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A set of consecutive integers were written down: $1, 2, 3, 4, 5, ... x-1, x, x+1, ... n-2, n-1, n$ One number, $x$, is deleted. The arithmetic mean of the remaining numbers is $602/17$. What does "$x$" and "$n$" equal?

I managed to get two equations down:

$x = (\dfrac12)n^2 - (\dfrac{1187}{34})n + \dfrac{602}{17}$
$n = \dfrac{(1187 ± \sqrt(2312x + 1327097))}{(34)}$

After that I got stuck but I managed these conditions: Since n is an integer Then... The numerator, $1187$ ± $\sqrt(2312x + 1327097)$, must be dividable by $34$ The square root must be odd The radicand, $2312x + 1327097$, must be a perfect square

How would you mathematically prove the answer from here on

xxxxxxxxx
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user113864
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1 Answers1

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For a given $n$, the lowest possible average after removing a value is $n/2$ and the largest possible average is $(n+2)/2$. Since $602/17$ must be between these values, $n=68,69$ or $70$. Since $\frac{\text{total}}{n-1} = \frac{602}{17}$, and the right hand side is in lowest terms, $17 \mid n-1$. Thus, $n=69$, and $x=7$.

ronno
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