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What is the smallest number whose prime factorization has: a) Three different prime numbers?
To find the smallest number with three distinct prime factors, multiply the smallest primes: 2,3, and 5. Their product is 2 × 3 × 5 = 30, which is the smallest composite number formed by these factors. Including additional primes or larger primes would increase the result, making 30 the minimal soluRead more
To find the smallest number with three distinct prime factors, multiply the smallest primes: 2,3, and 5. Their product is 2 × 3 × 5 = 30, which is the smallest composite number formed by these factors. Including additional primes or larger primes would increase the result, making 30 the minimal solution that satisfies the requirement of having three different prime factors.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Use prime factorization to check if 242 and 195 are co-prime.
Using prime factorization: • 242 = 2 × 11², with prime factors 2 and 11. • 195 = 3 × 5 × 13, with prime factors 3, 5, and 13. Since there are no common factors between the two sets of primes, the greatest common divisor (GCD) is 1. This confirms that 242 and 195 are co-prime, as they share no primeRead more
Using prime factorization:
• 242 = 2 × 11², with prime factors 2 and 11.
• 195 = 3 × 5 × 13, with prime factors 3, 5, and 13.
Since there are no common factors between the two sets of primes, the greatest common divisor (GCD) is 1. This confirms that 242 and 195 are co-prime, as they share no prime factors beyond the trivial case of 1.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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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:
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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/