Find the smallest even natural number $a$ such that $a+1$ is divisible by $3, a+2$ by $5, a+3$ by $7, a+4$ by $11, a+5$ by 13.
First, let's find the numbers that satisfy the first two conditions: $a$ is even and $a+1$ is divisible by 3.

The smallest such number is -2. Therefore, the numbers that satisfy these conditions are of the form $6k+2$ (GCD $(2,3)=6)$.

Adding the third condition: we are looking for numbers of the form $6k+4$ that are divisible by 5. The smallest such number is -10, so the numbers that satisfy all three conditions are of the form $30l+8$.

We are looking for a number of the form $30l+11$ that is divisible by 7. Then, the number $2l-3$ is also divisible by 7. The smallest $l$ is 5, so the numbers that satisfy all four conditions are of the form $210m+161-3=210m+158$.

We are looking for a number of the form $210m+162$ that is divisible by 11. Then, the number $m+8$ is also divisible by 7. The smallest $m$ is 3, so the numbers that satisfy all five conditions are of the form $2310n+788$.

The number 788 also satisfies the last condition: 793 is divisible by 13.

## Answer

788.