1. Using prime factorizations: • (18, 35): Factors are 2,3 and 5,7; GCD = 1, co-prime. • (15, 37): Factors are 3,5 and 37; GCD = 1, co-prime. • (30, 415): Factors are 2,3,5 and 5, 83; GCD = 5, not co-prime. • (17, 69): Factors are 17 and 3, 23; GCD = 1, co-prime. • (81, 18): Factors are 3⁴ and 2,3²; GCRead more

    Using prime factorizations:
    • (18, 35): Factors are 2,3 and 5,7; GCD = 1, co-prime.
    • (15, 37): Factors are 3,5 and 37; GCD = 1, co-prime.
    • (30, 415): Factors are 2,3,5 and 5, 83; GCD = 5, not co-prime.
    • (17, 69): Factors are 17 and 3, 23; GCD = 1, co-prime.
    • (81, 18): Factors are 3⁴ and 2,3²; GCD = 9, not co-prime.
    Not co-prime numbers are two numbers that share common factors other than 1. Their greatest common divisor (GCD) is greater than 1. Examples include (8, 12) and (14, 28), sharing factors like 2 or 7.

    For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
    https://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/

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  2. To find the smallest number that is a multiple of all numbers from 1 to 10 except 7, calculate their least common multiple (LCM). The LCM is 2520, derived from the primes 2³, 3², 5, 7. Exclude 7 by dividing: 2520 ÷ 7 = 360. The resulting number, 360, contains all factors of the remaining numbers 1 tRead more

    To find the smallest number that is a multiple of all numbers from 1 to 10 except 7, calculate their least common multiple (LCM). The LCM is 2520, derived from the primes 2³, 3², 5, 7. Exclude 7 by dividing: 2520 ÷ 7 = 360. The resulting number, 360, contains all factors of the remaining numbers 1 to 10, making it the smallest valid multiple without 7.

    For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
    https://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/

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  3. To find the smallest number that is a multiple of all numbers from 1 to 10, compute the least common multiple (LCM) using their prime factorizations. The primes involved are 2³ (for 8), 3² (for 9), 5, and 7. Multiplying these gives 2³ × 3² × 5 × 7 = 2520. This result ensures divisibility by each numRead more

    To find the smallest number that is a multiple of all numbers from 1 to 10, compute the least common multiple (LCM) using their prime factorizations. The primes involved are 2³ (for 8), 3² (for 9), 5, and 7. Multiplying these gives 2³ × 3² × 5 × 7 = 2520. This result ensures divisibility by each number in the range, confirming it as the smallest valid number.

    For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
    https://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/

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  4. Leap years are determined by divisibility rules: years divisible by 4 are leap years unless divisible by 100 but not 400. If someone was born in 2000, leap years since then are 2004, 2008, 2012, 2016, 2020, and 2024. Each year in this list satisfies the leap year criteria, making February 29 a validRead more

    Leap years are determined by divisibility rules: years divisible by 4 are leap years unless divisible by 100 but not 400. If someone was born in 2000, leap years since then are 2004, 2008, 2012, 2016, 2020, and 2024. Each year in this list satisfies the leap year criteria, making February 29 a valid date in those years. Such intervals occur every four years, excluding exceptions at centuries not divisible by 400.

    For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
    https://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/

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  5. Leap years from 2024 to 2099 occur every four years, starting from 2024 and ending at 2096. To find the count: Sequence: 2024, 2028, 2032, ..., 2096. Use the formula n = (2096-2024)/4+1, yielding 19 leap years. These satisfy the rule of being divisible by 4 and not excluded by the century rule (as 2Read more

    Leap years from 2024 to 2099 occur every four years, starting from 2024 and ending at 2096. To find the count:
    Sequence: 2024, 2028, 2032, …, 2096.
    Use the formula n = (2096-2024)/4+1, yielding 19 leap years.
    These satisfy the rule of being divisible by 4 and not excluded by the century rule (as 2099 is not divisible by 4), confirming the count.

    For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
    https://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/

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