The best way to do phone verification is to stop facebook from giving it the first place. Suggestion below. ------------------- Facebook are forcing people to put in their personal info to their facebook account specially phone numbers. I tried creating 100 facebook accounts with random emails. I use 1 IP address per account and I clear history of browser and ccleaner after each account. I am not a noob and know what I am doing but the results looks bad. Out of 100 registered accounts, I was only able to successfully make 2 accounts. All the other 98 is roadblock and require phone verification. The only way I can see how facebook will stop doing this is that we need an account creator bot which will allow us to put in phone numbers. We need to compile a real listing of cell phone numbers of establishments such as restaurants, news services, reporters, customer support services, hotels, your boss and other biz services that uses a cellphone. We use the bot to create accounts and once facebook starts sending 50,000 to 100,000 a day for phone text verification messages to random people, facebook will get lot of complaints from these establishments. Facebook will have no choice but to stop their phone verification and it will eventually lift the roadblock for creating new accounts. Facebook will become a big cellphone spam machine and a lot of people will be annoy for privacy issues. If this project already exist pls let me know and will buy it. If someone is willing to make this project, PM me and I am willing to donate. If you have suggestions on how to do this better post here. I am just tired of facebook roadblocks. Thank you.
For how much do you sell them? And how much do you sell fan pages? What are the niches? What does "facebook phone codes" mean? Tks!
Your other option is to buy a phone number and forward it to Skype. http://buyacircle.com/craigslistpva.html http://www.inumbr.com/ http://www.attlines.com/
I have not test yet. But It can help: Code: freevoipnews.com/virtual_phone_line/did_asia.htm I could see one for 2.5$ I am trying to find the 0.2$ ones.
