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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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    Is the $n$ on the right hand side of the second equation the same as the $n$ on the left hand side of that equation? Because from $n = 14$ it then follows that $m = 189/53 \notin \mathbb{N}$.2012-03-18
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    @Siddhi: Is the $n$ in $m+n$ also supposed to be $p$, or no?2012-03-18
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    I also had to assume $(m,p)=1$. Otherwise, this question does not make sense.2012-03-18
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    @Kannappan: Actually, if $m$ and $p$ are (positive) integers, then you can always answer the question with "No.", regardless of $(m,p) \stackrel{?}{=} 1$.2012-03-18
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    @TMM I fail to see what you say. Please elaborate.2012-03-18
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    @Kannappan: If $(m,p) = d$ then $m + p$ is also not prime, since $d | m + p$. So in both cases ($d > 1, d = 1$) the answer to "Is $m + p$ a prime?" is no.2012-03-18
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    @TMM I fail to see why that should be the case when $(m,p)=1$. $(4,3)=1$ and $4+3=7$ is prime.2012-03-18
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    For everyone confused with one notation: It does make sense, when you look at $\frac{4}{11}$ so for the other expression as well it is assumed to be in the simplest form, i.e. obviously $gcd(m,p)=1$2012-03-18
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    @Kannappan: In this question, if we assume that $(m,p) = 1$, then $m = 27$ and $p = 106$ and $m + p = 133 = 7 \cdot 19$ is not prime.2012-03-18
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    @TMM All I wanted to point out was to say that one does not have to work at all and claim this problem is ambiguous and say the answer is NO, while it would be a slightly non-trivial question given that $(m,p)=1$. I thought I always understood that a number which has non-trivial divisors was not prime.2012-03-18
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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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    (+1) and I agree that $n = 14$, but then suddenly $m + n = 27 + 106$? (for the problem as stated, $m$ is not integral...)2012-03-18
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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 $(m,p)=1$