1. **Problem Statement:** Find the least number of complete years required for a sum of money to more than double when invested at 20% compound interest per annum. 2. **Define variables:** Let the principal amount be $P$. After $n$ years at compound interest rate $r = 20\% = 0.2$, the amount $A$ is given by: $$ A = P(1 + r)^n = P(1.2)^n $$ 3. **Condition to double the money:** The amount must be more than twice the principal, i.e., $$ A > 2P $$ Substitute $A$: $$ P(1.2)^n > 2P $$ Divide both sides by $P$ (assuming $P > 0$): $$ (1.2)^n > 2 $$ 4. **Solve inequality:** Take the natural logarithm on both sides: $$ n \ln(1.2) > \ln(2) $$ So, $$ n > \frac{\ln(2)}{\ln(1.2)} $$ Calculate the values (using approximations): $$ \ln(2) \approx 0.6931 $$ $$ \ln(1.2) \approx 0.1823 $$ Thus, $$ n > \frac{0.6931}{0.1823} \approx 3.803 $$ 5. **Interpretation:** Since $n$ must be a complete number of years, the least integer $n$ satisfying the inequality is the smallest integer greater than 3.803, which is 4. **Final answer: 4 years.**