## Task 1 - 080731

We are looking for natural numbers that, when divided by 7, leave a remainder of 4, when divided by 4, leave a remainder of 3, and when divided by 3, leave a remainder of 1.

a) Determine the smallest such natural number!

b) How can one obtain further natural numbers from the number found in a) that satisfy the same conditions?
a) The sequence of natural numbers that leave a remainder of 4 when divided by 7 begins: 4; 11; 18; 25; 32; 39; 46; 53; 60; 67; 74; 81; 88; 95; ...

Of these numbers, those that leave a remainder of 3 when divided by 4 are exactly the numbers: 11; 39; 67; 95; ... (every fourth number of the previous sequence starting from 11).

The smallest of these numbers that also leaves a remainder of 1 when divided by 3 is the number 67.

b) Further natural numbers that meet the conditions can be obtained by adding common multiples of 3, 4, and 7 to 67.

In particular, the next larger such numbers are obtained by repeatedly adding the least common multiple (LCM) of 3, 4, and 7 (i.e., since 3, 4, and 7 are pairwise coprime, the number \(3 \cdot 4 \cdot 7 = 84\) to 67. The next such numbers are: 151; 235; 319; ...