The product $55\times 60\times 65$ is written as the product of five distinct positive integers. What is the least possible value of the largest of these integers?

Step 1: Find the prime factorization of the product:

Step 2: We want to write $P$ as the product of five distinct positive integers $x_1<x_2<x_3<x_4<x_5$ such that the largest integer $x_5$ is minimized.

Step 3: Since 13 is a prime factor of $P$, at least one of the five integers must be a multiple of 13. Since we want to minimize the largest element $x_5$, this multiple of 13 should be as small as possible. The smallest positive integer multiple of 13 is 13 itself.
Similarly, 11 is a prime factor, so one of the integers must be a multiple of 11. The smallest multiple is 11 itself.

Step 4: Let's set two of our integers to be $11$ and $13$. The remaining product is:

We need to factor $1500$ into three distinct positive integers $a<b<c$ such that $c$ is minimized and all five integers $\{11,13,a,b,c\}$ are distinct.

Step 5: To minimize $c$, the three integers $a,b,c$ should be as close to each other as possible, which means they should be close to $\sqrt[3]{1500}\approx 11.45$.
Let's test close integers:

• Try $a=5,b=15,c=20$. The product is $5\times 15\times 20=1500$. The five integers are $\{5,11,13,15,20\}$. These are all distinct! The largest is $20$.

Step 6: Can we get a largest integer smaller than 20? Let's check if the largest can be 19 or 18. If the largest is $\le 19$, since $11$ and $13$ are in the set, the remaining three must be distinct integers from $\{1,2,\dots ,19\}\setminus \{11,13\}$. Their product must be 1500. But the maximum product of three distinct integers $\le 19$ excluding $11,13$ is $19\times 18\times 17=5814$, so it is possible in theory, but we must form 1500. Through systematic checking, no three distinct integers $\le 19$ multiply to 1500. Thus, 20 is the absolute minimum.