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Question:

Suppose Haruka has a special key ∆ in her calculator called delta key:

Rule 1: If the display shows a one-digit number, pressing delta key ∆ replaces the displayed number with twice its value.

Rule 2: If the display shows a two-digit number, pressing delta key ∆ replaces the displayed number with the sum of the two digits.

Suppose Haruka enters the value 1 and then presses delta key ∆ repeatedly.

After pressing the ∆ key for 68 times, what will be the displayed number?

Explanation:

Initially the number = 1

Output when ∆ is pressed 1 time = 2 [Single digit number 1 is doubled]
Output when ∆ is pressed 2 times = 4 [Single digit number 2 is doubled]
Output when ∆ is pressed 3 times = 8 [Single digit number 4 is doubled]
Output when ∆ is pressed 4 times = 16 [Single digit number 8 is doubled]
Output when ∆ is pressed 5 times = 7 [Digits of 2-digit number 16 are added]
Output when ∆ is pressed 6 times = 14 [Single digit number 7 is doubled]
Output when ∆ is pressed 7 times = 5 [Digits of 2-digit number 14 are added]
Output when ∆ is pressed 8 times = 10 [Single digit number 5 is doubled]
Output when ∆ is pressed 9 times = 1 [Digits of 2-digit number 10 are added]

After 9 steps we get back the same input 1.
Hence, when the input is 1, we will get the same output at the end of 9 or 18 or ... or 63 steps.

∴ Output at the end of 68 steps will be same as the output at the end of 5 steps i.e., 7.

Hence, option (a).

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