Question

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:
\[
\boxed{2}
\]