Entrance Exams
Govt. Exams
Prime factorization: 12 = 2² × 3, 18 = 2 × 3². LCM = 2² × 3² = 4 × 9 = 36 (taking highest powers of all prime factors).
Using prime factorization: 48 = 2⁴ × 3, 64 = 2⁶. HCF = 2⁴ = 16 (taking lowest powers of common prime factors).
For n = p₁^a × p₂^b × p₃^c, number of divisors = (a+1)(b+1)(c+1).
Here: (3+1)(2+1)(1+1) = 4×3×2 = 24
Divisors of 28: 1, 2, 4, 7, 14, 28.
Sum = 1+2+4+7+14+28 = 56. (Note: 28 is a perfect number where sum of proper divisors = 28)
A number has exactly 3 factors only when it is the square of a prime.
For p² where p is prime, factors are: 1, p, p².
Example: 4 has factors 1,2,4 (3 factors). 9 has factors 1,3,9 (3 factors).
Let smaller odd number = x.
Then x+(x+2)=56.
So 2x+2=56, 2x=54, x=27.
Check: 27+29=56 ✓
Numbers of form 7k+3: when k=5, number=7(5)+3=38.
Check: 38÷7=5 remainder 3 ✓.
Check others: 24÷7=3 rem 3 (close but let's verify 38 first), 38÷7 gives remainder 3 ✓
This question asks us to find the smallest positive number that is divisible by both 12 and 18.
Break down each number into its prime factors.
The LCM uses the highest power of each prime factor that appears.
Multiply the highest powers together.
The LCM of 12 and 18 is 36, which is the smallest number divisible by both numbers.
If a number is divisible by both 3 and 5, and 3 and 5 are coprime (HCF=1), then the number must be divisible by their product: 3×5 = 15
Factors of 48: 1,2,3,4,6,8,12,16,24,48.
Factors of 64: 1,2,4,8,16,32,64.
Common factors: 1,2,4,8,16. HCF = 16