factorization

I have made an observation about prime numbers.

All prime numbers - except 2 and 5 - have as their last digit, either
1 or 3 or 7 or 9. ( it can't be 5 because it would then be divisable
by 5 ).

We notice that the last digit of the multiplication of two prime
numbers is also 1,3,7 or 9. How is that ?

let's take the last digits combinations as they are produced from the
multiplication algorithm :

1x1=1
1x3=3
1x7=7
1x9=9
3x3=9
3x7=21->1
3x9=27->7
7x7=49->9
7x9=63->3
9x9=81->1

So for each number of the form P*Q where P and Q are primes, we know
with big accuracy what is the last digit of the 2 primes P and Q.

The possible combinations for each possibility of last digit are :

number's last digit 1, possible combinations of primes' last digits :

1 and 1
or
3 and 7
or
9 and 9

number's last digit 3, possible combinations of primes' last digits :

1 and 3
or
7 and 9

number's last digit 7, possible combinations of primes' last digits :

1 and 7
or
3 and 9

number's last digit 9, possible combinations of primes' last digits :

1 and 9
or
3 and 3
or
7 and 7

as an example let's the number 38293 ( which is 149*257 ). Its last
digit is 3, so the possible combinations for the last digits of the
primes ( which we don't know ) are :

1 and 3
or
7 and 9

this fact can possibly lead to a factorization in polynomial time or
even faster.

Endri Tasho
Athens, Greece
17/5/2007

.

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