Entrance Exams
Govt. Exams
Powers of 2 mod 5: 2¹≡2, 2²≡4, 2³≡3, 2⁴≡1, 2⁵≡2...
Pattern repeats every 4. 50 = 4×12 + 2, so 2⁵⁰ ≡ 2² ≡ 4 (mod 5).
Perfect squares between 100 and 200: 11² = 121, 12² = 144, 13² = 169, 14² = 196, 15² = 225 (exceeds 200).
So there are 4 perfect squares.
Wait: 10² = 100 (not between), so squares are 121, 144, 169, 196.
Count = 4.
Actually checking: we need squares from 11² to 14², which is 4 numbers.
Using HCF × LCM = Product of two numbers: 15 × 180 = 45 × x.
Therefore, 2700 = 45x, so x = 60.
Digital root = 9 + 8 + 7 + 6 = 30 → 3 + 0 = 3.
Alternatively, since the sum is divisible by 9, the digital root is 9.
Wait: 9+8+7+6 = 30, 3+0 = 3.
So digital root is 3.
32 ÷ 7 = 4 remainder 4 (No). 33 ÷ 7 = 4 remainder 5 (No). 34 ÷ 7 = 4 remainder 6 (No). 32 = 7×4 + 5 = 28 + 5 (Yes). 32 ÷ 7 gives remainder 5.
20 = 2² × 5.
Sum of divisors = (1 + 2 + 4)(1 + 5) = 7 × 6 = 42.
Divisors are: 1, 2, 4, 5, 10, 20.
LCM(9, 11) = 99 (since 9 and 11 are coprime).
Smallest three-digit multiple of 99 is 99 × 2 = 198.
Odd numbers between 10 and 50: 11, 13, 15, ..., 49.
This is an AP with first term 11, last term 49, and common difference 2.
Number of terms = (49-11)/2 + 1 = 19 + 1 = 20.
∛512 = ∛(8³) = 8, since 8 × 8 × 8 = 512.
12 = 2² × 3, 18 = 2 × 3², 24 = 2³ × 3. LCM = 2³ × 3² = 8 × 9 = 72.