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Old 2017-05-21, 16:27   #1
Godzilla
 
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Default Do you know this method to factorize?

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Old 2017-05-21, 22:00   #2
CRGreathouse
 
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I can't make heads or tails of what's being proposed.
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Old 2017-05-21, 23:44   #3
science_man_88
 
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Quote:
Originally Posted by CRGreathouse View Post
I can't make heads or tails of what's being proposed.
it fails in step 1 anyways in my books you need to know one of the factors to use it and it states in step 4 you know the two factors but if you know there are only two factors you can just do the division to find the other one.
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Old 2017-05-22, 12:41   #4
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Quote:
Originally Posted by Godzilla View Post
Let's see here...
Quote:
First you have to know the exact number of digits of one of the two factors, if you do not know you have to try it by trial.
If there's a way to determine the size of a factor within an order of magnitude without actually finding that factor, I've never heard of it. If there were such a method, it would work better in binary, since you would have the factor within a factor of two instead of a factor of ten
;-)

Beyond that, the idea seems to be to refine the estimate based on which next refinement gives the smallest remainder. Uhh....

Perhaps our would-be Ruler of Mathematics would care to have a go at

N = 1176987252744371552614042899595067330158641674468716458480731390244611805610729474397547682062674215668721

N is the product of two prime factors.

One of the factors is between 5.85987448204883 x 10^50 and 5.859874482048834 x 10^50.

The other is between 2.00855369231876 x 10^54 and 2.00855369231877 x 10^54.

Last fiddled with by Dr Sardonicus on 2017-05-22 at 12:44
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Old 2017-05-22, 13:42   #5
Godzilla
 
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Quote:
Originally Posted by Dr Sardonicus View Post
Let's see here...
If there's a way to determine the size of a factor within an order of magnitude without actually finding that factor, I've never heard of it. If there were such a method, it would work better in binary, since you would have the factor within a factor of two instead of a factor of ten
;-)

Beyond that, the idea seems to be to refine the estimate based on which next refinement gives the smallest remainder. Uhh....

Perhaps our would-be Ruler of Mathematics would care to have a go at

N = 1176987252744371552614042899595067330158641674468716458480731390244611805610729474397547682062674215668721

N is the product of two prime factors.

One of the factors is between 5.85987448204883 x 10^50 and 5.859874482048834 x 10^50.

The other is between 2.00855369231876 x 10^54 and 2.00855369231877 x 10^54.
Hello,

In this case a factor has only one or two digits ... is it correct?

I have a method to understand the number of digits of a factor ...
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Old 2017-05-22, 14:00   #6
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Quote:
Originally Posted by Godzilla View Post
Hello,

In this case a factor has only one or two digits ... is it correct?
Er, no. I said:

Quote:
One of the factors is between 5.85987448204883 x 10^50 and 5.859874482048834 x 10^50.
That "x 10^50" means something. As does the "x 10^54" for the other factor...

Quote:
I have a method to understand the number of digits of a factor ...
Not on the evidence...
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Old 2017-05-22, 14:25   #7
Godzilla
 
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Quote:
Originally Posted by Dr Sardonicus View Post

Not on the evidence...
Right, I'm wrong, doing calculations with the method, a factor has 50 digits, the method works, can we try with another number please?
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Old 2017-05-22, 14:43   #8
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Quote:
Originally Posted by Godzilla View Post
Right, I'm wrong, doing calculations with the method, a factor has 50 digits, ... the method works, ...
Quote:
Originally Posted by Godzilla View Post

Do you know this method to factorize?
Yes, we know that one. It is called BS.

Quote:
Originally Posted by Godzilla View Post
... can we try with another number please?
If you insist:
519836782432571974626653699345320222861224472490026537370861305872053880051248110990757446907
The factor has ...eh, maybe 2 digits. Maybe 12. You will tell us, obviously.
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Old 2017-05-22, 14:47   #9
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Quote:
Originally Posted by Godzilla View Post
Right, I'm wrong, doing calculations with the method, a factor has 50 digits, the method works, can we try with another number please?
Sure. Take your pick:

10-digit times 10-digit:
17208259488206089057

20-digit times 20-digit:
768990307312430256760320658890412146149

30-digit time 30-digit:
309540903855485534899825306555081807764736360354385769798451

35-digit times 40-digit:
408951906146635774193681294678449630155609891027073698412735684902411405169

40-digit times 40-digit:
78231521636601247875336457900784367599128092389512275942729748682585263442515309

40-digit times 45-digit:
1250015611758285449780677590159693571294585219474798791002535377094492021165857842403

45-digit times 45-digit:
300210011371785203667563068531111049887669890967777805984194291037075428494660807769014091

50-digit times 50-digit:
537921660353926414896302666235982529116250251251160084488854041470524855208690432807766929132905963

50-digit times 100-digit:
34909135056115217061649401039290460511254141223373683287880568463749454374891016444864822929356360310066154225950337205088302480731655345147821877409

75-digit times 75-digit:
810218089621942204409278127804621588719598980295415395004109335746529975694728213544068302872840848666331339803093362360079108228255193625773167313627
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Old 2017-05-22, 14:58   #10
Godzilla
 
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The method is this , Try it if you want.


Then, to find the number of digits of a factor, proceed as follows:

I take in example the number or rather the product N= 2699*4999 = 13492301.


13492301/900000 = 14 Here I split up to six digits and number 9 and it is always
thus, practically two digits less than the product that has eight digits,
Assuming a factor has six digits in this case 900000.


Then

14 * 900000 = 12600000 the result has exactly eight digits and 5 Zeri, but in
Common with the product has only the first or the 1 and as disturbance elements
26.

Repeat the same procedures with a five-digit number or 90000


13492301/90000 = 149
149 * 90000 = 13410000 now has 134 shared, then three digits and 4 Zeri and the
The result has eight digits as the 1th disturbance element.


Then I repeat the four digit 9000 procedure.

13492301/9000 = 1499
1499 * 9000 = 13491000 has now in common 1349 four-digit and 3 zeros and one
1 as a disturbance element and it is always eight digits the result.

I repeat again with three digits 900.


13492301/900 = 14991
14991 * 900 = 13491900 has 1349 common and only 2 Zeros and 19 ie two elements of
disturbance and therefore is lower than the previous one that has 3 zeros and one element of disturbance.

I go with 90.

13492301/90 = 149914
149914 * 90 = 13492260 has 13492 in common and one Zero plus 26 two elements
of disturbance.

Then proceed with the 9.

13492301/9 = 1499144
1499144 * 9 = 13492296 has in common 13492, well three elements of disturbance and
None Zero.

So I chose 9000 because there are so many Zeros and the only element of
Disturbance and of course the common numbers. Now I know it has four digits one
factor.
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Old 2017-05-22, 15:22   #11
Dr Sardonicus
 
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Quote:
Originally Posted by Godzilla View Post
Right, I'm wrong, doing calculations with the method, a factor has 50 digits,
First, you still got the number of digits wrong. Hint: Does 10^1 have 1 digit?

Second:
Quote:
the method works, can we try with another number please?
No. You say you have a method. Let's see you apply it. I gave you an example. I spotted you the size of the factors, and quite a few of the significant digits of each. If, given all that, you can't even apply your own method, then please go away.
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