Claudine has $p$ packages containing 19 candies each.

If Claudine divides all of her candies equally among 7 friends, there are 4 candies left over. If Claudine divides all of her candies equally among 11 friends, there is 1 candy left over. What is the minimum possible value of $p$ ?
Suppose that Claudine has $N$ candies in total.

When Claudine divides all of her candies equally among 7 friends, she has 4 candies left over. This means that $N$ is 4 more than a multiple of 7 , or $N=7 a+4$ for some non-negative integer $a$.

When Claudine divides all of her candies equally among 11 friends, she has 1 candy left over. This means that $N$ is 1 more than a multiple of 11 , or $N=11 b+1$ for some non-negative integer $b$.

Note then that $N+10=7 a+14=7(a+2)$ and $N+10=11 b+11=11(b+1)$.

In other words, $N+10$ is a multiple of 7 and a multiple of 11 .

This means that $N+10$ is a multiple of 77 , or $N$ is 10 less than a multiple of 77 .

Since Claude has $p$ packages containing 19 candies each, then Claudine has $19 p$ candies.

Putting these pieces together, we want to find the smallest non-negative integer $p$ for which $19 p$ is 10 less than a multiple of 77 .

To do this efficiently, we list, in increasing order, the positive integers that are 10 less than a multiple of 77 until we reach one that is a multiple of 19:

$$
67,144,221,298,375,452,529,606,683,760
$$

Since 760 is the first number in this list that is a multiple of 19 , then 760 is the smallest multiple of 19 that is 4 more than a multiple of 7 and 1 more than a multiple of 11 .

This means that the smallest possible value of $p$ is $\frac{760}{19}$ or 40 .