A school's lockers are numbered 1 to 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, and so on. After all the students have passed the lockers, which lockers are open?
How can this be solved????
2006-10-22 15:00:16 · 6 answers · asked by Shonque 1 in Science & Mathematics ➔ Mathematics
If the pattern repeats, such that the fourth student would follow the pattern of the first, the fifth the pattern of the second, and the sixth the pattern of the third, then the 100th student would follow the first pattern and open all the lockers.
If the pattern continues as the first opens all, the second closes every other one, the third either closes or opens every third, and the fourht either opens or closes every fourth, then the number of factors is the answer. Locker numbers with an odd number of factors would be open, and locker numbers with a even number of factors would be closed.
2006-10-22 16:47:17 · answer #1 · answered by Leprechaun08 1 · 1⤊ 0⤋
All the lockers get opened.
Then, there are two options, the second student closes the locker, or it remains open.
Then even numbered lockers are now closed, and the odd numbered lockers remain open.
Now, everylocker can have another 2 outcomes from here.
If the locker is a multiple of 3, it will change its state again... So...
If the locker number is even and divisible by 3, it was closed, not it is open.
If the locker number is even, but not divisible by 3, it was closed, and remains closed.
If the locker number is odd, and divisible by 3, it was open, now it is closed.
If the locker number is odd, and t is not divised by 3, it was open and will remain open
so, the lockers that are open are the lockers which are even and divisible by 3, which would be the multiples of 6. as well *** the lockers that are odd and are not divisible by 3.
1+6X 5+6x
1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59,61,,65,67,71,73,,77,79,83,85,89,91,95,97,
so the locker numbers are
l=6x+0
l=6x+1
l=6x+2
2006-10-22 15:49:06 · answer #2 · answered by Amis 1 · 0⤊ 0⤋
The first locker is opened by student #1 and will remain open.
The second locker (a prime number) is opened by st. #1, but closed by st. #2, where it will remain. Indeed, ALL prime numbers (besides 1) will be closed by the student with that number.
Locker 4 is opened by st.#1, closed by #2, and opened by #4.
Locker 6 is opened by 1, closed by 2, opened by 3, closed by 6.
Locker 8: op by 1, cl by 2, op by 4, cl by 8.
L9: op 1, cl 3, op 9.
L12: op 1, cl 2, op 3, cl 4, op 6, cl 12.
Then trend here is that any number with an EVEN number of UNIQUE (including non-prime) factors, will be closed. All others are open.
Thus 24 has factors 1,2,3,4,6,8,12 & 25 - closed. 25 has factors 1, 5 & 25 - open. 31 is closed. 72 has factors 1,2,3,4,6,8,9,12,24 & 36 - an even number of factors.
2006-10-22 15:46:36 · answer #3 · answered by warmspirited 3 · 0⤊ 0⤋
Hi Shonque
In addition to the first answer, there is an easier way to determine which numbers have an odd number of factors: perfect squares (ask yourself why).
So lockers 1,4,9,16,25,36,49,64,81,100 will be open at the end. Others have an even number of factors, meaning every time they are opened they will be closed again.
Hope this helps!
The Chicken
2006-10-22 15:12:09 · answer #4 · answered by Magic Chicken 3 · 0⤊ 0⤋
if this pattern keeps repeating, the student number 100 will open all lockers, so all lockers will be open.
2006-10-22 15:08:50 · answer #5 · answered by arleth m 2 · 0⤊ 0⤋
Go through all of the numbers, 1 to 100, and see how many factors each one has. If it has an odd number of factors, it will be open. If it has an odd number of factors, it will be closed.
2006-10-22 15:07:08 · answer #6 · answered by DavidK93 7 · 0⤊ 0⤋