Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Given:
- n leaves remainder 2 when divided by 3
- n leaves remainder 3 when divided by 4
- n leaves remainder 4 when divided by 5
Key Insight: n ≡ 2 (mod 3) → n + 1 ≡ 0 (mod 3) n ≡ 3 (mod 4) → n + 1 ≡ 0 (mod 4) n ≡ 4 (mod 5) → n + 1 ≡ 0 (mod 5)
∴ n + 1 is divisible by 3, 4, and 5.
Calculation: LCM(3, 4, 5) = 60 n + 1 = 60 n = 59
Answer: The smallest such number is 59.
Why it is the smallest: The number n must be one less than a common multiple of 3, 4, and 5. The LCM of 3, 4, and 5 is 60, so the smallest positive number satisfying all conditions is 60 - 1 = 59.
Explanation
The chapter discusses finding numbers with specific remainder conditions. If n leaves remainder 2 when divided by 3, then n + 1 is divisible by 3. Similarly, n + 1 is divisible by 4 and 5. So n + 1 must be a common multiple of 3, 4, and 5. The LCM is 60, so n + 1 = 60, giving n = 59.
Solution Steps
n ≡ 2 (mod 3) → n + 1 ≡ 0 (mod 3)
n ≡ 3 (mod 4) → n + 1 ≡ 0 (mod 4)
n ≡ 4 (mod 5) → n + 1 ≡ 0 (mod 5)
n + 1 must be divisible by LCM(3, 4, 5) = 60
Smallest positive: n + 1 = 60, so n = 59