For Mathmaticians and other smart people?

Question

Ok. So there are a hundred lockers in a school. One hundred kids. The first kid comes and opens all the lockers. Then, a second kid closes every other one. Then a third kid changes every third locker. Like if it was open, you would close it, closed, you would open it... Then the fourth kid changes every fourth locker like the third kid only every fourth one. then the fifth kid cames and he changes every fifth one. Get it? Go on until you get to the hundreth kid. There are ten lockers open. Which ten lockers are open? Hint: one of them is the first locker. If you don't understand im me at [e-mail]. not jerry.

Answer 1

all the square numbers. 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
this is because they have an odd number of factors because they're squares.
each time one of their factors comes along, they open/close the locker. only when this happens a odd number of times is the locker open at the end.

Answer 2

The open lockers are in the following positions:
1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Answer 3

This only happens when x is an even-power, for proof see at bottom.

Choosing the notation will be important:

Define f(x,n) = 1 if n|x ("n divides x"), and 0 otherwise
f(x,n) says whether the locker door gets affected (opened/closed) on the nth pass

Then define s(n) = sum_x=1_x=100 [ f(x,n) ]
This is the final state of the locker door.

So locker n will be open if s(n) is odd.

Prime numbers x have 2 factors: x and 1.
Hence s(x) = 0 for prime x and they will all be CLOSED.

Squares x=p^2 will have 3 factors total: 1, p and x.
Hence s(x) = 1 for squares and they will all be OPEN.

Fourth powers x=p^4 will have 3 factors total: 1, p, p^2, p^3 and p^4(=x).
Hence s(x) = 1 for fourth powers (16, 81) and they will all be OPEN.

Sixth powers x=p^6 will have 7 factors total: 1, p, p^2, p^3, p^4, p^5 and p^6(=x).
Hence s(x) = 1 for fourth powers (64) and they will all be OPEN.

Numbers x with two distinct factors x=p*q (where p!=q) will have 3 factors total: 1, p, q, and x.
Hence s(x) = 0 for numbers with 2 distinct factors and they will all be CLOSED.

Numbers x with two distinct factors, one of which is repeated, x=p^2*q (where p!=q) will have 6 factors total: 1, p, p^2, q, pq and p^2q=x.
Hence s(x) = 0 for numbers with 2 distinct factors, one of which is repeated, and they will all be CLOSED.
Note that the repeated factor added 2 factors in total.

Numbers x with two distinct factors, both of which are repeated, x=p^2*q^2 (where p!=q) will have 8 factors total: 1, p, p^2, q, q^2, pq, p^2.q and p^2.q^2=x.
Hence s(x) = 0 for numbers with 2 distinct factors, both repeated, and they will all be CLOSED.
Note that the second repeated factor added 2 factors in total.

Numbers with three distinct factors x=p*q*r will have 8 factors total: 1, p, q, r, pq, pr, qr, and pqr=x.
Hence s(x) = 0 for numbers with three distinct factors and they will all be CLOSED.

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Here's a proof why only pure even-powers (squares, 4th-powers, 6th-powers) satisfy this:
Do a prime decomposition of x:
x = p_1^alpha_1 . p_2^alpha_2 ... p_i^alpha_i
where its distinct prime factors are p_1, p_2, ... p_i and their multiplicities are alpha_i

Now each of the different factors of x is given by:
p_1^a_1 . p_2^a_2 ... p_i^a_i
where 0<= a_i <= alpha_i
There are (alpha_i + 1) different integers from 0 to alpha_i.
Then the number of total different factors of x :
s(x) = (alpha_1 + 1)*(alpha_2 + 1)*...(alpha_i + 1), and the question is when is s(x) odd?

s(x) can only be odd if all the (alpha_i + 1) are odd, i.e. the multiplicities alpha_i are all even
This only happens when x is an even-power.

Answer 4

lockers 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100

Answer 5

1, 11, 22, 33, 44, 55, 66, 77, 88, 99 am i right