How many integers from 1 to 2024 can be written simultaneously as 4 a + 19 and 6 b + 21, where a, b are integers?

To solve this, we need to determine the integers from \( 1 \) to \( 2024 \) that can be represented in both forms \( 4a + 19 \) and \( 6b + 21 \), where \( a \) and \( b \) are integers. Let's go through the steps:

1. **Identify the forms:** 
   - \( x = 4a + 19 \): This implies that \( x \equiv 19 \pmod{4} \).
   - \( x = 6b + 21 \): This implies that \( x \equiv 21 \pmod{6} \).

2. **Rewrite the congruences:**
   - Since \( 21 \equiv 3 \pmod{6} \), we can rewrite the second congruence as:
     \[
     x \equiv 3 \pmod{6}
     \]

   We now have the two conditions:
   \[
   x \equiv 19 \pmod{4} \quad \text{and} \quad x \equiv 3 \pmod{6}.
   \]

3. **Use the Chinese Remainder Theorem (CRT):**
   We need to find an integer \( x \) that satisfies both conditions. This will provide a single congruence modulo the least common multiple of \( 4 \) and \( 6 \), which is \( 12 \).

   - Rewrite the first condition as:
     \[
     x \equiv 3 \pmod{6}
     \]

  

How many integers from 1 to 2024 can be written simultaneously as 4a + 19 and 6b + 21, where a, b are integers?

Learn how to find integers between 1 and 2024 that satisfy both the equations x = 4a + 19 and x = 6b + 21 using modular arithmetic and the Chinese Remainder Theorem. This problem involves key concepts in number theory and algebra, and is suitable for students in Grades 11-12.

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