A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:
Question:
A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:
A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is:
Options
Answer: 150
Explanation:
Given: If divided into teams of 9, 10, 12, or 25, exactly 4 are left out: N ≡ 4 (mod 9) N ≡ 4 (mod 10) N ≡ 4 (mod 12) N ≡ 4 (mod 25) If divided into teams of 11, no one is left out: N ≡ 0 (mod 11) We are asked: maximum number of teams of 12 each → floor(N / 12). Step 2: Combine congruences N ≡ 4 (mod 9,10,12,25) → N - 4 is divisible by LCM of 9,10,12,25 Compute LCM(9,10,12,25): 9 = 3² 10 = 2 × 5 12 = 2² × 3 25 = 5² LCM = 2² × 3² × 5² = 4 × 9 × 25 = 900 So: N - 4 ≡ 0 (mod 900) → N = 900k + 4 Step 3: Apply N ≡ 0 (mod 11) 900k + 4 ≡ 0 (mod 11) 900 mod 11 = 900 - 11×81 = 900 - 891 = 9 → 900 ≡ 9 (mod 11) 9k + 4 ≡ 0 (mod 11) → 9k ≡ 7 (mod 11) Inverse of 9 modulo 11 is 5 (since 9 × 5 = 45 ≡ 1 mod 11) Multiply both sides: k ≡ 5 × 7 = 35 ≡ 2 (mod 11) So k = 2 + 11t, t ≥ 0 Step 4: Find N < 5000 N = 900k + 4 = 900(2 + 11t) + 4 = 1804 + 9900 t N < 5000 → 9900 t < 3196 → t = 0 So k = 2 → N = 900×2 + 4 = 1804 Step 5: Maximum number of teams of 12 each Maximum teams = floor(N / 12) = floor(1804 / 12) = 150 Answer: 150
Explanation:
Given: If divided into teams of 9, 10, 12, or 25, exactly 4 are left out: N ≡ 4 (mod 9) N ≡ 4 (mod 10) N ≡ 4 (mod 12) N ≡ 4 (mod 25) If divided into teams of 11, no one is left out: N ≡ 0 (mod 11) We are asked: maximum number of teams of 12 each → floor(N / 12). Step 2: Combine congruences N ≡ 4 (mod 9,10,12,25) → N - 4 is divisible by LCM of 9,10,12,25 Compute LCM(9,10,12,25): 9 = 3² 10 = 2 × 5 12 = 2² × 3 25 = 5² LCM = 2² × 3² × 5² = 4 × 9 × 25 = 900 So: N - 4 ≡ 0 (mod 900) → N = 900k + 4 Step 3: Apply N ≡ 0 (mod 11) 900k + 4 ≡ 0 (mod 11) 900 mod 11 = 900 - 11×81 = 900 - 891 = 9 → 900 ≡ 9 (mod 11) 9k + 4 ≡ 0 (mod 11) → 9k ≡ 7 (mod 11) Inverse of 9 modulo 11 is 5 (since 9 × 5 = 45 ≡ 1 mod 11) Multiply both sides: k ≡ 5 × 7 = 35 ≡ 2 (mod 11) So k = 2 + 11t, t ≥ 0 Step 4: Find N < 5000 N = 900k + 4 = 900(2 + 11t) + 4 = 1804 + 9900 t N < 5000 → 9900 t < 3196 → t = 0 So k = 2 → N = 900×2 + 4 = 1804 Step 5: Maximum number of teams of 12 each Maximum teams = floor(N / 12) = floor(1804 / 12) = 150 Answer: 150
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