The population of a town in 2020 was 100,000. The population decreased by $$ y\% $$ from the year 2020 to 2021 and increased by $$ x\% $$ from the year 2021 to 2022, where $$ x $$ and $$ y $$ are two natural numbers. If the population in 2022 was greater than the population in 2020 and the difference between $$ x $$ and $$ y $$ is 10, then the lowest possible population of the town in 2021 was:

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To solve this problem:

#### Step 1: Define the conditions
The population decreased by $$ y\% $$ from 2020 to 2021 and increased by $$ x\% $$ from 2021 to 2022. The condition is:

$$
100,000 \cdot \left(1 - \frac{y}{100}\right) \cdot \left(1 + \frac{x}{100}\right) > 100,000
$$

Simplify this inequality:
$$
\left(1 - \frac{y}{100}\right) \cdot \left(1 + \frac{x}{100}\right) > 1
$$

#### Step 2: Substitute $$ x = y + 10 $$
Since $$ x - y = 10 $$, replace $$ x $$ with $$ y + 10 $$:
$$
\left(1 - \frac{y}{100}\right) \cdot \left(1 + \frac{y + 10}{100}\right) > 1
$$

Expand and simplify:
$$
(100 - y)(100 + y + 10) > 10,000
$$
$$
100^2 + 100y - y^2 - y^2 - 10y > 10,000
$$
$$
y^2 + (-10y - y^2) + (25 - y^2) < 1025
$$

#### Step 3: Solve for largest $$ y $$
To minimize the population in 2021, maximize $$ y $$. The largest value of $$ y = 32 - 5 = 27 $$.

#### Step 4: Calculate the minimum population in 2021
Substitute $$ y = 27 $$:
$$
100,000 \cdot (1 - \frac{27}{100}) = 73,000
$$

Thus, the lowest possible population of the town in **2021** is **73,000**, corresponding to option **D**.

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