From a set of positive consecutive integers starting with $1$, one number is erased and the AM of the remaining numbers is $\frac{602}{17}$ A set of positive consecutive integers starting with one is written on a blackboard. One number is erased and the AM of the remaining numbers is $\frac{602}{17}$. The erased number is 


*

*6

*7

*8

*9


Initially, $$\text{AM}=\frac{n+1}{2}$$
After removing the number $x$, $$\text{AM}=\frac{\frac{n(n+1)}{2}-x}{n-1}=\frac{602}{17}$$
How can I solve the above equation?
 A: Finally I came up with a solution.
$$\frac{\frac{n(n+1)}{2}-n}{n-1}\le \frac{602}{17}\le\frac{\frac{n(n+1)}{2}-1}{n-1}$$
$$\frac{n}{2}\le\frac{602}{17}\le\frac{n+2}{2}$$
$$n\le70+\frac{14}{17}\le n+2$$
$$n=69 \text{ or } 70$$
Substituting $n=69$ the actual equation, $x=7$
$n=70$ is not possible since I got a non integral $x$
A: Outline
The $17$ in the denominator is a give away. There are $17k + 1$ numbers to begin with. 
The second give away is the average which is slightly above $35$ when a number is reduced so there are about double the numbers. 
Put these two facts together and then one can quickly see that the sum of the first $69$ numbers is $2415$ and if we remove $\color{blue}{7}$, then $2408 / 68 $ gives you the required average.
A: Let's see.
Average of remaining numbers is something over 17 --> n-1 is a multiple of 17.
Assume n = 35.  The complete set of numbers adds up to 630, the remaining 34 numbers have to add up to 2×602 = 1204.  That does not work, we need bigger numbers, so we need a bigger n.
Assume n = 86.  The complete set of numbers adds up to 3741, after taking one away the remaining sum has to be 5×602 = 3010.  We have to take away more than the biggest number, we have too many big numbers, we need n to be smaller.
One value of n that's one more than a multiple of 17 will work.  The value of x will then follow.
