1. Determine the four smallest consecutive natural numbers such that the first is divisible by 2, the second by 3, the third by 7, and the fourth by 5.
First solution.

The first number is even because it is divisible by 2, and since the numbers are consecutive, the second is odd, the third is even, and the fourth is odd.

The fourth number is divisible by 5 and is odd, so its last digit is $5. \quad 1$ POINT

This means that the last digit of the third number is 4, and since it is divisible by 7, it could be the numbers

$2 \cdot 7=14, 12 \cdot 7=84, 22 \cdot 7=154, \ldots$

2 POINTS

Indeed, the last digits of 10 consecutive multiples of 7, starting from the smallest natural number, are 7, 4, 1, 8, 5, 2, 9, 6, 3, 0.

1 POINT

In this case, the second number could be $13, 83, 153, \ldots$

1 POINT

The smallest among them that is divisible by 3 is the number 153.

2 POINTS

Therefore, the required numbers are 152, 153, 154, and 155.

1 POINT

TOTAL 10 POINTS