15. (15 points) 4 consecutive natural numbers, from smallest to largest, are multiples of 11, 7, 5, and 3, respectively. Find the minimum sum of these 4 natural numbers.
【Analysis】Let the smallest number be $11k$, from smallest to largest they are $11k+1$, $11k+2$, $11k+3$; $11k+1$ is a multiple of 7, $k=5,12,19,26,33,40 \cdots$; $11k+2$ is a multiple of 5, $k=3,8,13,18,23,28,33,38 \cdots$; $11k+3$ is a multiple of 3, $k=3,6,9,12,15,18,24,27,30,33 \cdots$, thus the smallest value of $k$ is 33, the four numbers from smallest to largest are $363$, $364$, $365$, $366$, and the minimum sum of the four numbers can be found.

【Solution】Solution: Let the smallest number be $11k$, from smallest to largest they are $11k+1$, $11k+2$, $11k+3$;
$11k+1$ is a multiple of 7, $k=5,12,19,26,33,40 \cdots$
$11k+2$ is a multiple of 5, $k=3,8,13,18,23,28,33,38 \cdots$
$11k+3$ is a multiple of 3, $k=3,6,9,12,15,18,24,27,30,33 \cdots$
Obviously, the smallest value of $k$ is 33, the four numbers from smallest to largest are $363$, $364$, $365$, $366$, the minimum sum of the four numbers is $363+364+365+366=1458$.