Question

We define a function \(f\) on the integers: \(f(x) = \dfrac{x}{10}\) if \(x\) is divisible by 10, and \(f(x) = x+1\) 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 \(A_n = 2\)?

• A. 9
• B. 18
• C. 128
• D. 1993

Hint:

Split the journey into an 'add 1 until a multiple of 10' phase and a 'divide by 10' phase.

Answer 1

The Correct Option is A

Step 1: Understand the rule.
At every step, check whether the current number is a multiple of 10. If it is, divide it by 10. If it is not, add 1 to it. Keep applying this from \(A_0 = 1994\), and count how many steps it takes to first land on 2.

Step 2: Add 1 repeatedly until we hit a multiple of 10.
\(A_0 = 1994\) (not divisible by 10) \(\to A_1 = 1995\)
\(A_1 = 1995 \to A_2 = 1996\)
\(A_2 = 1996 \to A_3 = 1997\)
\(A_3 = 1997 \to A_4 = 1998\)
\(A_4 = 1998 \to A_5 = 1999\)
\(A_5 = 1999 \to A_6 = 2000\)
It took 6 steps of add 1 to climb from 1994 to the next multiple of 10, which is 2000.

Step 3: Now divide by 10 repeatedly.
\(A_6 = 2000\) is divisible by 10, so \(A_7 = 2000/10 = 200\).
\(A_7 = 200\) is divisible by 10, so \(A_8 = 200/10 = 20\).
\(A_8 = 20\) is divisible by 10, so \(A_9 = 20/10 = 2\).

Step 4: Read off the smallest n.
\(A_9 = 2\) for the first time, and no smaller index gives 2, since every earlier term is either bigger than 2 or a different value entirely.

Final Answer:
\[ \boxed{n = 9} \]