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One number is removed from a set of integers from 1 to n,the average of the remaining numbers is $\large{\frac{163}{4}}$. Which number was removed? I tried to find the mean of $$\frac{1+2+....+n-1}{n-1}=\frac{163}{4}$$ but I can`t find the value from there.

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4 Answers 4

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Hint $$1 + 2 + ............... + n = \frac{n(n+1)}{2}$$

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If $1$ is removed then the average is $\frac{2+\dots+n}{n-1}=\frac{1}{2}n+1$

If $n$ is removed then the average is $\frac{1+\dots+n-1}{n-1}=\frac{1}{2}n$

So $$\frac{1}{2}n\leq\frac{163}{4}\leq\frac{1}{2}n+1$$

Not much candidates for $n$ anymore. Try them out.

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Let $n \in \mathbb{N}^*$, and let $k$ be the number between $1$ and $n$ that has been removed. We have that the average of $\{ 1, \ldots, n \} \setminus \{k\}$ is $163/4$, hence

$$ \frac{n(n+1)}{2(n-1)} = \frac{163}{4}. $$

This gives $4k = 2n^2 - 161n + 163$. Since $k \leq n$ we have $4n \geq 2n^2 - 161n + 163$, so in particular $165 n \geq 2n^2$. Thus, $n \leq 82$.

Since $k \geq 1$ we have $4 \leq 2n^2 - 161n + 163$, so $0 \leq 2n^2 - 161n + 159 = 2(n-1)(n-79.5)$. So $n = 1$ or $n \geq 80$.

This shows that $n \in \{ 1, 80, 81, 82 \}$. Now the solution integer solution $k \in [1, n]$ to the equation $4k = 2n^2 - 161n + 163$ is $k = 61$ (and $n = 81$).

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From @drhab's equation $n \le 81.5 \le n+2$, you can find n to be 80 or 81.

Since $(n-1) \times 163 /4 $ should be integer, it can only be 81.

Now find k = summation of n minus summation of n except k.

$k = \Sigma_{i=1}^{81}i - (81 -1)*(163/4) = 61$

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