Let N=11 ^ 2 \times 13 ^ 4 \times 17 ^ 6 . How many positive factors of N^2 are less than N but not a factor of N?
To find the number of positive factors of N^2 that are less than N but not a factor of N, we need to first find the prime factorization of N.
We have:
N = 11^2 * 13^4 * 17^6
Next, let's find the prime factorization of N^2:
N^2 = (11^2 * 13^4 * 17^6)^2
= 11^4 * 13^8 * 17^12
Now, we need to find the positive factors of N^2 that are less than N but not a factor of N. To do this, we need to consider the prime factors 11, 13, and 17.
For factor 11:
Since 11 is raised to the power of 4 in N^2 and 11^2 in N, the possible powers of 11 in the factors of N^2 will be 0, 1, 2, and 3.
For factor 13:
Since 13 is raised to the power of 8 in N^2 and 13^4 in N, the possible powers of 13 in the factors of N^2 will be 0, 1, 2, 3, 4, 5, 6, and 7.
For factor 17:
Since 17 is raised to the power of 12 in N^2 and 17^6 in N, the possible powers of 17 in the factors of N^2 will be 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11.
Now, let's calculate the total number of factors of N^2 that are less than N but not a factor of N:
Total number of factors = (possible powers of 11) * (possible powers of 13) * (possible powers of 17)
= (4) * (8) * (12)
= 384
Therefore, there are 384 positive factors of N^2 that are less than N but not a factor of N.
To solve this problem, we first need to find the prime factorization of the number N.
N is given as N = 11^2 * 13^4 * 17^6.
Now, to calculate the number of positive factors of N^2 that are less than N, but not a factor of N, we can break this down into two parts.
Part 1: Find the number of factors of N^2 that are less than N.
Since N^2 is the square of N, any factor of N^2 that is less than N must also be a factor of N. Therefore, for this part, we just need to find the number of factors of N.
The number of factors of N can be calculated using the formula: (a + 1)(b + 1)(c + 1)..., where "a", "b", "c", etc. are the exponents of the prime factors in the prime factorization of N. In our case, N = 11^2 * 13^4 * 17^6, so a = 2, b = 4, and c = 6.
Therefore, the number of factors of N is (2 + 1)(4 + 1)(6 + 1) = 3 * 5 * 7 = 105.
Part 2: Find the number of factors of N^2 that are factors of N.
Since any factor of N^2 is also a factor of N, we need to exclude these factors from the count. In other words, we need to find the number of factors of N that are not factors of N^2.
We can calculate this by subtracting the number of common factors of N and N^2 from the number of factors of N.
To find the common factors of N and N^2, we can simply look at the prime factorization of N. In this case, N = 11^2 * 13^4 * 17^6.
Since N^2 contains all the prime factors of N raised to twice their exponent, the common factors of N and N^2 are obtained by removing the squared exponents from the prime factorization of N.
So, the common factors of N and N^2 are 11^1, 13^2, and 17^3.
Now, we subtract the number of common factors (which is 3) from the number of factors of N (which is 105) to find the number of factors of N^2 that are factors of N:
105 - 3 = 102.
Therefore, there are 102 positive factors of N^2 that are less than N but not a factor of N.