What is the least positive integer value of $n$ such that $28n$ is divisible by $365$?
To solve this problem, I need to find the smallest multiple of 28 that is also a multiple of 365. I could try to list some multiples of 28 and see if any of them are multiples of 365, but that might take a long time. A better way to approach this is to use the prime factorization of both numbers. If I write 28 as $2^2 \cdot 7$ and 365 as $5 \cdot 73$, then I can see that for 28n to be divisible by 365, it must have at least one factor of 5 and one factor of 73 in its prime factorization. This means that n must be a multiple of $5 \cdot 73 = 365$. The smallest multiple of 365 is 365 itself, so the least positive integer value of n is 365. 