At Mathville Middle School, there were 50 students and 50 lockers. Each year for Pi Day (March 14), the students lined up in alphabetical order and performed the following strange ritual. The first student opened every locker. The second student went to every second locker and closed it. The third student went to every third locker and changed it (i. e. if the locker was open, the student closed it; if the locker was closed, the student opened it). In a similar manner, the fourth, fifth, sixth, . . . student changed every fourth, fifth, sixth, . . . locker. After all 50 students had passed by the lockers, which lockers were left open? PL help me T^T

Step-by-step explanation:

This problem is a bit tedious to work out all at once. The first student opens every locker so lockers 1 to 50 are open.

The second student closes every two lockers so every even locker is closed while every odd locker is open. This will repeat over and over and will take too long.

An alternative for this would be checking the factors of each number. For example, 4 has three unique factors, which are 1, 2 and 4. The number of unique factors determine whether the locker is open or closed. In this case, locker 4 will be open after all 50 students finished. Student 1 opens locker 4, student 2 closes locker 4, student 3 doesn't change locker 4 and student 4 opens locker 4. However, students 5 to 50 will never change the locker.

I can't help put all of the answers at once but this should help.

the answer to this question is yes

step-by-step explanation: