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: Find n such that: n ≡ 2 (mod 3) n ≡ 3 (mod 4) n ≡ 4 (mod 5)
Step 1: Rewrite conditions: n + 1 ≡ 0 (mod 3) n + 1 ≡ 0 (mod 4) n + 1 ≡ 0 (mod 5)
Step 2: n + 1 must be a common multiple of 3, 4, and 5.
Step 3: LCM of 3, 4, and 5 = 60
Step 4: n + 1 = 60 n = 60 - 1 = 59
Verification: 59 ÷ 3 = 19 R2 59 ÷ 4 = 14 R3 59 ÷ 5 = 11 R4
Answer: The smallest such number is 59.
Why it is the smallest: The conditions require n to be 1 less than a common multiple of 3, 4, and 5. Since 60 is the LCM (smallest common multiple), n = 60 - 1 = 59 is the smallest positive solution.
Explanation
The chapter discusses finding numbers with specific remainder conditions. The conditions n ≡ 2 (mod 3), n ≡ 3 (mod 4), n ≡ 4 (mod 5) can be rewritten as n + 1 ≡ 0 (mod 3), n + 1 ≡ 0 (mod 4), n + 1 ≡ 0 (mod 5). This means n + 1 must be a common multiple of 3, 4, and 5. The LCM of 3, 4, and 5 is 60. Therefore, n + 1 = 60, giving n = 59. This is the smallest positive solution because 60 is the smallest common multiple. Verification: 59 ÷ 3 = 19 R2, 59 ÷ 4 = 14 R3, 59 ÷ 5 = 11 R4.
Solution Steps
Step 1: Conditions: n ≡ 2 (mod 3), n ≡ 3 (mod 4), n ≡ 4 (mod 5)
Step 2: Rewrite as: n + 1 ≡ 0 (mod 3), n + 1 ≡ 0 (mod 4), n + 1 ≡ 0 (mod 5)
Step 3: Find LCM of 3, 4, 5 = 60
Step 4: Therefore n + 1 = 60, so n = 59
Step 5: Verify: 59 ÷ 3 = 19 R2, 59 ÷ 4 = 14 R3, 59 ÷ 5 = 11 R4