Hello,
My buyer account was blocked 10 days ago because, according to them, it was linked to a previous blocked one which abused of their refunding policy ! Obviously, I've only one. But I remembered that I moved about a month ago without having updated my personal information. But it is likely that the new owner - therefore using my old billing address - had some issues with his package as I've had in the past (theft in the mailbox by a gang of neighborhood that apparently still has not worried by justice!). I had received a warning at the time of the famous CIS. That was a few months ago.
I have sent a copy of my ID card + proof of address (my address), I have received no return ... Obviously, every time I want to create a new account, it is blocked, as stipulated in their email.
My account was created in 2010 and I make regular payments on this site because it is handy it must be said. I therefore create a false account, but only to buy. I bought the ebook but I stayed on my hunger. So I'd like your opinion on my way to create a customer account that is not linked to the old (and grandchildren).
1) hire a VPS server with
Dedicated IP address (therefore not blacklisted) is in France but far from the billing/shipping adress that are also in France
=> Fingerprint fresh but maybe it can block if the distance between the 2 addresses (which are in the same town) and the location of the IP is too big?
2) VCC + + false name + false first name + prepaid SIM card.
3) Regarding the billing address, the street is real but the number is wrong, as no risk of it being already blacklisted. The shipping address will always be a pick-up station anyway.
With regard to the verification stage of the address during the first command, it was easy to bypass during my trial by sending a fax using a PSD file that I found on the Net for making a false utility bills + a screenshot of the account of the VCC with the CC number and my name / first name highlighted
Do you think this set-up is enough clean to do not be linked to my previous account ?
Thanks.