painful algebra, $n=14$. And, $m= 27$ and $p=106$, if I assume $(m,p)=1$