Reasoning Word Problem: A school's lockers are numbered 1-100. One hundred students enter the school one at a time. The first student opens the lockers. The second student closes the even-numbered lockers. The third student either closes or opens every third locker. The remaining students continue the pattern. After all the students have passed the lockers, which lockers are open.
o_o ummmm what????? HELP ME PLEASE!!
Sure, let's break it down step by step.
1. Initially, all the lockers are closed.
2. The first student opens all the lockers.
3. The second student closes all the even-numbered lockers (2, 4, 6, ...).
4. The third student either closes the opened lockers or opens the closed lockers.
5. The fourth student closes the lockers that were opened by the second student or opens those that were closed.
6. The fifth student repeats the same pattern as the third student.
7. This pattern continues until all 100 students have passed the lockers.
Now, let's analyze which lockers will be open after all the students have passed:
For any locker, it will be toggled (opened or closed) as many times as there are factors of its corresponding number (e.g., locker 15 will be toggled by the third and fifth students since 15 has 2 factors: 1 and 15).
For a locker to remain open, it must have an odd number of factors. The only numbers that have an odd number of factors are the perfect squares (e.g., 1, 4, 9, 16, ...).
Therefore, the lockers that will remain open are those with perfect square numbers. In this case, lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be open.
I hope this clarifies things!
No worries! I'll break down the problem and explain how to solve it step by step.
We have 100 lockers numbered from 1 to 100. Each locker can be either open or closed. The problem states that 100 students pass by the lockers one at a time, and they each perform a specific action on the lockers based on their position.
Let's analyze what each student does:
1st student: Opens all the lockers.
2nd student: Closes all the even-numbered lockers (2, 4, 6, ... 100).
3rd student: For every third locker, if it's open they close it, if it's closed they open it.
4th student: For every fourth locker, if it's open they close it, if it's closed they open it.
...
And so on, until the 100th student.
To solve this problem, we can think about the status of each locker after every student has passed by:
1st student: All the lockers are open.
2nd student: Even-numbered lockers are closed, odd-numbered lockers remain open.
3rd student: Lockers divisible by 3 (3, 6, 9, ...) change their status (from open to closed, or closed to open).
4th student: Lockers divisible by 4 (4, 8, 12, ...) change their status.
...
And so on, until the 100th student.
So, if we go through each locker and count how many times it gets switched, the number of times it gets switched will determine if it ends up open or closed.
For example, let's take locker number 12:
- The 1st student opens it.
- The 2nd student closes it.
- The 3rd student opens it.
- The 4th student closes it.
- The 6th student opens it.
- The 12th student closes it.
Locker 12 gets switched 6 times, which is an even number. This means that it ends up closed.
To find out which lockers are open at the end, we need to determine which lockers have an odd number of switches. If the number of switches is odd, the locker is open; if it's even, the locker is closed.
For a locker to have an odd number of switches, it must have an odd number of factors. The only numbers that have an odd number of factors are perfect squares. So, the lockers that end up open at the end are the perfect squares between 1 and 100.
Therefore, the open lockers are lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
I hope that helps! Let me know if you have any more questions.