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.
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.
For any two numbers: Product = HCF × LCM. Product = 12 × 144 = 1728.
Prime numbers between 10 and 30: 11, 13, 17, 19, 23, 29. Sum = 11 + 13 + 17 + 19 + 23 + 29 = 112
Let the number be x. According to problem: (8x)/2 = 64. Simplifying: 4x = 64. Therefore x = 16
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
A number that is both a perfect square and perfect cube must be a perfect sixth power. Checking options: 64 = 8^2 = 4^3, and 64 = 2^6. It satisfies both conditions
Total numbers formed = 3! = 6. Each digit appears in each position (units, tens, hundreds) exactly 2 times. Sum = 2(2+3+5)(100+10+1) = 2(10)(111) = 2220. This is incorrect. Correct: Each digit appears in each position 2 times. Sum = (2+3+5) × 2 × (1+10+100) = 10 × 2 × 111 = 2220. Actually for 6 numbers: sum = (100+10+1) × 2 × (2+3+5) = 111 × 2 × 10 = 2220. Recalculating: Each of 6 permutations. Each digit 2,3,5 appears in hundreds place twice: 2(200+300+500) = 2(1000) = 2000. Each in tens place twice: 2(20+30+50) = 2(100) = 200. Each in units place twice: 2(2+3+5) = 2(10) = 20. Total = 2000+200+20 = 2220. Given answer D is 3996, need verification of question intent.
Sum of first n natural numbers = n(n+1)/2. For n=50: Sum = 50(51)/2 = 2550/2 = 1275
Using division algorithm: Dividend = (Divisor × Quotient) + Remainder. Number = 7 × 12 + 5 = 84 + 5 = 89