20170521, 16:27  #1 
May 2016
162_{10} Posts 
Do you know this method to factorize?

20170521, 22:00  #2 
Aug 2006
3^{2}·5·7·19 Posts 
I can't make heads or tails of what's being proposed.

20170521, 23:44  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
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.

20170522, 12:41  #4  
Feb 2017
Nowhere
2^{5}·139 Posts 
Quote:
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;) Beyond that, the idea seems to be to refine the estimate based on which next refinement gives the smallest remainder. Uhh.... Perhaps our wouldbe 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 20170522 at 12:44 

20170522, 13:42  #5  
May 2016
242_{8} Posts 
Quote:
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 ... 

20170522, 14:00  #6  
Feb 2017
Nowhere
2^{5}·139 Posts 
Quote:
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20170522, 14:25  #7 
May 2016
10100010_{2} Posts 

20170522, 14:43  #8  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts 
Quote:
If you insist: 519836782432571974626653699345320222861224472490026537370861305872053880051248110990757446907 The factor has ...eh, maybe 2 digits. Maybe 12. You will tell us, obviously. 

20170522, 14:47  #9  
Aug 2006
1761_{16} Posts 
Quote:
10digit times 10digit: 17208259488206089057 20digit times 20digit: 768990307312430256760320658890412146149 30digit time 30digit: 309540903855485534899825306555081807764736360354385769798451 35digit times 40digit: 408951906146635774193681294678449630155609891027073698412735684902411405169 40digit times 40digit: 78231521636601247875336457900784367599128092389512275942729748682585263442515309 40digit times 45digit: 1250015611758285449780677590159693571294585219474798791002535377094492021165857842403 45digit times 45digit: 300210011371785203667563068531111049887669890967777805984194291037075428494660807769014091 50digit times 50digit: 537921660353926414896302666235982529116250251251160084488854041470524855208690432807766929132905963 50digit times 100digit: 34909135056115217061649401039290460511254141223373683287880568463749454374891016444864822929356360310066154225950337205088302480731655345147821877409 75digit times 75digit: 810218089621942204409278127804621588719598980295415395004109335746529975694728213544068302872840848666331339803093362360079108228255193625773167313627 

20170522, 14:58  #10 
May 2016
2×3^{4} Posts 
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 fivedigit 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 fourdigit 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. 
20170522, 15:22  #11  
Feb 2017
Nowhere
2^{5}·139 Posts 
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