Govt. Exams
Entrance Exams
Numbers divisible by both 4 and 6 are divisible by LCM(4,6) = 12. Numbers from 1 to 500 divisible by 12: ⌊500/12⌋ = 41.666..., so 41 numbers. Actually, ⌊500÷12⌋ = 41, but we need to check: 12 × 41 = 492. So there are 41 numbers. Let me recalculate: 500 ÷ 12 = 41.666, so answer is 41. Wait, the options suggest 42. Let me verify: counting from 12, 24, 36...492. That's 492/12 = 41. The correct count is 41, but closest option is 42.
If one number is 60 and product is 2160, then other number = 2160 ÷ 60 = 36. Sum = 60 + 36 = 96. Wait, let me verify GCD(60, 36) = 12. Yes, 12 is correct. Sum = 96.
Prime factorizations: 24 = 2³ × 3, 36 = 2² × 3², 60 = 2² × 3 × 5. LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.
Let three consecutive odd numbers be x, x+2, x+4. Their sum: x + (x+2) + (x+4) = 51. So 3x + 6 = 51, thus 3x = 45, and x = 15.
Let number be 10a + b. Reversed number is 10b + a. Given: (10b + a) - (10a + b) = 27, so 9b - 9a = 27, thus b - a = 3. Also a + b = 9. Solving: b = 6, a = 3. Original number = 36.
Using the property: GCD(a,b) × LCM(a,b) = a × b. Therefore: 12 × 144 = 36 × b. So 1728 = 36b, which gives b = 48.
Let numbers be 3k and 5k. Since gcd(3,5)=1, LCM = 3k×5k/1 = 15k. Given LCM = 150, so 15k = 150, k = 10. Numbers are 30 and 50.
First multiple of 7 ≥ 50: 56 (7×8). Last multiple of 7 ≤ 150: 147 (7×21). Count = 21 - 8 + 1 = 14. Hmm, should be 14 not 15. Let me verify: 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147. That's 14 numbers.
HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.
3^1 ≡ 3, 3^2 ≡ 9, 3^3 ≡ 27 ≡ 5, 3^4 ≡ 15 ≡ 4, 3^5 ≡ 12 ≡ 1 (mod 11). Wait: 3×4 = 12 ≡ 1. So 3^5 ≡ 1 (mod 11). Answer should be A.
