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Which of the following numbers is the product of exactly three distinct prime numbers: 45, 60, 91, 105, 330?
Among the given numbers, 105 and 330 are products of exactly three distinct prime numbers. Their factorizations are: • 105 = 3 × 5 × 7. • 330 = 2 × 3 × 5 × 11, but considering three distinct primes (2, 3, 5), this is valid. Other numbers, like 45 (3² × 5) and 91 (7 × 13), do not meet the condition.Read more
Among the given numbers, 105 and 330 are products of exactly three distinct prime numbers. Their factorizations are:
• 105 = 3 × 5 × 7.
• 330 = 2 × 3 × 5 × 11, but considering three distinct primes (2, 3, 5), this is valid.
Other numbers, like 45 (3² × 5) and 91 (7 × 13), do not meet the condition. The count excludes repetitions and confirms distinct prime contributions.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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How many three-digit prime numbers can you make using each of 2, 4, and 5 exactly once?
Using 2, 4, and 5, the six three-digit permutations are: 245, 254, 425, 452, 524, and 542. Checking their primality: • 245 is divisible by 5. • 254 is divisible by 2. • 425 is divisible by 5. • 452 is divisible by 2. • 524 is divisible by 2. • 542 is divisible by 2. Thus, none of these numbers are pRead more
Using 2, 4, and 5, the six three-digit permutations are: 245, 254, 425, 452, 524, and 542. Checking their primality:
• 245 is divisible by 5.
• 254 is divisible by 2.
• 425 is divisible by 5.
• 452 is divisible by 2.
• 524 is divisible by 2.
• 542 is divisible by 2.
Thus, none of these numbers are prime. Despite using each digit exactly once, all numbers are divisible by either 2 or 5.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Observe that 3 is a prime number, and 2 × 3 + 1 = 7 is also a prime. Are there other primes for which doubling and adding 1 gives another prime?
Several primes satisfy the condition where doubling and adding 1 yields another prime: • For 3, 2 × 3 + 1 = 7 (prime). • For 5, 2 × 5 + 1 = 11 (prime). • For 11, 2 × 11 + 1 = 23 (prime). • For 13, 2 × 13 + 1 = 27 (prime). The sequence demonstrates how doubling primes can yield new primes, though excRead more
Several primes satisfy the condition where doubling and adding 1 yields another prime:
• For 3, 2 × 3 + 1 = 7 (prime).
• For 5, 2 × 5 + 1 = 11 (prime).
• For 11, 2 × 11 + 1 = 23 (prime).
• For 13, 2 × 13 + 1 = 27 (prime).
The sequence demonstrates how doubling primes can yield new primes, though exceptions exist. Testing higher primes verifies these conditions.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Which of the following pairs are co-prime: (30, 45), (57, 85), (121, 1331), (343, 216)?
To determine co-primality, find the greatest common divisor (GCD): • (30, 45): Not co-prime, GCD = 15. • (57, 85): Co-prime, GCD = 1, as they share no factors other than 1. • (121, 1331): Not co-prime, GCD = 11. • (343, 216): Co-prime, GCD = 1, as no common factors exist. Pairs are co-prime only ifRead more
To determine co-primality, find the greatest common divisor (GCD):
• (30, 45): Not co-prime, GCD = 15.
• (57, 85): Co-prime, GCD = 1, as they share no factors other than 1.
• (121, 1331): Not co-prime, GCD = 11.
• (343, 216): Co-prime, GCD = 1, as no common factors exist.
Pairs are co-prime only if their GCD is 1, ensuring no shared prime factors.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Is the first number divisible by the second? Use prime factorization. a) 225 and 27
The prime factorization of 225 is 3² x 5² and that of 27 is 3³. For divisibility, the prime factorization of the divisor must be included in the dividend’s factorization. Here, 27 has 3³, while 225 only has 3², making divisibility impossible. Thus, 225 is not divisible by 27 because the power of 3 iRead more
The prime factorization of 225 is 3² x 5² and that of 27 is 3³. For divisibility, the prime factorization of the divisor must be included in the dividend’s factorization. Here, 27 has 3³, while 225 only has 3², making divisibility impossible. Thus, 225 is not divisible by 27 because the power of 3 in 225’s factorization is insufficient to accommodate the full factorization of 27.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/