The relation between GCF and LCM of 30 and 35 is given as, LCM(30, 35) × GCF(30, 35) = Product of 30, 35 Prime factorization of 30 and 35 is given as, 30 = (2 × 3 × 5) = 21 × 31 × 51 and 35 = (5 × 7) = 51 × 71 LCM(30, 35) = 210 GCF(30, 35) = 5 LHS = LCM(30, 35) × GCF(30, 35) = 210 × 5 = 1050 RHS = Product of 30, 35 = 30 × 35 = 1050 ⇒ LHS = RHS = 1050 Hence, verified.

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To calculate the LCM of 30 and 35 by the division method, we will divide the numbers(30, 35) by their prime factors (preferably common). The product of these divisors gives the LCM of 30 and 35.

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The value of LCM of 30, 35 is the smallest common multiple of 30 and 35. The number satisfying the given condition is 210.

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The LCM of two non-zero integers, x(30) and y(35), is the smallest positive integer m(210) that is divisible by both x(30) and y(35) without any remainder.

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LCM(35, 30) × GCF(35, 30) = 35 × 30 Since the LCM of 35 and 30 = 210 ⇒ 210 × GCF(35, 30) = 1050 Therefore, the greatest common factor = 1050/210 = 5.

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To find the LCM of 30 and 35 using prime factorization, we will find the prime factors, (30 = 2 × 3 × 5) and (35 = 5 × 7). LCM of 30 and 35 is the product of prime factors raised to their respective highest exponent among the numbers 30 and 35. ⇒ LCM of 30, 35 = 21 × 31 × 51 × 71 = 210.

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LCM of 30 and 35 is the smallest number among all common multiples of 30 and 35. The first few multiples of 30 and 35 are (30, 60, 90, 120, 150, 180, . . . ) and (35, 70, 105, 140, . . . ) respectively. There are 3 commonly used methods to find LCM of 30 and 35 - by division method, by listing multiples, and by prime factorization.

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Prime factorization of 30 and 35 is (2 × 3 × 5) = 21 × 31 × 51 and (5 × 7) = 51 × 71 respectively. LCM of 30 and 35 can be obtained by multiplying prime factors raised to their respective highest power, i.e. 21 × 31 × 51 × 71 = 210. Hence, the LCM of 30 and 35 by prime factorization is 210.

The LCM of 30 and 35 is 210. To find the least common multiple of 30 and 35, we need to find the multiples of 30 and 35 (multiples of 30 = 30, 60, 90, 120 . . . . 210; multiples of 35 = 35, 70, 105, 140 . . . . 210) and choose the smallest multiple that is exactly divisible by 30 and 35, i.e., 210.

The LCM of 30 and 35 is the product of all prime numbers on the left, i.e. LCM(30, 35) by division method = 2 × 3 × 5 × 7 = 210.

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Let the other number be z. ∵ GCD × LCM = 30 × z ⇒ z = (GCD × LCM)/30 ⇒ z = (5 × 210)/30 ⇒ z = 35 Therefore, the other number is 35.

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The least number divisible by 30 and 35 = LCM(30, 35) LCM of 30 and 35 = 2 × 3 × 5 × 7 [Incomplete pair(s): 2, 3, 5, 7] ⇒ Least perfect square divisible by each 30 and 35 = LCM(30, 35) × 2 × 3 × 5 × 7 = 44100 [Square root of 44100 = √44100 = ±210] Therefore, 44100 is the required number.

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