Suppose Haruka has a special key $$\triangle$$ in her caculator called delta key:

Rule 1: If the display shows a one-digit number, pressing delta key $$\triangle$$ replace the displayed number with twice its value.
Rule 2: If the display shows a two-digits number, pressing delta key $$\triangle$$ replace the displayed number with the number sum of two digits.
Suppose Haruka enters the value 1 and then presses delta key $$\triangle$$ repeated.
After pressing the key for 68 times, what will be the displayed number?

The first number Haruka entered is 1.

From Rule-1, the series will be 2, 4, 8, 16

From Rule 2, 16 will be followed by 7

Like-wise, the series will be 1, 2, 4, 8, 16, 7, 14, 5, 10, 1, 2, 4,......

The series is repeating itself for every 9 iterations.

Initially, input is 1 and after first iteration, result is 2. 

This implies,

9n = 1

9n + 1 = 2

9n + 2 = 4

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9n + 7 = 5

9n + 8 = 10

In the question, it is given that key is pressed 68 times. 

68 = 9(7) + 5

68 is in the form of 9n + 5

This implies, 7 is the output.

The answer is option A.