Govt. Exams
Entrance Exams
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).
This question asks us to find the average value of the numbers 1 through 15.
The first 15 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
Use the formula for sum of first n natural numbers: \[\text{Sum} = \frac{n(n+1)}{2}\]
Average is the sum divided by the count of numbers.
The average of the first 15 natural numbers is 8.
