20160711, 23:57  #23 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 
because he got 8 for the maximum ( technically that's only if converted to whole values you get 8.21.... otherwise) so plugging the maximum in got the minimum as 6.
Last fiddled with by science_man_88 on 20160711 at 23:58 
20160711, 23:58  #24 
Apr 2010
Over the rainbow
100110000010_{2} Posts 
he substract 9.. (9 repating until number of digit product1) . technically that would give 5.19, so 5... but well...

20160712, 00:01  #25 
May 2016
161_{10} Posts 
no no I say the same number of digits and little difference
example 2111*2399 
20160712, 00:05  #26 
"Forget I exist"
Jul 2009
Dumbassville
8,369 Posts 

20160712, 00:08  #27 
Apr 2010
Over the rainbow
2×1,217 Posts 
One factor HAVE to be close to sqrt(product) otherwise it won't work. wich we could assume to work with RSA100
Last fiddled with by firejuggler on 20160712 at 00:22 
20160712, 00:16  #28 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{3}×241 Posts 
Thank you science man
To maximize the sum the primes should be as far apart as possible. Ignoring the 2, let's work with 3 and 101. The product is 303. The sum is 104. Substituting the product in the formula for the maximum sum we get: Sqrt(2178*303)/22=36.9 37 is a maximum well below the actual sum of 104. Last fiddled with by a1call on 20160712 at 00:20 
20160712, 03:33  #29 
Romulan Interpreter
Jun 2011
Thailand
2^{2}·3·739 Posts 
How close? as in "pollard rho" close? in that case we won't need a new method... baby step, giant step, blah blah, repeat, we found the factors....
If you are able to find the sum of the factors, knowing the product, then you are able to factor any number instantly, using the quadratic equation and Vieta's formula. (sometime we miss RDS. Seriously!) 
20160712, 06:10  #30  
"Robert Gerbicz"
Oct 2005
Hungary
2602_{8} Posts 
Quote:
Quote:


20160712, 06:18  #31 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1011010101111_{2} Posts 
So in summary this method/formula is totally useless. Not only doesn't it provide any useful information when it does work, it is also not guaranteed to give the correct bounds anyway.
1 
20160712, 07:50  #32  
May 2016
7×23 Posts 
Quote:
Why 6534 ? test the program . lets take sqrt(2178*6534*208509803)/ 22 = 2476048 / 22 = 112547 there is a big interval but near sqrt(2178*6534*108509804)/ 22 = 1786200 / 22 = 81190 there is a big interval but near or take 101*3 sqrt(2178*6534*303) / 22= 2984 / 22 = 135 sqrt(2178*6534*(30399)) / 22= 2449 / 22 = 111 is near but over the sum 104 what does it mean . #include<cstring> #include<iostream> using namespace std; long long int inversione(long long int numero) { long long int ris = 0; while(numero) { ris = ris * 10 + numero % 10; numero /= 10; } return ris; } int main () { long long int aa = 0; long long int ii = 0; long long int num1 = 0; long long int a = 0; long long int primo = 0; long long int num = 0; while (aa == 0) { cout<<"Inserire numero superiore a 99 : "; cin >>primo; num = primo; while (num != num1) { if (ii == 0 ) { // num = (6534 * (inversione(num )  num ) ) / 22; // produce serie di numeri che si annullano o ritornano , con qualsiasi numero , test this //num = ((15657 * (inversione(num )  num ) ))/22; // num = (2178 * (inversione(num )  num ) ) /22; // produce 2178 o 21....9....78 e l'opposto 6534 o 65...9...34 con qualsiasi numero , test this num = (2178 * ( num ) )/22; ii++; cout<<"Numero invertito : "<<num; if (num < 0) { num = num * 1; cout<<num; } } if (ii > 0) { num = inversione(num)  num ; cout<<"\n"<<"Numero invertito : "; if (num < 0) { num = num * 1; cout<<num; } else cout<<num; } } cin>>a; } } Last fiddled with by Godzilla on 20160712 at 07:51 

20160712, 08:51  #33 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
13257_{8} Posts 

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