How many consequtive numbers are there in the set of positive integers in which powers of all prime factors in their prime factorizations are odd numbers?

$ 
\textbf{(A)}\ 3
\qquad\textbf{(B)}\ 7
\qquad\textbf{(C)}\ 8
\qquad\textbf{(D)}\ 10
\qquad\textbf{(E)}\ 15
$
1. To solve this problem, we need to understand the properties of numbers whose prime factors have only odd powers. Such numbers are called "square-free" numbers, meaning they are not divisible by any perfect square greater than 1.
2. Consider any set of consecutive integers. Among any set of 8 consecutive integers, at least one of them must be divisible by 4. This is because every fourth number in the sequence of integers is divisible by 4.
3. If a number is divisible by 4, it can be written as \(4k\) for some integer \(k\). The prime factorization of \(4k\) includes \(4 = 2^2\), which has an even power of 2. Therefore, this number cannot have all prime factors with odd powers.
4. Hence, in any set of 8 consecutive integers, there will always be at least one number that does not meet the condition of having all prime factors with odd powers.
5. However, in any set of 7 consecutive integers, it is possible that none of the numbers are divisible by 4. For example, consider the set \(\{37, 38, 39, 40, 41, 42, 43\}\). In this set, none of the numbers are divisible by 4, and thus it is possible for all numbers to have prime factors with odd powers.

Therefore, the maximum number of consecutive integers in which all numbers have prime factors with odd powers is 7.

The final answer is \(\boxed{7}\).