## Task 4 - 180714

Determine the smallest prime number that leaves a remainder of 1 when divided by 5, 7, and 11!
If $p$ is the sought prime number, then $p-1$ is divisible by 5, 7, and 11. Furthermore, since the only even prime number $p=2$ does not satisfy the required properties, $p$ must be odd, and thus $p-1$ is also divisible by 2. Therefore, and because 2, 5, 7, and 11 are pairwise coprime, the only candidates for $p$ are numbers that are one more than a multiple of $2 \cdot 5 \cdot 7 \cdot 11 = 770$.

The number 771 is divisible by 3, and the number $2 \cdot 770 + 1 = 1541$ is divisible by 23. The next such number is $3 \cdot 770 + 1 = 2311$. Since it is not divisible by 2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37, 41, 43, or 47, and since $53^2 = 2809 > 2311$, 2311 is a prime number.

2311 leaves a remainder of 1 when divided by 5, 7, and 11, so it is the sought number.

Solutions of the I. Round 1978 taken from [5]

### 4.20.2 II. Round 1978, Class 7