If two consecutive numbers are removed from the series $1+2+3+\ldots+n$ the average becomes $99/4$. Find the two numbers.

Question

The initial average will be $\frac{n+1}{2}$. If the two numbers are $k$ and $k+1$ then the new average will be $\frac{n(n+1)/2-(2k+1)}{n-2}$. I couldn't figure further even though I got the relation between $n$ and $k$ in many different ways.

If the question is not clear, here is an example to explain it. If $n=10$, the initial average will be $5\cdot 5$ {$(1+2+\cdots + 10)/10$} Now if two consecutive numbers like $2,3$ or $8,9$ are removed from this series, the new average changes, and this new average has been given to be $99/4$, however we also don't know the value of $n$, so the question seems to be pretty difficult.

Answer 1

As in Claude Leibovici's answer, removing $k+(k+1)$ from $1+2+\cdots+n$ to leave an average of $99/4$ implies, after a bit of algebra, that

$$k={2n^2-97n+194\over8}$$

is an integer between $1$ and $n-1$. It's easy to see that $k$ being an integer implies $n\equiv2$ mod $8$. If we write $n=8m+2$ (with $m\gt0$, since $n=2$ is obviously not possible), we find, after a tad more algebra, that $k=16m^2-89m+1$. The inequality constraints are thus now

$$1\le16m^2-89m+1\le8m+1$$

or

$$89\le16m\le97$$

There is clearly only one integer solution: $m=6$, corresponding to $n=50$ and $k=43$.

Answer 2

The new average will be $$\frac{\frac{n(n+1)}2-(2k+1)}{n-2}=\frac {99}4$$ Solving for $k$ gives $$k=\frac{2 n^2-97 n+194}{8} $$ which must be a positive integer lower or equal to $n$.

Solving for $n$ gives $$n= \frac{97+\sqrt{64 k+7857}}{4} $$ which must be an integer.

Now consider the extreme cases $k=1$ and $k=n-1$; this gives very narrow bounds for $n$. From algebra, $k=1\to n=\frac{93}2$ and $k=n-1\to n=\frac{101}2$. In the worst case, only four values of $n$ would need to be tested.

Does this help you ?

Answer 3

This is a calculated guessing approach. The average of $ n $consecutive integers will be. $ (n+1)/2$. $99/4$ is a little less than$ 25$. So $n $can be taken to be$ 50$. Not$ 49,$ because two less than$ 49$, which is $47 $is not divisible by $4$. So first$ 50 $numbers sum will be$ 1275$. $99/4 * 48$ will be$ 1188.$ Here $48 $is divisible by $4$ is the reason I took $ n=50. $. $ 1275-1188$ will be$ 87$. two consecutive numbers giving $87$ as sum are$ 43,44$. I hope this is of some help

Answer 4

We have, that the sum of $n+1$ terms, excluding the $m$-th and $m+1$-th, is: $$ \begin{gathered} S(n + 1,m) = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{m + 2\, \leqslant \,k\, \leqslant \,n + 1} k = \sum\limits_{1\, \leqslant \,k\, \leqslant \,m - 1} k + \sum\limits_{1\, \leqslant \,k\, \leqslant \,n - m} {\left( {k + m + 1} \right)} = \hfill \\ = \left( \begin{gathered} m \\ 2 \\ \end{gathered} \right) + \left( {m + 1} \right)\left( {n - m} \right) + \left( \begin{gathered} n + 1 - m \\ 2 \\ \end{gathered} \right) = \hfill \\ = \frac{1} {2}m\left( {m - 1} \right) + \left( {m + 1} \right)\left( {n - m} \right) + \frac{1} {2}\left( {n + 1 - m} \right)\left( {n - m} \right) = \hfill \\ = \frac{1} {2}\left( {m\left( {n - 1} \right) + \left( {n + 3} \right)\left( {n - m} \right)} \right) = \hfill \\ = \frac{1} {2}\left( {n\left( {n - 1} \right) + 4\left( {n - m} \right)} \right) = \frac{{n\left( {n + 3} \right)}} {2} - 2m \hfill \\ \end{gathered} $$ So we shall have: $$ \begin{gathered} \frac{{S(n + 1,m)}} {{n - 1}} = \frac{{99}} {4}\quad \Rightarrow \quad \left\{ \begin{gathered} n - 1 = 4\,q \hfill \\ S(n + 1,m) = \frac{{n\left( {n + 3} \right)}} {2} - 2m = 99\;q \hfill \\ \end{gathered} \right.\quad \Rightarrow \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} 1 \leqslant m \leqslant n = 4\,q + 1 \hfill \\ n\left( {n + 3} \right) = 198\;q + 4m \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$ The last gives: $$ \begin{gathered} 4 \leqslant 4\left( {4q + 1} \right)\left( {q + 1} \right) - 198\;q = 4m \leqslant 4\left( {4\,q + 1} \right) \hfill \\ 0 \leqslant \left( {4q + 1} \right)\left( {q + 1} \right) - \frac{{198}} {4}\;q - 1 \leqslant 4\,q \hfill \\ 0 \leqslant q^{\,2} - \frac{{178}} {{16}}\;q \leqslant \,q \hfill \\ 0 \leqslant q - \frac{{178}} {{16}} \leqslant \,1 \hfill \\ \end{gathered} $$ i.e. $$ \frac{{178}} {{16}} \leqslant q \leqslant \,\frac{{194}} {{16}}\quad \Rightarrow \quad \left\lceil {\frac{{178}} {{16}}} \right\rceil \leqslant q \leqslant \,\left\lfloor {10 + \frac{{34}} {{16}}} \right\rfloor \quad \Rightarrow \quad 12 \leqslant q \leqslant 12 $$ In conclusion, so we have: $$ \left\{ \begin{gathered} q = 12 \hfill \\ n = 4\,q + 1 = 49 \hfill \\ m = \frac{1} {4}\left( {n\left( {n + 3} \right) - 198\;q} \right) = 43 \hfill \\ \end{gathered} \right. $$ which in fact gives: $$ \frac{{S(n + 1,m)}} {{n - 1}} = \frac{{\frac{{n\left( {n + 3} \right)}} {2} - 2m}} {{n - 1}} = \frac{{\frac{{49 \cdot 52}} {2} - 86}} {{48}} = \frac{{1188}} {{48}} = \frac{{99}} {4} $$

Answer 5

This approach finds that the new average lies within $\pm 1$ of the original average. This significantly narrows down possibilities, and the solution can then be found easily by elimination.

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After removing the two numbers, the new average, $a$, is given by $$\begin{align} \frac {99}4=a&=\frac{\frac{n(n+1)}2-(2k+1)}{n-2}\\ &=\underbrace{\frac {n+1}2}_{\text{original average, $a_0$}}+ \underbrace{\frac {n-2k}{n-2}}_{\in [-1,1] \text{ for } 1\le k\le n-1} \end{align}$$

Hence $a$ lies within $\pm 1$ of the original average $a_0$ before removal, i.e. $$a_0-1\;\le\; a=\frac {99}4=24.75\;\le\; a_0+1$$

As $a_0=\frac {n+1}2$, it can only be either an integer or an integer and a half, hence $24\le a_0\le 25.5$.

$$\begin{array} {lrrrr} \hline{a_0(n)=\frac{n+1}2} &24&24.5&25&25.5\\ n &47 &48 &49 &\color{red}{50}\\ n-2 &45 &46 &47 &\boxed{48} \\ a-a_0(n)=\frac {n-2k}{n-2}&\frac 34&\frac 14&-\frac14&-\frac34\\ \hline \end{array}$$ Also, the sum of the remaining numbers $\frac {99}4 (n-2)$ must be integer, so $(n-2)$ must a multiple of $4$, the only candidate for which is $48$.

Hence we conclude that $\color{red}{n=50, k=43}\qquad \blacksquare$