Find three consecutive numbers such that the first number is a multiple of 2, the second number is a multiple of 3, and the third number is a multiple of 4. Are there more such numbers? How often do they occur?
Given: Find three consecutive numbers such that:
- First number is a multiple of 2
- Second number is a multiple of 3
- Third number is a multiple of 4
Let the three consecutive numbers be: n, n+1, n+2
Conditions:
- n divisible by 2 → n is even
- n+1 divisible by 3 → n ≡ 2 (mod 3)
- n+2 divisible by 4 → n ≡ 2 (mod 4)
Solution: From n ≡ 2 (mod 3) and n ≡ 2 (mod 4) LCM(3, 4) = 12 ∴ n ≡ 2 (mod 12) ∴ n = 12k + 2 for k = 0, 1, 2, ...
Answer: The smallest such triple is (2, 3, 4).
These triples occur every 12 numbers: (2, 3, 4), (14, 15, 16), (26, 27, 28), (38, 39, 40), ...
The pattern is (12k + 2, 12k + 3, 12k + 4) for k = 0, 1, 2, 3, ...
Explanation
The chapter uses algebraic reasoning. Let the three consecutive numbers be n, n+1, n+2. We need: n divisible by 2 (n even), n+1 divisible by 3 (n ≡ 2 mod 3), n+2 divisible by 4 (n ≡ 2 mod 4). From n ≡ 2 (mod 3) and n ≡ 2 (mod 4), we get n ≡ 2 (mod 12) by the Chinese Remainder Theorem. So n = 12k + 2 for k = 0, 1, 2, ...
Solution Steps
Let numbers be n, n+1, n+2
n divisible by 2 → n even
n+1 divisible by 3 → n ≡ 2 (mod 3)
n+2 divisible by 4 → n ≡ 2 (mod 4)
LCM(3, 4) = 12, so n ≡ 2 (mod 12)
n = 12k + 2 for k = 0, 1, 2, ...
Triples: (2, 3, 4), (14, 15, 16), (26, 27, 28), ...