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Use prime factorization to check if 40 and 231 are co-prime.
To verify co-primality, find the prime factorization: • 40 = 2³ × 5, with prime factors 2 and 5. • 231 = 3 × 7 × 11, with prime factors 3, 7, and 11. Since there are no overlapping prime factors between 40 and 231, their greatest common divisor (GCD) is 1. Therefore, they are co-prime, satisfying thRead more
To verify co-primality, find the prime factorization:
• 40 = 2³ × 5, with prime factors 2 and 5.
• 231 = 3 × 7 × 11, with prime factors 3, 7, and 11.
Since there are no overlapping prime factors between 40 and 231, their greatest common divisor (GCD) is 1. Therefore, they are co-prime, satisfying the condition of having no shared factors except for 1.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Check if 168 is divisible by 12 using prime factorization.
Prime factorization is the process of expressing a number as a product of prime numbers. For example, 24 = 2³ × 3. It breaks down the number into its smallest divisible prime components. To check divisibility, find the prime factorization: • 168 = 2³ × 3 × 712 • 12 = 2² × 3 All prime factors of 12 aRead more
Prime factorization is the process of expressing a number as a product of prime numbers. For example, 24 = 2³ × 3. It breaks down the number into its smallest divisible prime components.
To check divisibility, find the prime factorization:
• 168 = 2³ × 3 × 712
• 12 = 2² × 3
All prime factors of 12 are included in 168’s factorization, with 2² and 3 present. Dividing confirms this: 168 ÷ 12 = 14, with no remainder. Since 12’s prime factors are fully represented in 168’s decomposition, it is divisible by 12.
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/
Check if 75 is divisible by 21 using prime factorization.
To determine divisibility, use prime factorizations: • 75 = 3 × 5² • 21 = 3 × 7 While both share the factor 3, 75 lacks the factor 7, which is required for divisibility. As a result, the prime factorization of 21 is not fully included in that of 75. Dividing confirms this: 75 ÷ 21 leaves a remainderRead more
To determine divisibility, use prime factorizations:
• 75 = 3 × 5²
• 21 = 3 × 7
While both share the factor 3, 75 lacks the factor 7, which is required for divisibility. As a result, the prime factorization of 21 is not fully included in that of 75. Dividing confirms this: 75 ÷ 21 leaves a remainder, proving that 75 is not divisible by 21.
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/
The teacher asked if 14560 is divisible by all of 2, 4, 5, 8, and 10. Guna checked for divisibility by only two of these numbers and declared it was divisible by all. What could those two numbers be?
Guna tested divisibility for 8 and 5 to conclude that 14560 is divisible by 2, 4, 5, 8, and 10. Divisibility by 8 ensures the number is divisible by 2 and 4, as 8 encompasses their factors. Similarly, divisibility by 5 guarantees divisibility by 10, since 10 = 2 × 5. These two tests are sufficient tRead more
Guna tested divisibility for 8 and 5 to conclude that 14560 is divisible by 2, 4, 5, 8, and 10. Divisibility by 8 ensures the number is divisible by 2 and 4, as 8 encompasses their factors. Similarly, divisibility by 5 guarantees divisibility by 10, since 10 = 2 × 5. These two tests are sufficient to confirm that the number is divisible by all five conditions without additional checks.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Which of the following numbers are divisible by all of 2, 4, 5, 8, and 10: 572, 2352, 5600, 6000, 77622160?
To be divisible by 2, 4, 5, 8, and 10, a number must be a multiple of their least common multiple (LCM), which is 40. Checking each: • 572: Not divisible by 8 or 40. • 2352: Not divisible by 5. • 5600: Divisible by 40. • 6000: Divisible by 40. • 77622160: Divisible by 40. Thus, 5600, 6000, and 77622Read more
To be divisible by 2, 4, 5, 8, and 10, a number must be a multiple of their least common multiple (LCM), which is 40. Checking each:
• 572: Not divisible by 8 or 40.
• 2352: Not divisible by 5.
• 5600: Divisible by 40.
• 6000: Divisible by 40.
• 77622160: Divisible by 40.
Thus, 5600, 6000, and 77622160 satisfy all conditions for divisibility by the given numbers.
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/