Govt Exams
Using the formula: HCF × LCM = Product of two numbers.
Therefore, 12 × LCM = 2160, so LCM = 2160 ÷ 12 = 180.
Prime factorization of 360: 360 = 2³ × 3² × 5.
The unique prime factors are 2, 3, and 5.
Product = 2 × 3 × 5 = 30.
Let the number be n.
Given: n = 8k + 5 for some integer k.
When divided by 4: n = 8k + 5 = 4(2k) + 4 + 1 = 4(2k + 1) + 1.
Therefore, remainder = 1.
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.
36 = 2² × 3².
Number of factors = (2+1)(2+1) = 3 × 3 = 9.
The factors are: 1, 2, 3, 4, 6, 9, 12, 18, 36.
51 = 3 × 17 (not prime), 53 is prime (only divisible by 1 and 53), 55 = 5 × 11 (not prime), 57 = 3 × 19 (not prime).
Therefore, 53 is the smallest prime greater than 50.
Using dividend = divisor × quotient + remainder: Number = 11 × 9 + 5 = 99 + 5 = 104
This question asks us to find the remainder when 527 is divided by 15 using the division algorithm.
We need to divide 527 by 15 and find what's left over.
Determine how many times 15 goes into 527 completely.
Subtract the product from the original number to find the remainder.
The remainder when 527 is divided by 15 is 2, so the answer is (A).