Question

Suppose Haruka has a special key Δ in her calculator called delta key:
Rule 1: If the display shows a one-digit number, pressing delta key
Δ replaces the displayed number with twice its value.
Rule 2: If the display shows a two-digit number, pressing delta key
Δ replaces the displayed number with the sum of the two digits.
Suppose Haruka enters the value 1 and then presses delta key
Δ repeatedly. After pressing the Δ key for 68 times, what will be the displayed number?

• A. 7
• B. 4
• C. 10
• D. 2
• E. 8

Answer 1

The Correct Option is A

To solve the problem, we need to understand the behavior of the special key Δ on Haruka's calculator as per the given rules.

Step-by-step Solution:

• Initial Value: Haruka starts with the value 1 on the display.
• Rule 1: If the number is a one-digit number, pressing the Δ key doubles the displayed value.
• The sequence starts with 1. Upon pressing the Δ key, it becomes 2.
• Continuing to apply Rule 1:
  • 2 becomes 4 (2 * 2 = 4)
  • 4 becomes 8 (4 * 2 = 8)
  • 8 becomes 16 (8 * 2 = 16), at this point Rule 2 becomes applicable.
• Rule 2: If the number is a two-digit number, the Δ key changes the displayed number to the sum of its digits.
• 16 becomes 7 (1 + 6 = 7), now we revert to Rule 1 as 7 is a single-digit number.
• Continuing under Rule 1:
  • 7 becomes 14 (7 * 2 = 14), then apply Rule 2: 14 becomes 5 (1 + 4 = 5)
  • 5 becomes 10 (5 * 2 = 10), and apply Rule 2: 10 becomes 1 (1 + 0 = 1)
  • 1 becomes 2 (since we're back to a single digit)

From here, the sequence 2, 4, 8, 16, 7, 14, 5, 10, 1 begins to repeat its pattern every 9 steps, cycling through: [1, 2, 4, 8, 7, 5, 1]. Importantly, the cycle restarts when we reach the value 1.

Since the cycle length is 9, and we've pressed the Δ key 68 times, we calculate the final position after 68 presses:

• \( 68 \mod 9 = 5 \). Thus, the 5th number in the repetitive cycle [1, 2, 4, 8, 7, 5, 1] is our answer.

The 5th number in the cycle is 7.

Therefore, after pressing the Δ key for 68 times, the displayed number will be 7.

Answer 2

To solve the problem of determining the number displayed after pressing the delta key (Δ) 68 times, follow these steps:

1. Initially, Haruka enters the number 1. This is a one-digit number.
2. Applying Rule 1: If the display shows a one-digit number, pressing delta key Δ replaces the displayed number with twice its value.
3. First Press: The number changes from 1 to 2 (twice of 1).
4. Second Press: The number changes from 2 to 4 (twice of 2).
5. Third Press: The number changes from 4 to 8 (twice of 4).
6. Fourth Press: Since 8 is a one-digit number, apply Rule 1 again, changing the number from 8 to 16 (twice of 8).
7. Fifth Press: The number is now 16, a two-digit number. Apply Rule 2: If the display shows a two-digit number, pressing delta key Δ replaces it with the sum of the two digits.
8. Sixth Press: The sum of digits in 16 is 1 + 6 = 7.
9. Seventh Press: The number is now 7, a one-digit number. Applying Rule 1, it changes to 14 (twice of 7).
10. Eighth Press: Apply Rule 2 on 14, and the sum of digits is 1 + 4 = 5.
11. Ninth Press: The number is now 5; applying Rule 1 changes it to 10 (twice of 5).
12. Tenth Press: Apply Rule 2 on 10, and the sum of digits is 1 + 0 = 1.
13. This cycle of transformations continues:
14. Cycle Repeats: After obtaining 1 again, the next values are 2, 4, 8, 16, 7, 14, 5, 10, and then back to 1. This cycle has a length of 9 transformations.
15. Calculate Total Cycles: With 68 presses, determine how many complete cycles and extra presses occurred. Dividing 68 by 9 gives a quotient of 7 (representing full cycles) with a remainder of 5 (extra presses into the next cycle).
16. Performing 5 extra presses beginning from 1 results in the sequence: 1 → 2 → 4 → 8 → 16 → 7.

Thus, after pressing the delta key 68 times, the displayed number is 7.