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Explore and find out if each statement is always true, sometimes true, or never true: a) The sum of two even numbers gives a multiple of 4.
The sum of two even numbers is sometimes a multiple of 4. While even numbers are divisible by 2, their sums may or may not be divisible by 4. For example, 2 + 2 = 4, which is divisible by 4, but 2 + 6 = 8, which is not. Divisibility by 4 depends on whether both numbers are divisible by 4 or their coRead more
The sum of two even numbers is sometimes a multiple of 4. While even numbers are divisible by 2, their sums may or may not be divisible by 4. For example, 2 + 2 = 4, which is divisible by 4, but 2 + 6 = 8, which is not. Divisibility by 4 depends on whether both numbers are divisible by 4 or their combined properties align to meet this condition.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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The sum of two odd numbers gives a multiple of 4.
The sum of two odd numbers is always a multiple of 4. Odd numbers can be expressed as 2n + 1, where n is an integer. Adding two odd numbers, (2n + 1) + (2m + 1) = 2(n + m + 1), yields an even number divisible by 2. Additionally, the remainders modulo 4 sum to 4, confirming divisibility by 4. For exaRead more
The sum of two odd numbers is always a multiple of 4. Odd numbers can be expressed as 2n + 1, where n is an integer. Adding two odd numbers, (2n + 1) + (2m + 1) = 2(n + m + 1), yields an even number divisible by 2. Additionally, the remainders modulo 4 sum to 4, confirming divisibility by 4. For example, 3 + 5 = 8, 7 + 9 = 16, both multiples of 4.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Four different prime numbers?
To determine the smallest number with four distinct prime factors, use the smallest primes: 2,3,5, and 7. Their product, 2 × 3 × 5 × 7 = 210, is the minimal result. This ensures that all four factors are primes, and no other smaller combination meets the criteria. Including additional primes or largRead more
To determine the smallest number with four distinct prime factors, use the smallest primes: 2,3,5, and 7. Their product, 2 × 3 × 5 × 7 = 210, is the minimal result. This ensures that all four factors are primes, and no other smaller combination meets the criteria. Including additional primes or larger primes would result in a larger product, confirming 210 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/
Use prime factorization to check if 40 and 231 are co-prime.
To verify co-primality, find the prime factorization: • 40 = 2³ × 5, with prime factors 2 and 5. • 231 = 3 × 7 × 11, with prime factors 3, 7, and 11. Since there are no overlapping prime factors between 40 and 231, their greatest common divisor (GCD) is 1. Therefore, they are co-prime, satisfying thRead more
To verify co-primality, find the prime factorization:
• 40 = 2³ × 5, with prime factors 2 and 5.
• 231 = 3 × 7 × 11, with prime factors 3, 7, and 11.
Since there are no overlapping prime factors between 40 and 231, their greatest common divisor (GCD) is 1. Therefore, they are co-prime, satisfying the condition of having no shared factors except for 1.
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/
Check if 168 is divisible by 12 using prime factorization.
Prime factorization is the process of expressing a number as a product of prime numbers. For example, 24 = 2³ × 3. It breaks down the number into its smallest divisible prime components. To check divisibility, find the prime factorization: • 168 = 2³ × 3 × 712 • 12 = 2² × 3 All prime factors of 12 aRead more
Prime factorization is the process of expressing a number as a product of prime numbers. For example, 24 = 2³ × 3. It breaks down the number into its smallest divisible prime components.
To check divisibility, find the prime factorization:
• 168 = 2³ × 3 × 712
• 12 = 2² × 3
All prime factors of 12 are included in 168’s factorization, with 2² and 3 present. Dividing confirms this: 168 ÷ 12 = 14, with no remainder. Since 12’s prime factors are fully represented in 168’s decomposition, it is divisible by 12.
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/