From a set of positive consecutive integers starting with $1$, one number is erased and the AM of the remaining numbers is $\frac{602}{17}$

Question

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

1. 6
2. 7
3. 8
4. 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?

Answer 1

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 $x$

Answer 2

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.

Answer 3

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.