When a prime is divided by 60, the remainder is a composite number. When a second prime is divided by 60, the remainder is a prime. Find the smallest possible value of the second prime.

Answer 1

We need to find primes that satisfy the conditions given. Let's break it down step by step.

1. First Prime: The first prime \( p_1 \) must give a composite remainder when divided by 60.

   The composite numbers less than 60 are: \[ 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60 \] We select \( p_1 \) such that \( p_1 \mod 60 \) returns one of these numbers. However, the specific value of \( p_1 \) is not initially needed to find \( p_2 \).

2. Second Prime: The second prime \( p_2 \) must give a prime remainder when divided by 60.

   The prime numbers less than 60 are: \[ 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59 \]

   We need to find the smallest prime \( p_2 \) that has a remainder of one of these primes when divided by 60.

3. Finding the smallest prime \( p_2 \): Let’s start testing the smallest prime numbers in ascending order to find the remainders:

   • \( p_2 = 2 \): \[ 2 \mod 60 = 2 \quad (\text{which is prime}) \]
   • \( p_2 = 3 \): \[ 3 \mod 60 = 3 \quad (\text{which is prime}) \]
   • \( p_2 = 5 \): \[ 5 \mod 60 = 5 \quad (\text{which is prime}) \]
   • \( p_2 = 7 \): \[ 7 \mod 60 = 7 \quad (\text{which is prime}) \]
   • \( p_2 = 11 \): \[ 11 \mod 60 = 11 \quad (\text{which is prime}) \]
   • \( p_2 = 13 \): \[ 13 \mod 60 = 13 \quad (\text{which is prime}) \]
   • \( p_2 = 17 \): \[ 17 \mod 60 = 17 \quad (\text{which is prime}) \]
   • \( p_2 = 19 \): \[ 19 \mod 60 = 19 \quad (\text{which is prime}) \]
   • \( p_2 = 23 \): \[ 23 \mod 60 = 23 \quad (\text{which is prime}) \]
   • \( p_2 = 29 \): \[ 29 \mod 60 = 29 \quad (\text{which is prime}) \]
   • \( p_2 = 31 \): \[ 31 \mod 60 = 31 \quad (\text{which is prime}) \]
   • Check \( p_2 = 37 \): \[ 37 \mod 60 = 37 \quad (\text{which is prime}) \]
   • Check \( p_2 = 41 \): \[ 41 \mod 60 = 41 \quad (\text{which is prime}) \]
   • Check \( p_2 = 43 \): \[ 43 \mod 60 = 43 \quad (\text{which is prime}) \]
   • Check \( p_2 = 47 \): \[ 47 \mod 60 = 47 \quad (\text{which is prime}) \]
   • Check \( p_2 = 53 \): \[ 53 \mod 60 = 53 \quad (\text{which is prime}) \]
   • Check \( p_2 = 59 \): \[ 59 \mod 60 = 59 \quad (\text{which is prime}) \]

From our list, we noted that \( 2 \) is the smallest prime whose remainder when divided by 60 is also prime.

4. Answering the question: The smallest possible value of the second prime \( p_2 \) that leaves a prime remainder when divided by 60 is: \[ \boxed{2} \]