Govt Exams
We need to find x where x² = 169. Taking square root: x = √169 = 13.
12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.
Let the number be x. According to the problem: 5x - 3 = 47. Therefore, 5x = 50, so x = 10.
For divisibility by 6, number must be divisible by both 2 and 3. 104÷6 = 17.33... (not divisible). Others: 72, 84, 90 are all divisible by 6
12 = 2²×3, 18 = 2×3². LCM = 2²×3² = 4×9 = 36
Sum of first n natural numbers = n(n+1)/2. For n=15: 15×16/2 = 240/2 = 120
6^x ÷ 216 = 6. 216 = 6³. So 6^x ÷ 6³ = 6. Therefore 6^(x-3) = 6¹. Thus x - 3 = 1, so x = 4.
Digit sum = 9 + 9 + 9 + 9 = 36. Note: A number and its digit sum have the same remainder when divided by 9.
72 = 2³ × 3². Number of divisors = (3+1)(2+1) = 4 × 3 = 12.
61 is only divisible by 1 and itself. 51 = 3 × 17, 57 = 3 × 19, 63 = 9 × 7 are composite numbers.