use a ultra/great ball
Apps can be removed at any time, but all apps that are on the iPod Touch will remain there unless you delete them.
sure you can download an app called "apshopper" with that app you will know which paid apps are given for free just for today. each day thousands of paid apps are offered for free for a few hours. Another way is to register with http://appiration.com and get not only apps for free but earn a little bonus. All for reviewing the apps. You can also check appbox.tinyviews.com everyday to get paid apps for free without jailbreaking.
try swamperttools.webs.com its a pretty good place for all generations of Pokemon ROM hacking tools
Yes you can. Get the poke modifier.
people all over will give you poke apps also the Pokemon company in jublife city invents poke apps so go there offtan
hoenn
What game?
at the poke app store in jubilife city when you talk to the president on the right door.
i play that game and asked the same question but forgot to share
You cannot get all apps for free without jailbreaking an iPod Touch. It is not possible.
try a more brod qestion like how do you get all the poke apps.
no.
Poke Tech 23,24,and 25 are all obtained through Nintendo Events. For a list of all the Apps go to: http://www.serebii.net/diamondpearl/poketch.shtml
No it does not. All apps from installous are free but you will not be able to sync the apps into itunes. you will not be able to sync unless you delete the apps. If you get appsync from cydia you can sync apps without deleting.
ALL and i mean ALL types of poke-mon unless your poke-mon is like at least i dont know mab 20-30 levels highr which is kinda hard but dragon, dark, flying, fire, electric all tyepes without all of them its kinda hard even if the poke-mon your battleing iz like 40 levels below you if its a poke-mon that your weak to your gonna need a poke-mon that that poke-mon is weak to so all types of poke-mon will help you along the way. GOOD LUCK!!!!!!
It is not named. It's all built in, you can't change it without tons of hacking.