# is this useful to help break RSA?

So, I'm kinda newbie in this cryptography stuff, I just like to understand complex things.A couple of months ago I was learning and I maybe found a pattern in prime numbers and their multiplications, lets check out.

So, it's high demanding to multiplicate random prime numbers, correct? I thought this: Instead of multiplicate the numbers, you can just isolate the last number of each one and sum them up, for example: 23518 and 619741, we can just take the last digit of both and get 2, right? I found out that, if the sum result equals to 2, the ending number in the multiplication it will ALWAYS be 1 (tested with 20k prime numbers) This also holds true for all sums (2, 4, 6, 8, 10, 12, 14, 16 and 18) Table: 1 [(1, 1), (3, 7), (9, 9)] 3 [(1, 3), (7, 9)] 5 [(1, 5), (3, 5), (5, 5), (5, 7), (5, 9)] 7 [(1, 7), (3, 9)] 9 [(1, 9), (3, 3), (7, 7)]

So in the RSA-260 challenge number

( 2211282552952966643528108525502623092761208950247001539441374831912882294140 2001986512729726569746599085900330031400051170742204560859276357953757185954 2988389587092292384910067030341246205457845664136645406842143612930176940208 46391065875914794251435144458199) we know that the last digit sum MUST BE, 6, 10 or 14, right? reducing our possibilities by a lot.

Can anyone confirm this for me? I'm really dumb in programming but I think this is a start. Check out the spreadsheet I did. https://docs.google.com/spreadsheets/d/1rcAIz5094WGqVZLGgCiTZotzrHg8TeQ2kQSB6lhAEaU/edit?usp=sharing