1. For which eleven consecutive integers is the sum of the first six numbers equal to the sum of the last five?
Answer: $25,26,27,28,29,30,31,32,33,34,35$.

## Solution:

There are no other solutions, because shifting these values along the number line by $k$ units changes the sum of the first six numbers by $6k$, and the sum of the last five by $5k$.

How can we find this solution? First, sum the first 6 numbers (from 1 to 6) and the next 5 numbers (from 7 to 11). The sums will not be equal, but their difference will indicate the shift of all numbers along the number line. Each such shift by one unit reduces the difference of the found sums by 1.

Another solution method, which simultaneously allows finding the required set and justifying its uniqueness:

Let the first number in this sequence be $x$.

Then we have the following numbers:

![](https://cdn.mathpix.com/cropped/2024_05_06_507ced5c03720ddcb08bg-1.jpg?height=49&width=1248&top_left_y=2557&top_left_x=267)

The sum of the first six is $5x + 15$, and the sum of the last five is $6x + 40$.

By the condition, these sums are equal: $5x + 15 = 6x + 40$, from which $x = 25$ (and the required 11 numbers: 25, 26, ...).

## Instructions for Checking:

An answer without justification that there are no other numbers - 3 points.