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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/
Find the smallest number that is divisible by all numbers from 1 to 12.
To find the smallest number divisible by all numbers from 1 to 12, calculate their least common multiple (LCM): • Prime factorizations: 2³ (8), 3² (9), 5, 7, 11 • Multiply: 2³ × 3² × 5 × 7 × 11 = 27720. This result ensures divisibility by each number, as it incorporates the highest powers of all priRead more
To find the smallest number divisible by all numbers from 1 to 12, calculate their least common multiple (LCM):
• Prime factorizations: 2³ (8), 3² (9), 5, 7, 11
• Multiply: 2³ × 3² × 5 × 7 × 11 = 27720.
This result ensures divisibility by each number, as it incorporates the highest powers of all primes within the range, making 27720 the minimal valid solution.
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 all multiples of 3 and 7 that lie between 100 and 200.
To find multiples of 3, divide the range 100–200 by 3: • Start at 102, end at 198, step 3 (33 multiples). For multiples of 7, divide by 7: • Start at 105, end at 196, step 7 (14 multiples). Common multiples are those divisible by both 3 and 7, i.e., their LCM = 21. From 100–200, these are 105, 126,Read more
To find multiples of 3, divide the range 100–200 by 3:
• Start at 102, end at 198, step 3 (33 multiples).
For multiples of 7, divide by 7:
• Start at 105, end at 196, step 7 (14 multiples).
Common multiples are those divisible by both 3 and 7, i.e., their LCM = 21. From 100–200, these are 105, 126, 147, 168, and 189, forming the overlap of both sequences.
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/