The numbers 4725 and 126, written as the products of their prime factors, are 4725=3^3 x 5^2 x 7. The smallest positive integer value of m for which 126m, is a multiple of 4725. Find the answer
To find the smallest positive integer value of m for which 126m is a multiple of 4725, we need to ensure that 126m has the prime factors 3, 5, and 7 with at least as many occurrences as in 4725.
The prime factorization of 126 is 2 x 3^2 x 7. To make 126m a multiple of 4725, we need to increase the powers of 3 and 5 in the prime factorization of 126m.
From the prime factorization of 4725, we can see that the highest power of 3 is 3^3 and the highest power of 5 is 5^2. Therefore, to match or exceed these powers in 4725, we need to include at least 3^3 and 5^2 in the prime factorization of 126m.
m needs to have a factor of 3^3 and 5^2. The smallest positive integer value for m that satisfies this condition is lcm(3^3, 5^2), which is the least common multiple of 27 and 25.
Calculating the least common multiple:
Prime factorization of 27: 3^3
Prime factorization of 25: 5^2
To find lcm(27, 25), we take the maximum power of each prime factor: lcm(3^3, 5^2) = 3^3 x 5^2 = 3^3 x 25 = 27 x 25 = 675.
Therefore, the smallest positive integer value of m for which 126m is a multiple of 4725 is m = 675. Answer: \boxed{675}