A young girl counted in the following way on the fingers of her left hand. She started calling the thumb $1$, the index finger $2$, middle finger $3$, ring finger $4$, little finger $5$, then reversed direction, calling the ring finger $6$, middle finger $7$, index finger $8$, thumb $9$ and then back to the index figure for $10$, middle finger for $11$, and so on. She counted upto $1994$. She ended on her

### Concept & Logic This is a classic periodic sequence (or cyclic pattern) problem. To find the position of a large number in a repeating sequence, you must identify the length of one complete cycle and then use modular arithmetic (finding the remainder) to locate the target number's position within that cycle. ### Step-by-Step Solution * **Given:** * Target number to reach: $1994$. * The counting pattern across $5$ fingers, reversing at the ends. * **Calculation / Deduction:** 1. Map the numbers to the fingers to find the full repeating cycle: $1 \rightarrow$ Thumb $2 \rightarrow$ Index $3 \rightarrow$ Middle $4 \rightarrow$ Ring $5 \rightarrow$ Little (Reverse happens here) $6 \rightarrow$ Ring $7 \rightarrow$ Middle $8 \rightarrow$ Index $9 \rightarrow$ Thumb (Reverse happens here, cycle repeats) 2. Notice that the pattern begins anew on the Thumb at $9$, $17$, $25$, etc. The numbers assigned to the Thumb are $1, 9, 17...$ which follow the format $8k + 1$. 3. This means the entire counting sequence operates on a cycle of exactly $8$ steps. 4. To find where $1994$ lands, divide it by the cycle length ($8$) and find the remainder. $$1994 \div 8$$ 5. Execute the division: $8 \times 200 = 1600$. $1994 - 1600 = 394$. $8 \times 40 = 320$. $394 - 320 = 74$. $8 \times 9 = 72$. $74 - 72 = 2$. So, $1994 = (249 \times 8) + 2$. 6. The remainder is $2$. 7. Match the remainder to the mapped cycle in step 1. A remainder of $2$ corresponds to position $2$ in the cycle, which is the Index finger. ### Exam Strategy & Shortcut Instead of long division, use the divisibility rule for $8$ to find the remainder instantly. A number is divisible by $8$ if its last three digits are divisible by $8$. Look at the last three digits of $1994$, which are $994$. $994 \div 8 = 124$ with a remainder of $2$ ($124 \times 8 = 992$). Since the remainder is $2$, look at the 2nd step of the cycle: $1$ is thumb, $2$ is index finger. You arrive at the answer in seconds. ### Common Pitfall A very common mistake is assuming the cycle length is $10$ (because we count "back and forth" across $5$ fingers) or $5$ (the number of unique fingers). Mapping out the numbers explicitly until you hit the thumb a second time (at $9$) is crucial to proving the cycle length is actually $8$. ### Final Answer Therefore, the correct answer is index finger.