Is there a way to differentiate between numbers

epsi00
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like N=5053=163*31 and N=169*37=6253 if we do not know the factors and if we do not want to factor them. They both have the same last two digits.
I can't think of any test to apply to this kind of numbers ( it's in fact a family of numbers that share the same last two digits but are product of (last digit only ) 3*1 or 7*9 ).
 
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Here's what I understand from your post. You have two disjoint sets of numbers, A and B. You want a method to determine whether a number is a member of set A or set B under the assumption that it is in one of those, without factoring the number.

If that's right, would you define A and B more precisely?
 
CRGreathouse said:
Here's what I understand from your post. You have two disjoint sets of numbers, A and B. You want a method to determine whether a number is a member of set A or set B under the assumption that it is in one of those, without factoring the number.

If that's right, would you define A and B more precisely?

That's correct. A and B both are of the form 6*k+1. and multiplication ( within each set ) of two numbers ( whose last digits can be a 3*1 or a 7*9) (from the two different sets) always produces two numbers with the last two digits ( not necessarily 53 and not necessarily the same) 13 or 33 or 93 or 73).

example:
163*31 = 5053 a number produced by a 3*1 multiplication ( last digits only )
169*37 = 6253 a number produced by a 9*7 multiplication ( last digits only )

another example would be:
133*61 = 8113
139*67 = 9313

but also
103*31 = 3193
109*37 = 4033

I am interested in finding out a test by which, given a number, we can say it's a "3*1" product or a 7*9 ( without having to factor the number ).

thanks
 
What if you can't? 2673 = 81*33 = 27*99
 
hamster143 said:
What if you can't? 2673 = 81*33 = 27*99


your number is not a product of a 6k+1 by another 6k+1. Those numbers obey different rules.
I am only interested in numbers of the form 6*k+1.

multiplication of numbers of the form (6j+1)(6i+1) have periodic "last two digits" like I wrote in my earlier post. So just looking at the last two digits of a number will not tell us if that number was a product of a 3*1 or a 7*9 because these two products produce the same last two digits ( 13, 99, 73 and 53 ). Here 3*1 and 7*9 refer only to the last digit of the factors making up a number. ( 163*31 is a 3*1 product ).
 
epsi00 said:
your number is not a product of a 6k+1 by another 6k+1. Those numbers obey different rules.
I am only interested in numbers of the form 6*k+1.

Okay then, 31*343 =(6*5+1)*(6*57+1) = 49*217 = (6*8+1)*(6*36+1) = 10633.
 
hamster143 said:
Okay then, 31*343 =(6*5+1)*(6*57+1) = 49*217 = (6*8+1)*(6*36+1) = 10633.

true again but so what? I am talking about the general case where the 3*1 number is not of the same value as a 7*9. I consider pairs of numbers, one of each kind.
to 31*343 can be associated 37*349 and I don't think these two are the same.
to 49*217 can be associated 43*211 and I again don't think that they have the same value.

the original question was about a test to tell the difference between a 3*1 and 7*9 number without having to factor them. In some cases the value is the same and my question loses its meaning but like I said I am talking about pairs such that:

N = p*q = a 3*1 number
M = (p+6)*(q+6) = a 7*9 number.

and all the examples I gave are of pairs ( in one of my previous posts )

so is there a test to tell the difference between the two.
 
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