Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
What is the smallest number that is a multiple of all the numbers from 1 to 10 except 7?
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
Find the smallest number that is a multiple of all numbers from 1 to 10.
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
Which leap years occurred from the year you were born till now?
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
From the year 2024 till 2099, how many leap years are there?
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
What is the prime factorization of 141, 1728, and 1024?
Prime factorizations are: • 141: Divisible by 3, 141 ÷ 3 = 47, so 141 = 3 × 47. • 1728: Divisible by 2, 1728 = 2⁶ × 3³, as confirmed through successive divisions. • 1024: All factors are 2, 1024 = 2¹⁰ Prime factorization breaks down numbers into the smallest building blocks, revealing these decomposRead more
Prime factorizations are:
• 141: Divisible by 3, 141 ÷ 3 = 47, so 141 = 3 × 47.
• 1728: Divisible by 2, 1728 = 2⁶ × 3³, as confirmed through successive divisions.
• 1024: All factors are 2, 1024 = 2¹⁰
Prime factorization breaks down numbers into the smallest building blocks, revealing these decompositions through iterative division.
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/