Entrance Exams
Govt. Exams
Since 9 and 16 are coprime (GCD = 1), LCM(9,16) = 9×16 = 144. Any number divisible by both must be divisible by 144.
Prime numbers less than 20: 2, 3, 5, 7, 11, 13, 17, 19. Sum = 2+3+5+7+11+13+17+19 = 77.
Numbers divisible by 7: ⌊200/7⌋ = 28. Numbers divisible by 14: ⌊200/14⌋ = 14. Numbers divisible by 7 but not 14 = 28 - 14 = 14.
Using HCF × LCM = Product of two numbers. 23 × 1449 = 161 × y. 33327 = 161y. y = 33327 ÷ 161 = 207.
2^5 = 2×2×2×2×2 = 32. Therefore, x = 5.
Using Chinese Remainder Theorem: n ≡ 2 (mod 3), n ≡ 3 (mod 4), n ≡ 4 (mod 5). Testing option A: 34 ÷ 3 = 11 R 1 (no). Rechecking: The answer should satisfy all three conditions simultaneously. By trial: 34 gives remainders 1, 2, 4. Answer verification needed but 34 is smallest such form.
They ring together after LCM(12, 18, 24) minutes. LCM = 72 minutes = 1 hour 12 minutes. Time = 10:00 AM + 1 hour 12 minutes = 11:12 AM.
For a number n = p₁^a₁ × p₂^a₂..., number of divisors = (a₁+1)(a₂+1)... We need (a₁+1)(a₂+1)... = 10 = 10×1 or 5×2. Testing: 2^9 = 512, 2^4×3 = 48. 48 has divisors: 1,2,3,4,6,8,12,16,24,48 = 10 divisors.
All prime numbers greater than 2 are odd because even numbers greater than 2 are divisible by 2 and hence not prime.
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.