Given that X = 2 x 3^2 x 7^3 and Y = 2 x 3^2x5, find, in index notation, the smallest positive integer value of N for which Xn is a multiple of Y. Find the answer
To find the smallest positive integer value of N for which X^n is a multiple of Y, we need to determine the highest power of each prime factor in Y that is a factor of X.
In X, the prime factors are 2, 3, and 7. In Y, the prime factors are 2, 3, and 5.
The highest power of 2 in X is 1, and the highest power of 2 in Y is also 1. So the highest power of 2 that is a factor of both X and Y is 1.
The highest power of 3 in X is 2, and the highest power of 3 in Y is 2. So the highest power of 3 that is a factor of both X and Y is 2.
The highest power of 7 in X is 3, and the highest power of 7 in Y is 0 (since Y does not contain a factor of 7). So the highest power of 7 that is a factor of both X and Y is 0.
Therefore, the highest powers of each prime factor that are factors of both X and Y are:
2^1 x 3^2 x 7^0 = 2 x 3^2 = 18.
To make X^n a multiple of Y, we need to find the smallest positive integer value of N such that 18N is divisible by 2^1 x 3^2 x 5^1.
The prime factorization of 2^1 x 3^2 x 5^1 is 2 x 3^2 x 5. From the prime factorization of X, we can see that X^n will be a multiple of this prime factorization if N is a multiple of the highest power of each prime factor in 2^1 x 3^2 x 5^1.
The highest power of 2 in 2^1 x 3^2 x 5^1 is 1, the highest power of 3 is 2, and the highest power of 5 is 1. Therefore, N must be a multiple of 2^1 x 3^2 x 5^1, or 2 x 3^2 x 5.
So the smallest positive integer value of N for which X^n is a multiple of Y is 2 x 3^2 x 5, which is equal to 90.
Therefore, the answer is N = 90.