Two numbers under 10 with an LCM exceeding 50 are 7 and 8. Their LCM is calculated using the formula LCM(a, b) = (a × b) ÷ GCD(a, b). Here, 7 and 8 have no common factors besides 1, making their GCD 1. The LCM is thus (7 × 8) ÷ 1 = 56. This exceeds 50, meeting the condition while ensuring the indiviRead more
Two numbers under 10 with an LCM exceeding 50 are 7 and 8. Their LCM is calculated using the formula LCM(a, b) = (a × b) ÷ GCD(a, b). Here, 7 and 8 have no common factors besides 1, making their GCD 1. The LCM is thus (7 × 8) ÷ 1 = 56. This exceeds 50, meeting the condition while ensuring the individual numbers are less than 10.
For ‘idli-vada’ to first occur after 50, the two numbers must have an LCM exceeding 50 but less than 60. Numbers under 10 with this property are 6 and 9, whose LCM is 54. The LCM is calculated as (6 × 9) ÷ 3 = 54, where 3 is their greatest common divisor (GCD). Thus, the first shared multiple afterRead more
For ‘idli-vada’ to first occur after 50, the two numbers must have an LCM exceeding 50 but less than 60. Numbers under 10 with this property are 6 and 9, whose LCM is 54. The LCM is calculated as (6 × 9) ÷ 3 = 54, where 3 is their greatest common divisor (GCD). Thus, the first shared multiple after 50 is 54, fulfilling the game’s condition for ‘idli-vada.’
To reach both treasures, the jump size must divide both 28 and 70. The factors of 28 are 1, 2, 4, 7, 14, and 28, and the factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70. The common factors are 1, 2, 7, and 14. Among these, 14 is the greatest common divisor (GCD), ensuring it is the smallest jump siRead more
To reach both treasures, the jump size must divide both 28 and 70. The factors of 28 are 1, 2, 4, 7, 14, and 28, and the factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70. The common factors are 1, 2, 7, and 14. Among these, 14 is the greatest common divisor (GCD), ensuring it is the smallest jump size that successfully lands on both numbers.
The least common multiple (LCM) of all numbers from 1 to 10 is 2520, found by considering their prime factorizations. To exclude 7, divide 2520 by 7, yielding 360. This number is divisible by all integers from 1 to 10 except 7, satisfying the condition. The prime factorization of 360 further confirmRead more
The least common multiple (LCM) of all numbers from 1 to 10 is 2520, found by considering their prime factorizations. To exclude 7, divide 2520 by 7, yielding 360. This number is divisible by all integers from 1 to 10 except 7, satisfying the condition. The prime factorization of 360 further confirms this: 360 = 2 × 2 × 2 × 3 × 3 × 5, excluding 7 as a factor.
To determine the smallest number that is a multiple of all numbers from 1 to 10, compute their least common multiple (LCM). Using prime factorizations, we find that 2, 3, 5, and 7 must be included with appropriate powers. The LCM is 2520, derived from 2³ × 3² × 5 × 7. This number is divisible by eveRead more
To determine the smallest number that is a multiple of all numbers from 1 to 10, compute their least common multiple (LCM). Using prime factorizations, we find that 2, 3, 5, and 7 must be included with appropriate powers. The LCM is 2520, derived from 2³ × 3² × 5 × 7. This number is divisible by every integer from 1 to 10, confirming it as the smallest common multiple for this range.
Find two numbers smaller than 10 such that their LCM exceeds 50.
Two numbers under 10 with an LCM exceeding 50 are 7 and 8. Their LCM is calculated using the formula LCM(a, b) = (a × b) ÷ GCD(a, b). Here, 7 and 8 have no common factors besides 1, making their GCD 1. The LCM is thus (7 × 8) ÷ 1 = 56. This exceeds 50, meeting the condition while ensuring the indiviRead more
Two numbers under 10 with an LCM exceeding 50 are 7 and 8. Their LCM is calculated using the formula LCM(a, b) = (a × b) ÷ GCD(a, b). Here, 7 and 8 have no common factors besides 1, making their GCD 1. The LCM is thus (7 × 8) ÷ 1 = 56. This exceeds 50, meeting the condition while ensuring the individual numbers are less than 10.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Anshu and his friends play the idli-vada game with two numbers smaller than 10. The first time anybody says ‘idli-vada’ is after the number 50. What could these two numbers be?
For ‘idli-vada’ to first occur after 50, the two numbers must have an LCM exceeding 50 but less than 60. Numbers under 10 with this property are 6 and 9, whose LCM is 54. The LCM is calculated as (6 × 9) ÷ 3 = 54, where 3 is their greatest common divisor (GCD). Thus, the first shared multiple afterRead more
For ‘idli-vada’ to first occur after 50, the two numbers must have an LCM exceeding 50 but less than 60. Numbers under 10 with this property are 6 and 9, whose LCM is 54. The LCM is calculated as (6 × 9) ÷ 3 = 54, where 3 is their greatest common divisor (GCD). Thus, the first shared multiple after 50 is 54, fulfilling the game’s condition for ‘idli-vada.’
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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What jump sizes will land on both treasures at 28 and 70?
To reach both treasures, the jump size must divide both 28 and 70. The factors of 28 are 1, 2, 4, 7, 14, and 28, and the factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70. The common factors are 1, 2, 7, and 14. Among these, 14 is the greatest common divisor (GCD), ensuring it is the smallest jump siRead more
To reach both treasures, the jump size must divide both 28 and 70. The factors of 28 are 1, 2, 4, 7, 14, and 28, and the factors of 70 are 1, 2, 5, 7, 10, 14, 35, and 70. The common factors are 1, 2, 7, and 14. Among these, 14 is the greatest common divisor (GCD), ensuring it is the smallest jump size that successfully lands on both numbers.
For more NCERT Solutions for Class 6 Math Chapter 5 Prime Time Extra Questions and Answer:
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Find the smallest number that is a multiple of all the numbers from 1 to 10 except for 7.
The least common multiple (LCM) of all numbers from 1 to 10 is 2520, found by considering their prime factorizations. To exclude 7, divide 2520 by 7, yielding 360. This number is divisible by all integers from 1 to 10 except 7, satisfying the condition. The prime factorization of 360 further confirmRead more
The least common multiple (LCM) of all numbers from 1 to 10 is 2520, found by considering their prime factorizations. To exclude 7, divide 2520 by 7, yielding 360. This number is divisible by all integers from 1 to 10 except 7, satisfying the condition. The prime factorization of 360 further confirms this: 360 = 2 × 2 × 2 × 3 × 3 × 5, excluding 7 as a factor.
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 the numbers from 1 to 10.
To determine the smallest number that is a multiple of all numbers from 1 to 10, compute their least common multiple (LCM). Using prime factorizations, we find that 2, 3, 5, and 7 must be included with appropriate powers. The LCM is 2520, derived from 2³ × 3² × 5 × 7. This number is divisible by eveRead more
To determine the smallest number that is a multiple of all numbers from 1 to 10, compute their least common multiple (LCM). Using prime factorizations, we find that 2, 3, 5, and 7 must be included with appropriate powers. The LCM is 2520, derived from 2³ × 3² × 5 × 7. This number is divisible by every integer from 1 to 10, confirming it as the smallest common multiple for this range.
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/