6. On the paper, the first 66 natural numbers are written. One of these numbers needs to be erased so that the sum of the remaining 65 numbers is divisible by 7. Which number can be erased? Find all solutions.
First method:

We calculate the sum of the first 66 natural numbers using Gauss's method.

The sum is $(1+66) \cdot 66: 2=67 \cdot 66: 2=4422: 2=2211$. 4 POINTS

The quotient of the number 2211 and the number 7 is 315, with a remainder of 6, i.e., $2211=315 \cdot 7+6$. 2 POINTS

The sum will be divisible by 7 if we remove the number 6. 1 POINT

Similarly, the sum will be divisible by 7 if we remove all numbers of the form $7k+6$, i.e., those numbers that give a remainder of 6 when divided by 7.

These numbers are: 6, 13, 20, 27, 34, 41, 48, 55, and 62.

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