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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:
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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/
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/