The population of a country increased by an average of \(1\%\) per year from 2016 to 2019 . if the population of this country was 21 million on December 31, 2019, then the population of this country on January 1, 2016, to the nearest thousand had been ____________.

Explanation: Let the population on January 1, 2016 be x . The population increases by \(1\%\) 4 times. So, the population on December 31, 2019 will be \(x\times1.01\times1.01\times1.01\times1.01\), Or, \(x\times(1.01)^4\). So, \(x\times(1.01)^4=21,000,000\). As this calculation is very difficult without a calculator, we can use another method to get an approximation. Increasing \(1\%\) of each option 4 times in a linear way can be done to find the approximation which will be closest to 21 million. Again, 21 million expressed as thousands is 21,000 . So, for each option we calculate in the following manner: Option (A): \(20,400+(1\% \text{ of } 20,400\times4)=20,400+(204\times4)=21,216\). Option (B): \(20,200+(202\times4)=21,008\). Option (C): \(20,000+(200\times4)=20,800\). Options (D) and (E) will go further away from 21,000 . So, option (B) provides the best approximation.