Three numbers in an arithmetic progression have a sum of 39 and a product of 2145. Find the smallest number of that progression.

Sure, let's solve this step by step.

Let's denote the three numbers as a-d, a, a+d, where a is the middle number and d is the common difference.

From the problem, we know that:

1. The sum of the numbers is 39, so we can write the equation: (a-d) + a + (a+d) = 39. Simplifying this, we get 3a = 39, so a = 13.

2. The product of the numbers is 2145, so we can write the equation: (a-d)a(a+d) = 2145. Substituting a = 13 into this equation, we get (13-d)13(13+d) = 2145. Simplifying this, we get 169 - d^2 = 165. Solving for d, we get d = ±2.

Therefore, the three numbers are 11, 13, and 15. The smallest number is 11.