Entrance Exams
Govt. Exams
Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024
Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45
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
120 = 2³ × 3 × 5. Odd divisors come from 3 × 5 = 15. Divisors of 15: 1, 3, 5, 15. Sum = 1 + 3 + 5 + 15 = 24. Recalculate: sum of odd divisors = (1+3+5+15) = 24. Hmm, check options. Divisors: 1, 3, 5, 15 sum to 24. Not in options. Recheck: 120 = 8×15, odd divisors of 15 are 1,3,5,15. Sum = 24. Let me verify: divisors of form 3^a × 5^b where a∈{0,1}, b∈{0,1}.
6^x ÷ 216 = 6. 216 = 6³. So 6^x ÷ 6³ = 6. Therefore 6^(x-3) = 6¹. Thus x - 3 = 1, so x = 4.
Let number be x. x + 1/x = 2.5. Multiply by x: x² - 2.5x + 1 = 0. x = (2.5 ± √(6.25-4))/2 = (2.5 ± 1.5)/2. x = 2 or 0.5.
By Wilson's theorem, (p-1)! ≡ -1 (mod p) for prime p. So 16! ≡ -1 (mod 17). 16! = 13! × 14 × 15 × 16. -1 ≡ 13! × 14 × 15 × 16 (mod 17).
For divisibility by 11: alternate sum of digits. 12321: (1+3+1) - (2+2) = 5 - 4 = 1. Recheck: (1+3+1) - (2+2) = 5-4=1. Actually 1-2+3-2+1 = 1. Try: 1-2+3-2+1 = 1. Check: 12321/11 = 1120.09... Correct: (2+2) - (1+3+1) = 4-5 = -1, still divisible.