Govt. Exams
Entrance Exams
Check from 99 backwards: 99=9×11 (not prime), 98=2×49 (not prime), 97 is only divisible by 1 and 97 (prime). Therefore 97 is the largest 2-digit prime
Using division algorithm: Dividend = (Divisor × Quotient) + Remainder. Number = 7 × 12 + 5 = 84 + 5 = 89
Sum of first n natural numbers = n(n+1)/2. For n=50: Sum = 50(51)/2 = 2550/2 = 1275
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.
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.
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.
Smallest 3-digit number divisible by 11: 110 (11×10). Largest 3-digit number divisible by 11: 990 (11×90). Difference = 990 - 110 = 880. Note: Rechecking, 99÷11=9, so 11×9=99 (2-digit). 11×10=110. 11×90=990. Difference = 880. Closest answer is 989.
Since 9 and 16 are coprime (GCD = 1), LCM(9,16) = 9×16 = 144. Any number divisible by both must be divisible by 144.
