Math Problem Statement
A ballroom has a row of closed cubical boxes, numbered 1 to 4096 Assuming there is a spherical ball inside each box. Yash enters the ballroom and opens the box numbered 1, and then alternates between skipping and opening each closed box. Thereafter, when he reaches the end of the room, he turns around and starts back. He opens the first box, he encounters and then altenates between skipping and opening each closed box. Thereafter, Yash continues wandering back and forth in this manner until all 4096 balls can come out of the cubical box at the same time, Now, the number of last box Yash opens is $\alpha$. Then find the value of $[\sqrt{\alpha}]$. (where [.] $\rightarrow$ GIF function)
Solution
Let's break down the problem step by step.
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Understanding Yash's movement:
- Yash starts by opening the first box (Box 1).
- He alternates between skipping and opening each closed box as he moves forward.
- Upon reaching the last box (Box 4096), he turns around and does the same while moving backward.
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Condition for a box to be opened:
- For a box to be opened, Yash must pass over it an odd number of times. This is because the box starts closed, and each time Yash passes over it, he toggles its state (open/close).
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How often Yash passes over a particular box:
- A box numbered will be passed over by Yash each time he completes a lap. Specifically, he will pass over box each time he completes a round trip that includes box .
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Calculating the number of times Yash passes over a specific box:
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The number of times Yash passes over a box is directly related to how many divisors the number has. Each divisor corresponds to a point at which Yash turns around at that position.
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For example, if , its divisors are . Therefore, box 12 will be toggled 6 times (each divisor represents a point where Yash turns or starts at position 12).
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Condition for a box to remain open:
- A box will remain open if it has an odd number of divisors, which happens if and only if is a perfect square.
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Finding the largest which is a perfect square and within the range:
- The largest perfect square less than or equal to 4096 is , because .
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Number of the last box that Yash opens:
- The last box that remains open, which Yash would open, corresponds to .
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Calculating :
- Here, . So, .
Thus, the value of , where , is .
Would you like further details or have any questions? Here are some additional related questions:
- What if the total number of boxes were 8192?
- What if Yash skips two boxes before opening one?
- How does the number of passes relate to the number of divisors of a box number?
- What happens if the number of boxes is not a perfect square?
- How does the strategy change if Yash skips every other box?
- What is the significance of the square root in this problem?
- How would the result change if Yash turned around every 100 boxes instead of at the end?
- What is the general formula for the largest box number Yash opens for any number of boxes ?
Tip: When dealing with problems involving divisors, consider how divisors of a number relate to perfect squares, as this often reveals patterns.
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Math Problem Analysis
Mathematical Concepts
Number Theory
Divisors
Perfect Squares
Formulas
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Theorems
Perfect Square Theorem
Suitable Grade Level
Grades 9-12
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