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.
Find numbers between 1730 and 1740, and 2030 and 2040, that are divisible by 4.
To find numbers divisible by 4: • 1730 – 1740: Test divisibility: 1732 ÷ 4 = 433, 1736 ÷ 4 = 434, 1740 ÷ 4 = 435. Results are 1732, 1736, and 1740. • 2030 – 2040: Test divisibility: 2032 ÷ 4 = 508, 2036 ÷ 4 = 509, 2040 ÷ 4 = 510. Results are 2032, 2036, and 2040. These numbers meet the condition, beRead more
To find numbers divisible by 4:
See less• 1730 – 1740: Test divisibility: 1732 ÷ 4 = 433, 1736 ÷ 4 = 434, 1740 ÷ 4 = 435. Results are 1732, 1736, and 1740.
• 2030 – 2040: Test divisibility: 2032 ÷ 4 = 508, 2036 ÷ 4 = 509, 2040 ÷ 4 = 510. Results are 2032, 2036, and 2040.
These numbers meet the condition, being multiples of 4 within their respective ranges.
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/
Find the largest and smallest 4-digit numbers that are divisible by 4 and are also palindromes.
The smallest 4-digit palindrome divisible by 4 is 2002, found after testing smaller palindromes for divisibility. Similarly, the largest 4-digit palindrome divisible by 4 is 9999, meeting the symmetry and divisibility conditions. Verifying divisors ensures no false inclusions or exclusions occur, paRead more
The smallest 4-digit palindrome divisible by 4 is 2002, found after testing smaller palindromes for divisibility. Similarly, the largest 4-digit palindrome divisible by 4 is 9999, meeting the symmetry and divisibility conditions. Verifying divisors ensures no false inclusions or exclusions occur, particularly as the symmetry constraints require careful handling of the specific numeric properties concerning modularity and arithmetic parity.
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 of the following pairs of numbers are co-prime? (18, 35), (15, 37), (30, 415), (17, 69), (81, 18).
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
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/