Discussion

Question:

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, then back to the index finger for 10, middle finger for 11, and so on. She counted up to 1994. She ended on her.

Explanation:

The counting is done in the following manner:

thumb - index finger - middle finger - ring finger - little finger
   1 2 3 4 5
9 8 7 6
10 11 12 13
17 16 15 14

∴ The numbers, counted on thumb are 1, 9, 17 and so on. This is an arithmetic progression whose first term is 1 and common difference is 8.
⇒ A number counted on thumb can be represented as = 1 + (n - 1) × 8 = 8n - 7

Now, let us figure out highest number less than or equal to 1994 that will be counted on thumb.
⇒ 8n - 7 ≤ 1994
⇒ n ≤ 2001/8 = 125.125

Highest number counted on thumb is when n is 125 = 8 × 125 - 7 = 1993

∴ Next number 1994 will be counted on index finger.

Hence, option (b).

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