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Two consecutive numbers are removed from the progression 1, 2, 3..n. The arithmetic mean of the remaining numbers is 105/4. What is the smallest value of n? ?What are the two numbers removed?

Now many answers are given but they are given in a hit and trial method. Is there any other method to solve this problem?

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    $\begingroup$ Is the mean of the remaining numbers 261/4 or 105/4? $\endgroup$ Sep 20, 2017 at 18:16
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    $\begingroup$ Let $k$ denote the smaller of the two numbers removed then you know $k+1\le n$ and you can write an equation in $n$ and $k$ with whichever of the two fractions you decide is correct on the right hand side. You should make at least that much of an effort before asking for help. $\endgroup$ Sep 20, 2017 at 18:19
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    $\begingroup$ are you sure that we have enough information ? $\endgroup$ Sep 20, 2017 at 18:31
  • $\begingroup$ At the moment there is just ONE answer (mine) and it is fully algebraic. You did nothing to try and solve the problem and now you say that my answer is "hit and trial"? Come on! :) $\endgroup$
    – Raffaele
    Sep 20, 2017 at 19:03
  • $\begingroup$ yes....it came in our school ecam...@datodatuashvili $\endgroup$
    – Pole_Star
    Sep 21, 2017 at 4:14

2 Answers 2

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$\sum _{k=1}^n k= \dfrac{1}{2} n (n+1)$

Subtract $x+(x+1)$. The sum is then $\dfrac{1}{2} n (n+1) -2x-1$

The mean is then $\dfrac{\frac{1}{2} n (n+1) -2x-1}{n-2}=\dfrac{105}{4}$

$x=\dfrac{1}{8} \left(2 n^2-103 n+206\right)$ and $0<x<n$

$x>0 \to 2 n^2-103 n+206>0$ which means $n\ge 50$

And $x<n \to \dfrac{1}{8} \left(2 n^2-103 n+206\right)<n $ which means $2\le n\le 53$

so we have $50\le n\le 53$ but the only value that gives an integer $x$ is $n=50$

Finally the solution is $x=7$ when $n=50$

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  • $\begingroup$ I suspect that since the OP could not even obtain the equation in $x$ in $n$ that they will not understand how you obtained the solution to the equation. $\endgroup$ Sep 20, 2017 at 18:43
  • $\begingroup$ @JohnWaylandBales Do you think that now it can be more understandable? $\endgroup$
    – Raffaele
    Sep 20, 2017 at 19:01
  • $\begingroup$ Yes, I think so. Thanks. $\endgroup$ Sep 20, 2017 at 19:03
  • $\begingroup$ +1 nice job. N is even so 51 and 53 are excluded but any logical reason for 50 and not 52 ? $\endgroup$ Sep 20, 2017 at 19:11
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    $\begingroup$ @Isham $52$ gives $x=\dfrac{159}{4}$ while $x$ must be integer $\endgroup$
    – Raffaele
    Sep 20, 2017 at 19:26
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Since the given arithmetic mean in the title is $261/4$, we know that $(n-2)\times 261/4$ must be an integer, so $n\equiv 2 \bmod 4$.

The sum of the full set of integers from $1$ to $n$ is $n(n+1)/2$, with average $\color{blue}{(n+1)/2}$. Removing two numbers from that set cannot change the average by more than $1$, since removing the largest two makes it $(n-1)/2 = \color{blue}{(n+1)/2}-1$ and removing the two smallest makes it $(n-1)/2 + 2= \color{blue}{(n+1)/2}+1$.

So we know that $261/4 -1 \le \color{blue}{(n+1)/2} \le 261/4+1$ so $128.5\le n\le 132.5$. Now since only $130\equiv 2 \bmod 4$ in this range, we can be confident that $n=130$ (assuming a well-posed question).

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