Riddle that is really a math problem - help me solve!

[jackdaniels]

There are 100 on/off switches in a row. There are 100 people who can flick these switches. Each person is assigned a number (1-100) and each will flick a switch once using the following logic: 1st person will flick each of the 100 switches. 2nd person will flick only switches 2,4,6,8 etc... 3rd person will flick only switches 3,6,9 etc... up to 100th person who can only flick switch number 100.

Assuming all switches are OFF and each person goes through the sequence of switching ONE time, how many switches will be in the ON position when the sequence is complete (100th person flicks the 100th switch).

All people must go in order.

[Silent A]

A total of 10 switches will be on.

Only these switches will be on (in white) : <font color="white"> 1,4,9,16,25,36,49,64,81,100 </font>

[jackdaniels]

[ QUOTE ]
A total of 10 switches will be on.

Only these switches will be on (in white) : <font color="white"> 1,4,9,16,25,36,49,64,81,100 </font>

[/ QUOTE ]

Your answer has been verified as correct! TY!

[AWoodside]

Is it true in general silent A that only square numbers have an even number of factors? I had never heard this before.

[edit] It's easy enough to show that every square number must have an even number of factors. I'm wondering if you can show that non-square numbers wont?

[Silent A]

If we include the number itself as a factor then it should be stated this way:

All square numbers have an odd number of factors and all non-squares have an even number.

I'm not exactly sure if a mathematician would consider this a rigorous proof:

For every factor a number has it has to have a coresponding pair unless that pair is the same number, in which case we're dealing with a square number.

If the a given number 'N' is not a square it must have an even number of factors since every one will have a matching pair.

If the number is a perfect square then it will be missing one matching pair and therefore have an odd number of factors.

[Enrique]

It does count as a proof.

This problem is popular in the math olympiad.
I always teach it at the beginning.

There is another (famous) proof, but the pairing up one is prettier .