X is a series of five consecutive multiples of three whose average is 36. Y is a series of four consecutive multiples of four, where the second smallest number in Y is 13 more than the second highest number in X. Find the sum of the smallest number of the X series and the highest number of the Y series.

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**Solution:**

1. Let the five consecutive multiples of 3 in series X be:
   $$3n, 3(n+1), 3(n+2), 3(n+3), 3(n+4)$$

2. The average of these five numbers is given as 36:
   $$
   $$\frac{3n + 3(n+1) + 3(n+2) + 3(n+3) + 3(n+4)}{5} = 36$$
   $$

3. Simplify the numerator:
   $$
   3n + 3n + 3 + 3n + 6 + 3n + 9 + 3n + 12 = 15n + (3 + 6 + 9 + 12) = 15n + 30
   $$

4. So,
   $$
   $$\frac{15n + 30}{5} = 36 \implies 3n + 6 = 36 \implies 3n = 30 \implies n = 10$$
   $$

5. Therefore, the series X is:
   $$30, 33, 36, 39, 42$$

6. The second highest number in X is the 4th term: 39

7. Let the four consecutive multiples of 4 in series Y be:
   $$4m, 4(m+1), 4(m+2), 4(m+3)$$

8. The second smallest number in Y is the 2nd term: $4(m+1)$

9. According to the problem, the second smallest number in Y is 13 more than the second highest number in X:
   $$
   4(m+1) = 39 + 13 = 52
   $$

10. Solve for $m$:
    $$
    $$4m + 4 = 52 \implies 4m = 48 \implies m = 12$$
    $$

11. The highest number in Y is the 4th term:
    $$4(m+3) = 4(12 + 3) = 4 \times 15 = 60$$

12. The smallest number in X is 30.

13. Required sum = smallest number of X + highest number of Y = $30 + 60 = 90$

**Answer:** 90

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