Entrance Exams
Govt. Exams
We find the pattern of powers of 2 mod 7: 2^1≡2, 2^2≡4, 2^3≡1 (mod 7). The cycle repeats every 3 terms. Since 100 = 33×3 + 1, we have 2^100 ≡ 2^1 ≡ 2 (mod 7).
Divisors of 20: 1, 2, 4, 5, 10, 20. Proper divisors (excluding 20): 1, 2, 4, 5, 10. Sum = 1 + 2 + 4 + 5 + 10 = 22, not 48.
We need LCM(6, 8, 9). 6 = 2 × 3, 8 = 2³, 9 = 3². LCM = 2³ × 3² = 8 × 9 = 72. Smallest multiple of 72 greater than 100: 72 × 2 = 144.
For n ≡ 3 (mod 7): possible numbers are 3, 10, 17, 24, 31, 38, 45, 52, 59... For n ≡ 5 (mod 11): possible numbers are 5, 16, 27, 38, 49, 60... Common number is 58. Check: 58 = 7(8) + 2... Let me recheck: 58/7 = 8 rem 2, not 3. Try 38: 38/7 = 5 rem 3 ✓, 38/11 = 3 rem 5 ✓. Answer is A=38.
20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.
Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.
We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = 120/4 = 30.
144 = 2⁴ × 3². Odd divisors come only from 3² = (2+1) = 3 odd divisors: 1, 3, 9.
Let the integers be x, x+1, x+2. Then x + (x+1) + (x+2) = 45. So 3x + 3 = 45, thus 3x = 42, x = 14.
We need a number of the form 8k + 5. Check: 29 = 8(3) + 5 = 24 + 5. ✓