The minimum number of years in which a sum of money put at 20% interest compounded annually will become more than double is:

To solve this problem, we can use the formula for compound interest, which is:

A = P (1 + r/n)^(nt)

Where: A = the amount of money accumulated after n years, including interest. P = principal amount (the initial amount of money) r = annual interest rate (in decimal) n = number of times that interest is compounded per year t = time the money is invested for in years

We want to find the minimum number of years (t) it will take for a sum of money to double when it's put at 20% interest compounded annually. This means that A = 2P, r = 0.20, and n = 1 (since the interest is compounded annually).

Substituting these values into the formula, we get:

2P = P (1 + 0.20/1)^(1*t) 2 = (1 + 0.20)^t 2 = 1.20^t

To solve for t, we can take the natural logarithm (ln) of both sides:

ln(2) = t * ln(1.20)

Then, divide both sides by ln(1.20) to solve for t:

t = ln(2) / ln(1.20)

Using a calculator, we find that t ≈ 3.8 years. However, since we can't have a fraction of a year, we round up to the next whole number.

So, the minimum number of years in which a sum of money put at 20% interest compounded annually will become more than double is 4 years.