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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 $m = 189/53 \notin \mathbb{N}$. –  TMM Mar 18 '12 at 1:50
    
@Siddhi: Is the $n$ in $m+n$ also supposed to be $p$, or no? –  anon Mar 18 '12 at 2:01
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I also had to assume $(m,p)=1$. Otherwise, this question does not make sense. –  user21436 Mar 18 '12 at 2:09
    
@Kannappan: Actually, if $m$ and $p$ are (positive) integers, then you can always answer the question with "No.", regardless of $(m,p) \stackrel{?}{=} 1$. –  TMM Mar 18 '12 at 2:13
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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. –  TMM Mar 18 '12 at 2:17

3 Answers 3

up vote 6 down vote accepted

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...) –  TMM Mar 18 '12 at 1:53
    
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 $s=1$. –  Vadim Mar 18 '12 at 2:32

$k+(k+1)+...+kn=\frac{k}{2}(n+1)(kn-k+1)$ (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. –  Vadim Mar 18 '12 at 2:29

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 $(m,p)=1$

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