Suppose that at the end of any year, a unit of money has lost $10\%$ of the value it had at the beginning of that year. Find the smallest integer $n$ such that after $n$ years, the money will have lost at least $90\%$ of its value. (To the nearest thousandth $\log_{10}3=.477$.)

$\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 22$
1. Let the initial value of the money be \( V_0 \). After one year, the value of the money will be \( 0.9V_0 \) because it loses \( 10\% \) of its value.
2. After \( n \) years, the value of the money will be \( V_0 \times (0.9)^n \).
3. We need to find the smallest integer \( n \) such that the money has lost at least \( 90\% \) of its value. This means the remaining value should be at most \( 10\% \) of the initial value, i.e., \( 0.1V_0 \).
4. Set up the inequality:
   \[
   V_0 \times (0.9)^n \leq 0.1V_0
   \]
5. Divide both sides by \( V_0 \) (assuming \( V_0 \neq 0 \)):
   \[
   (0.9)^n \leq 0.1
   \]
6. Take the logarithm of both sides. Using the common logarithm (base 10):
   \[
   \log_{10}((0.9)^n) \leq \log_{10}(0.1)
   \]
7. Apply the power rule of logarithms, \( \log_{10}(a^b) = b \log_{10}(a) \):
   \[
   n \log_{10}(0.9) \leq \log_{10}(0.1)
   \]
8. We know that \( \log_{10}(0.1) = -1 \) because \( 0.1 = 10^{-1} \):
   \[
   n \log_{10}(0.9) \leq -1
   \]
9. Solve for \( n \):
   \[
   n \geq \frac{-1}{\log_{10}(0.9)}
   \]
10. We need to find \( \log_{10}(0.9) \). Recall that \( 0.9 = \frac{9}{10} \), so:
    \[
    \log_{10}(0.9) = \log_{10}\left(\frac{9}{10}\right) = \log_{10}(9) - \log_{10}(10) = \log_{10}(9) - 1
    \]
11. We know that \( 9 = 3^2 \), so:
    \[
    \log_{10}(9) = \log_{10}(3^2) = 2 \log_{10}(3)
    \]
12. Given \( \log_{10}(3) = 0.477 \):
    \[
    \log_{10}(9) = 2 \times 0.477 = 0.954
    \]
13. Therefore:
    \[
    \log_{10}(0.9) = 0.954 - 1 = -0.046
    \]
14. Substitute back into the inequality:
    \[
    n \geq \frac{-1}{-0.046} = \frac{1}{0.046} \approx 21.74
    \]
15. Since \( n \) must be an integer, we round up to the nearest whole number:
    \[
    n = 22
    \]

The final answer is \( \boxed{22} \)