Entrance Exams
Govt. Exams
Sum of digits = 21 means divisible by 3. But divisibility by 9 requires sum = 18, 27, etc. Without knowing if the number is even or odd, we cannot determine all properties.
Let smaller number be x. Then x(x+12) = 189. x² + 12x - 189 = 0. Factoring: (x+21)(x-9) = 0. Since x must be positive, x = 9.
Let the number be 7k + 3. When divided by 14: if k is even (k=2m), number = 14m + 3 (remainder 3); if k is odd (k=2m+1), number = 14m + 10 (remainder 10). So remainder is either 3 or 10.
We need a number of form LCM(12,15,20) + 7. LCM(12,15,20): 12=2²×3, 15=3×5, 20=2²×5. LCM = 2²×3×5 = 60. So the number = 60k + 7. For k=2: 120+7 = 127.
144 = 2⁴ × 3², 180 = 2² × 3² × 5, 216 = 2³ × 3³. GCD = 2² × 3² = 4 × 9 = 36.
Perfect cubes: 1³=1, 2³=8, 3³=27, 4³=64, 5³=125, 6³=216, 7³=343, 8³=512, 9³=729, 10³=1000. Between 1 and 1000 (inclusive) there are 10 perfect cubes.
Let the number be x. Then x + 1/x = 2.1. Multiply by x: x² + 1 = 2.1x. Rearranging: x² - 2.1x + 1 = 0. Using quadratic formula or testing: If x = 2.5, then 2.5 + 1/2.5 = 2.5 + 0.4 = 2.9 (not correct). Testing x = 2: 2 + 0.5 = 2.5. For x = 1.5: 1.5 + 2/3 ≈ 2.167. Actually solving properly gives 2 or 0.5.
For any two numbers a and b: a × b = HCF(a,b) × LCM(a,b). So 2160 = 12 × LCM. LCM = 2160/12 = 180.
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.