Entrance Exams
Govt. Exams
For any two numbers a and b: a × b = HCF(a,b) × LCM(a,b). So 2160 = 12 × LCM. LCM = 2160/12 = 180.
3¹ ≡ 3, 3² ≡ 2, 3³ ≡ 6, 3⁴ ≡ 4, 3⁵ ≡ 5, 3⁶ ≡ 1 (mod 7). Pattern repeats every 6. 100 = 16(6) + 4, so 3^100 ≡ 3⁴ ≡ 4 (mod 7). Wait, let me recalculate: 3⁶ ≡ 1 (mod 7), 100 ÷ 6 = 16 remainder 4. So 3^100 ≡ 3⁴ (mod 7). 3⁴ = 81 = 11(7) + 4, so remainder is 4. Correction: Answer should be C, but this seems wrong. Let me verify: 3¹=3, 3²=9≡2, 3³=27≡6, 3⁴=81≡4, 3⁵≡12≡5, 3⁶≡15≡1 (mod 7). So cycle = 6. 100 = 16×6 + 4, so 3^100 ≡ 3⁴ ≡ 4 (mod 7). However, given answer is A(1), let me recalculate the order. Actually, this needs verification.
Let numbers be 2k, 3k, 4k. LCM(2k, 3k, 4k) = 12k = 120, so k = 10. Numbers are 20, 30, 40. Largest = 40. (Note: Check - LCM = 120 means we need 12k = 120, k = 10, numbers 20, 30, 40. But ratio check: 20:30:40 = 2:3:4 ✓). Wait, recalculating: if ratio is 2:3:4 and k=10, numbers are 20, 30, 40. LCM(20,30,40) = 120 ✓. Largest = 40.
144 = 2⁴ × 3², 108 = 2² × 3³. HCF = 2² × 3² = 4 × 9 = 36.
Let number = 10a + b. (10a + b) - (10b + a) = 45. So 9a - 9b = 45, a - b = 5. Also a + b = 9. Solving: a = 7, b = 2. Number = 72.
Let number = 5a + 2 = 7b + 3. From 5a + 2 = 7b + 3, we get 5a = 7b + 1. Testing b = 2: 7(2) + 1 = 15, a = 3. Number = 5(3) + 2 = 17. Check: 17 ÷ 5 = remainder 2, 17 ÷ 7 = remainder 3. ✓
Let consecutive odd numbers be (2n-1) and (2n+1). (2n-1)(2n+1) = 323. 4n² - 1 = 323, so 4n² = 324, n² = 81, n = 9. Numbers are 17 and 19.
If number is x, then (x × 3/4) ÷ (2/3) = (3x/4) × (3/2) = 9x/8. The factor is 9/8.
24 = 2³ × 3, 36 = 2² × 3². LCM = 2³ × 3² = 8 × 9 = 72.
91 = 7 × 13, so it's composite. 97, 89, and 83 are all prime numbers.