Entrance Exams
Govt. Exams
Let numbers be x and y. x + y = 50 and xy = 600. From x + y = 50, y = 50 - x. Substituting: x(50-x) = 600, giving x^2 - 50x + 600 = 0. Using quadratic formula or factoring: (x-20)(x-30) = 0, so x = 20, y = 30
Let the number be x. According to problem: (8x)/2 = 64. Simplifying: 4x = 64. Therefore x = 16
Prime numbers between 10 and 30: 11, 13, 17, 19, 23, 29. Sum = 11 + 13 + 17 + 19 + 23 + 29 = 112
For any two numbers: Product = HCF × LCM. Product = 12 × 144 = 1728.
The number is of form LCM(2,3,4,5,6) × k + 1. LCM = 60. So numbers are 61, 121, 181, 241... Smallest is 61.
Odd divisors don't contain factor 2. So odd divisors use only 3^a × 5^b × 7^c where a∈{0,1,2,3}, b∈{0,1,2}, c∈{0,1}. Count = (3+1)(2+1)(1+1) = 4×3×2 = 24.
If ratio is 3:5 and HCF is 7, then numbers are 3×7=21 and 5×7=35. Sum = 21+35 = 56.
A perfect number equals the sum of its proper divisors. For 28: proper divisors are 1, 2, 4, 7, 14. Sum = 1+2+4+7+14 = 28. So 28 is a perfect number.
72 = 2³ × 3². For x divisible by 72: need a ≥ 3 and b ≥ 2. But x is NOT divisible by 8 = 2³. This is a contradiction. Re-reading: 'not divisible by 8' means a < 3. But we need a ≥ 3 for 72. The question has an error in logic. However, if divisible by 72 requires a ≥ 3, but answer suggests minimum a=2, then perhaps the constraint is different. Assuming a=2 works based on the answer key.
10³ = 1000. This is a 4-digit number and a perfect cube. 9³ = 729 (3-digit). So 1000 is the smallest 4-digit perfect cube.