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. Prime factorization helps verify divisibility: • 42 = 2 × 3 × 7 • 12 = 2² × 3 While 42 contains 2 and 3, it lacksRead 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.
Prime factorization helps verify divisibility:
• 42 = 2 × 3 × 7
• 12 = 2² × 3
While 42 contains 2 and 3, it lacks 2², which is necessary for divisibility by 12. Dividing 42 ÷ 12 leaves a remainder of 6, confirming that the factors of 12 are not fully represented in 42’s factorization, making it not divisible.
Remainders obtained when each of the following numbers is divided by 10, 5, 2 • 572: Rem. by 10: 2, by 5: 2, by 2: 0 • 980: Rem. by 10: 0, by 5: 0, by 2: 0 • 1111: Rem. by 10: 1, by 5: 1, by 2: 1 • 2345: Rem. by 10: 5, by 5: 0, by 2: 1 Each remainder is calculated by direct division and finding theRead more
Remainders obtained when each of the following numbers is divided by 10, 5, 2
• 572: Rem. by 10: 2, by 5: 2, by 2: 0
• 980: Rem. by 10: 0, by 5: 0, by 2: 0
• 1111: Rem. by 10: 1, by 5: 1, by 2: 1
• 2345: Rem. by 10: 5, by 5: 0, by 2: 1
Each remainder is calculated by direct division and finding the leftover.
To make 8560 divisible by 8, the last two digits must form a number divisible by 8. Replace 60 with 56, creating 8556. Dividing 8556 ÷ 8 = 1069 confirms divisibility without remainder. This adjustment ensures the number meets the criteria, as divisibility by 8 depends on the last two digits formingRead more
To make 8560 divisible by 8, the last two digits must form a number divisible by 8. Replace 60 with 56, creating 8556. Dividing 8556 ÷ 8 = 1069 confirms divisibility without remainder. This adjustment ensures the number meets the criteria, as divisibility by 8 depends on the last two digits forming a valid multiple of 8.
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.
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.
Check if 42 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. Prime factorization helps verify divisibility: • 42 = 2 × 3 × 7 • 12 = 2² × 3 While 42 contains 2 and 3, it lacksRead 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.
Prime factorization helps verify divisibility:
• 42 = 2 × 3 × 7
• 12 = 2² × 3
While 42 contains 2 and 3, it lacks 2², which is necessary for divisibility by 12. Dividing 42 ÷ 12 leaves a remainder of 6, confirming that the factors of 12 are not fully represented in 42’s factorization, making it not divisible.
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 remainders obtained when each of the following numbers is divided by: i) 10, ii) 5, iii) 2. Numbers: 78, 99, 173, 572, 980, 1111, 2345
Remainders obtained when each of the following numbers is divided by 10, 5, 2 • 572: Rem. by 10: 2, by 5: 2, by 2: 0 • 980: Rem. by 10: 0, by 5: 0, by 2: 0 • 1111: Rem. by 10: 1, by 5: 1, by 2: 1 • 2345: Rem. by 10: 5, by 5: 0, by 2: 1 Each remainder is calculated by direct division and finding theRead more
Remainders obtained when each of the following numbers is divided by 10, 5, 2
• 572: Rem. by 10: 2, by 5: 2, by 2: 0
• 980: Rem. by 10: 0, by 5: 0, by 2: 0
• 1111: Rem. by 10: 1, by 5: 1, by 2: 1
• 2345: Rem. by 10: 5, by 5: 0, by 2: 1
Each remainder is calculated by direct division and finding the leftover.
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/
Change the last two digits of 8560 so that the resulting number is a multiple of 8.
To make 8560 divisible by 8, the last two digits must form a number divisible by 8. Replace 60 with 56, creating 8556. Dividing 8556 ÷ 8 = 1069 confirms divisibility without remainder. This adjustment ensures the number meets the criteria, as divisibility by 8 depends on the last two digits formingRead more
To make 8560 divisible by 8, the last two digits must form a number divisible by 8. Replace 60 with 56, creating 8556. Dividing 8556 ÷ 8 = 1069 confirms divisibility without remainder. This adjustment ensures the number meets the criteria, as divisibility by 8 depends on the last two digits forming a valid multiple of 8.
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/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-5/