Govt. Exams
Entrance Exams
Since 9 and 11 are coprime (HCF = 1), the number must be divisible by 9 × 11 = 99.
We need LCM(12, 15, 20). 12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.
To find the remainder when \(2^{50}\) is divided by 5, we use the Fermat's Little Theorem approach by finding the pattern of powers of 2 modulo 5.
Step 1: Find the pattern of powers of 2 (mod 5)
Calculate successive powers of 2 and their remainders when divided by 5:
The pattern repeats every 4 powers.
Step 2: Apply the cycle to find \(2^{50} \pmod{5}\)
Since the remainder repeats with period 4, divide the exponent by 4:
This means \(2^{50}\) has the same remainder as \(2^2\).
Step 3: Calculate the remainder
Answer: The remainder is \(4\) (Option D)
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.
Sum of squares = 1² + 2² + 3² + 4² + 5² = 1 + 4 + 9 + 16 + 25 = 55.
Formula: n(n+1)(2n+1)/6 = 5×6×11/6 = 55
