I was just wondering how everyone solves puzzles like the 2x2x3, 2x2x4, 2x2x5, etc.

I've found the easiest way is to use Square-1 corner permutation algorithms.

I use the method shown in this video for the 2x2x3 and am able to apply it to all 2x2xn puzzles

http://www.youtube.com/watch?v=3UE_5PVwup0

The 2x2x3 tutorial part starts at around a minute in.

That's what I use on my 2x2x3 and 2x2x4.

With 2x2x3 I first finish the middle edges and the upper layer. Then I use the s-1 algorithm to 3-cycle the bottom layer corners (= use twice the algorithm that swaps two adjacent corners in the bottom layer and two diagonal corners on the upper layer, along with a twist of the bottom layer in between the two).

With 2x2x4 I first get it into shape, then pair up the outside layer pieces with the corresponding middle layer pieces (using the s-1 algorithm as above). Finally, I solve it like a 2x2x2.

I use these algorithms http://www.cubezone.be/square1step4.html

For 2x2x4/2x2x6 I solve shape first. Then middle layer with the algorithms. Then last ones.

For 2x3x3/3x3x4 I solve the corners with the algorithms then a series of R2/L2 U2 moves to switch edge pieces around. Then R2 U R2 U R2 U2 R2 U2 R2 U R2 U'R2

Also works to solve 2x2x2 Just solve 2 opposite sides like yellow and white (don't have to be with each other as long as there on opposite sides). Orient the yellow and white. The permute the corners. It works pretty well, I can average 8 sec. with this method.

solve top and bottom layers with ortega.
orient one layer, then the other. say u follow the BOY color scheme, either orient the yellow or the white face first and the other face would automatically be oriented...(ie make sure yellow and white are on their respective layers)
then permute the corners with an ortega algorithm that solves the corners when there are bars on both the top and the bottom layers...if it is not possible to do so, apply R2 U R2 U' R2 U R2 U' R2

for 2x2x3, the middle layer is just trivial stuff...
for 2x2x4 and up, I would just use simple commutators like r U2 l D2 l' U2 l D2 l' r' (after rotation such that the corners solved are on the right and left) to solve the (n-a)th and (a)th layer at the same time where a is an integer < n. repeat until all layers are done? XD for odd layered ones, the middle layer can be solved trivially and the rest solved as described above.

hope this helps...

Wasn't looking for help. Was just curious as how other solve because I've seen other solutions ,and was wondering how they worked. Thanks though .