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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/
Find three consecutive composite numbers that are divisible by 5.
Composite numbers divisible by 5 must include factors of 5 and additional divisors, making them non-prime. In the range of consecutive numbers, 90, 95, and 100 fit this criterion: • 90 = 2 × 3 × 3 × 5 • 95 = 5 × 19 • 100 = 2² × 5² These numbers are consecutive, divisible by 5, and composite, confirmRead more
Composite numbers divisible by 5 must include factors of 5 and additional divisors, making them non-prime. In the range of consecutive numbers, 90, 95, and 100 fit this criterion:
• 90 = 2 × 3 × 3 × 5
• 95 = 5 × 19
• 100 = 2² × 5²
These numbers are consecutive, divisible by 5, and composite, confirming they meet the required conditions.
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 smallest number whose prime factorization includes two 3s, one 2, and one 7?
To construct the smallest number with the specified factors 2, 3², and 7, multiply them together: • 2 × 3² × 7 = 2 × 9 × 7 = 126. This number includes exactly one 2, two 3s, and one 7 in its prime factorization. No smaller number contains all these factors simultaneously, confirming 126 as the minimRead more
To construct the smallest number with the specified factors 2, 3², and 7, multiply them together:
• 2 × 3² × 7 = 2 × 9 × 7 = 126.
This number includes exactly one 2, two 3s, and one 7 in its prime factorization. No smaller number contains all these factors simultaneously, confirming 126 as the minimal valid solution for the problem.
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 greatest 3-digit number that is divisible by 6, 9, and 12.
To find the greatest 3-digit number divisible by 6, 9, and 12, calculate their least common multiple (LCM): • 6 = 2 × 3, 9 = 3², 12 = 2² × 3 • LCM = 2² × 3² = 36 Divide the largest 3-digit number, 999, by 36: 999 ÷ 36 = 27.75 Round down to 27, multiply: 27 × 36 = 972. Thus, 972 is the largest 3-digiRead more
To find the greatest 3-digit number divisible by 6, 9, and 12, calculate their least common multiple (LCM):
• 6 = 2 × 3, 9 = 3², 12 = 2² × 3
• LCM = 2² × 3² = 36
Divide the largest 3-digit number, 999, by 36: 999 ÷ 36 = 27.75
Round down to 27, multiply: 27 × 36 = 972.
Thus, 972 is the largest 3-digit number divisible by 36.
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/