We know that GCD × LCM = Product of the two numbers. So 4 × LCM = 120. Therefore, LCM = 120/4 = 30.

Number of divisors = (3+1)(2+1)(1+1) = 4 × 3 × 2 = 24.

20 = 2² × 5. Divisors are: 1, 2, 4, 5, 10, 20. Sum = 1 + 2 + 4 + 5 + 10 + 20 = 42.

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.

To find the remainder when \(2^{100}\) is divided by 7, we use Fermat's Little Theorem, which states that if \(p\) is prime and \(\gcd(a,p) = 1\), then \(a^{p-1} \equiv 1 \pmod{p}\).

Step 1: Apply Fermat's Little Theorem

Since 7 is prime and \(\gcd(2,7) = 1\):

Step 2: Express the exponent in terms of 6

Divide 100 by 6:

Therefore:

Step 3: Simplify using the congruence

Step 4: Calculate \(2^4 \bmod 7\)

Therefore:

Answer: The remainder is \(2\) (Option B)

# Unit Digit of 7^2019

To find the unit digit of any power, we identify the cyclical pattern of unit digits for that base number.

Step 1: Find the Cyclical Pattern of Unit Digits for Powers of 7

The unit digit of powers of 7 repeats in a cycle. Let's calculate the first few powers:

The pattern repeats: 7, 9, 3, 1, 7, 9, 3, 1, ...

The cycle length is 4.

Step 2: Find the Position of 7^2019 in the Cycle

Divide the exponent 2019 by the cycle length 4 to find which position in the pattern it corresponds to:

Since the remainder is 3, we need the 3rd unit digit in our cycle pattern (7, 9, 3, 1).

The 3rd position corresponds to unit digit 3.

The unit digit of 7^2019 is 3.

Answer: (A) 3

Using the property: HCF × LCM = Product of two numbers. 6 × 60 = 12 × x. 360 = 12x. x = 30.

Numbers divisible by 3: floor(100/3) = 33. Numbers divisible by both 3 and 5 (i.e., by 15): floor(100/15) = 6. Numbers divisible by 3 but not by 5 = 33 - 6 = 27. Wait, that's option B. Let me verify: 27 is correct.

To find \(3^5 \bmod 11\), we calculate successive powers of 3 and reduce modulo 11 at each step to keep numbers manageable.

Step 1: Calculate \(3^2 \bmod 11\)

Since \(9 < 11\), we have \(3^2 \equiv 9 \pmod{11}\)

Step 2: Calculate \(3^4 \bmod 11\)

Dividing: \(81 = 7 \times 11 + 4\), so \(3^4 \equiv 4 \pmod{11}\)

Step 3: Calculate \(3^5 \bmod 11\)

Step 4: Final reduction

Answer: \(3^5 \equiv 1 \pmod{11}\) (Option A)

HCF takes minimum power of each prime: HCF = 2^min(4,3) × 3^min(3,2) × 5^min(1,2) = 2^3 × 3^2 × 5.