We define a function f on the integers `f(x) = x//10`, if x is divisible by 10, `f(x) = x + 1`, if x is not divisible by 10. If `A_(0) = 1994` and `A_(n+1) = f(A_(n))`.
What is the smallest n such that `A_(n) = 2` ?

Options

A. A) 9
B. B) 18
C. C) 128
D. D) 1993

Answer

To solve the problem, we need to apply the function \( f \) repeatedly starting from \( A_0 = 1994 \) until we reach \( A_n = 2 \). The function \( f \) is defined as follows:

- \( f(x) = \frac{x}{10} \) if \( x \) is divisible by 10
- \( f(x) = x + 1 \) if \( x \) is not divisible by 10

Let's calculate the values step by step:

1. **Calculate \( A_0 \)**:
\[ A_0 = 1994 \]

2. **Calculate \( A_1 \)**:
Since \( 1994 \) is not divisible by \( 10 \):
\[ A_1 = f(A_0) = 1994 + 1 = 1995 \]

3. **Calculate \( A_2 \)**:
Since \( 1995 \) is not divisible by \( 10 \):
\[ A_2 = f(A_1) = 1995 + 1 = 1996 \]

4. **Calculate \( A_3 \)**:
Since \( 1996 \) is not divisible by \( 10 \):
\[ A_3 = f(A_2) = 1996 + 1 = 1997 \]

5. **Calculate \( A_4 \)**:
Since \( 1997 \) is not divisible by \( 10 \):
\[ A_4 = f(A_3) = 1997 + 1 = 1998 \]

6. **Calculate \( A_5 \)**:
Since \( 1998 \) is not divisible by \( 10 \):
\[ A_5 = f(A_4) = 1998 + 1 = 1999 \]

7. **Calculate \( A_6 \)**:
Since \( 1999 \) is not divisible by \( 10 \):
\[ A_6 = f(A_5) = 1999 + 1 = 2000 \]

8. **Calculate \( A_7 \)**:
Since \( 2000 \) is divisible by \( 10 \):
\[ A_7 = f(A_6) = \frac{2000}{10} = 200 \]

9. **Calculate \( A_8 \)**:
Since \( 200 \) is divisible by \( 10 \):
\[ A_8 = f(A_7) = \frac{200}{10} = 20 \]

10. **Calculate \( A_9 \)**:
Since \( 20 \) is divisible by \( 10 \):
\[ A_9 = f(A_8) = \frac{20}{10} = 2 \]

Now we have reached \( A_9 = 2 \). Therefore, the smallest \( n \) such that \( A_n = 2 \) is:

\[ \boxed{9} \]