How do you balance this redox reaction using the oxidation-number change method?

KClO3(s)?KCl(s)+O2(g)

Cl in ClO3 = 5+ and becomes 1- in KCl, gains 6 e-, reduced
O in ClO3 = 2- and becomes 0 in O2, loses 2e-, oxidized

Cl 5+ + 6e- Cl-
O 2- O2 + 2e-

to cancel out the electrons, multiply by a multiples of a common factor, in this case, 6
Cl5+ + 6e- Cl-
3O 2- 3O2 + 6e-
cancel out the electrons and combine the elements/compounds

KClO3 KCl + 3O2

Except Caroline's equation isn't balanced.

The balanced equation, which you can easily balance by inspection, is...
2KClO3(s) ? 2KCl(s) + 3O2(g)

With all due respect, just a simple look at this one will allow you to balance it by inspection.  Clearly, the number of oxygen atoms on the right hand side must be in multiples of 2, which will always be an even number, and there are three oxygen atoms on the left.  Doubling that number will give you 6, also an even number.  Multiplying O2 by 3 will give you six oxygen atoms, and then balancing K and Cl is trivial.   Using the methods reserved for more complicated redox reactions will simply slow you down.

So, where did CM go wrong? Right here:

Cl 5+ + 6e- Cl-
O 2- O2 + 2e-

The second equation isn't balanced for atoms, so let's fix it:

Cl 5+ + 6e- Cl-
O 2- (1/2)O2 + 2e-

Let's use a factor of three to balance the electrons:

Cl 5+ + 6e- Cl-
3O 2- (3/2)O2 + 6e-

Let's combine the equations:

ClO3^- ---> Cl- + (3/2)O2

Notice how I made the chlorate. Now, let's put some potassium ion back in:

KClO3 ---> KCl + (3/2)O2

It's now balanced, but we need to get rid of the fraction, so multiply through by 2:

2KClO3 ---> 2KCl + 3O2

pisgachemist says:

"With all due respect, just a simple look at this one will allow you to balance it by inspection."

He's right, but the point of the question is to learn the oxidation-number change method and that is best done with a simple example. The mch more common method is the half-reaction method. I have lots of discussion about it here:

http://www.chemteam.info/Redox/Redox.html