Write a method to generate a random number between 1 and 7, given a method that generates a random number between 1 and 5 (i.e., implement rand7() using rand5()).

First, observe that we cannot do this in a guaranteed finite amount of time. Why? Let’s see bya parallel example: How would you use rand2() to create rand3()?
Observe that each call of rand2() and the corresponding decision you make can be represented by a decision tree. On each node, you have two branches. You take the left one when
rand2() equals 0 (which happens with 1/2 probability). You take the right one when rand2() equals 1 (which happens with 1/2 probability). You continue branching left and right as you continue to call 1/2. When you reach a leaf, you return a result of 1, 2 or 3 (your rand3() results).
»»    What’s the probability of taking each branch? 1/2.
»»    What’s the probability to reach a particular leaf node? 1/2^j (for some j).
»»    What the probability of returning 3 (for example)? We could compute this by summing up the probabilities of reaching each leaf node with value 3. Each of these paths has   probability 1/2^j, so we know that the total probability of returning 3 must be a series of terms of reciprocal powers of 2 (e.g., 1/2^x + 1/2^y + 1/2^z + ...).
We also know, however, that the probability of returning 3 must be 1/3 (because rand3() should be perfectly random). Can you find a series of reciprocal powers of 2 that sum to 1/3?

No, because 3 and 2 are relatively prime.
We can similarly conclude that to solve this problem, we will need to accept a small (infinitesimally small) chance that this process will repeat forever. That’s ok.
So, how do we solve this?
In order to generate a random number between 1 and 7, we just need to uniformly generate a larger range than we are looking for and then repeatedly sample until we get a number that is good for us. We will generate a base 5 number with two places with two calls to the RNG.

      public static int rand7() {
              while (true) {
                      int num = 5 * (rand5() - 1) + (rand5() - 1);
                      if (num < 21) return (num % 7 + 1);
              }
      }