The prime factors of 13195 are 5, 7, 13 and 29.

What is the largest prime factor of the number 600851475143 ?

format PE GUI 4.0
entry start

include '%fasminc%\win32a.inc'
include 'last.inc'
include 'factor.inc'

section '.data' data readable writeable

bignum  dd      0xE589EAC7
  dd      0x8B
        dd      0
temp       dd      0
   dd      0
   dd      0
hello      db      'Answer',0
sresult rb      20

section '.code' code readable writeable

start:   
    mov ebx, 2   ;divisor
       mov     esi,sresult

begin:   cmp     ebx,0xFFFFFFFF
      jae      exit

   mov     ecx,[bignum+8]
      mov     [temp+8],ecx
        mov     ecx,[bignum+4]
      mov     [temp+4],ecx
        mov     ecx,[bignum]
        mov     [temp],ecx

      mov edx, 0
  mov eax, [bignum+8]   ;EDX:EAX = number to divide
       div ebx
     mov [bignum+8], eax

     mov eax, [bignum+4]
 div ebx
     mov [bignum+4], eax

     mov eax, [bignum]
   div ebx
     mov [bignum], eax

       test    edx,edx
     jne     yes
 
    push    ebx
 push    esi
 stdcall int2a

   cmp     ebx,9
       jle     @f
  jmp     n1

@@:       add     esi,1
       jmp     next

n1:     cmp     ebx,99
      jle     @f
  jmp     n2

@@:       add     esi,2
       jmp     next

n2:     cmp     ebx,999
     jle     @f
  jmp     n3

@@:       add     esi,3
       jmp     next

n3:     cmp     ebx,9999
    jle     @f
  jmp     n4

@@:       add     esi,4
       jmp     next

n4:     cmp     ebx,99999
   jle     @f
  jmp     n5

@@:       add     esi,5
       jmp     next

n5:     cmp     ebx,999999
  jle     @f
  jmp     n6

@@:       add     esi,6
       jmp     next

n6:     cmp     ebx,9999999
 jle     @f
  jmp     n7

@@:       add     esi,7
       jmp     next

n7:     cmp     ebx,99999999
        jle     @f
  jmp     n8

@@:       add     esi,8
       jmp     next

n8:     cmp     ebx,999999999
       jle     @f
  jmp     more

@@:     add     esi,9
       jmp     next

more:   add     esi,10

next: mov     byte [esi],32
       add     esi,1
       jmp     begin

yes:
       add     ebx,1
       mov     ecx,[temp+8]
        mov     [bignum+8],ecx
      mov     ecx,[temp+4]
        mov     [bignum+4],ecx
      mov     ecx,[temp]
  mov     [bignum],ecx
        jmp     begin   


exit:
       invoke MessageBox,NULL,sresult,hello,MB_OK
  invoke ExitProcess,0

    ret

section '.idata' import data readable writeable
        library KERNEL32,       'KERNEL32.DLL',\
         USER32,         'USER32.DLL'

import    KERNEL32,\
                 ExitProcess,            'ExitProcess'

 import  USER32,\
                   MessageBox,             'MessageBoxA'

    

You mean if result overflows when the carried part is added? hmmm

Hmm, I still don't understand why it takes so long to display the MessageBox. I know I am doing lots of divisions and copying into memory etc, but I thought for such a small program that it would still run fast.

From project euler, other people's programs are running in milliseconds, and although I'm doing a brute force solution that isn't the best, I don't see how it's different from many of the other solutions that were posted.

Any ideas?

And lastly, Good job bitRake. I've seen your solutions to other problems, and they are always so elegant and efficient.

It is possible to factor a number without division or multiplcation! (shameless self reference) Oddly enough this method finds the large factors first as well.

There is an ever increasing volume of algorithms on my web site.

You might want to look at GMP once you're done.

proc factor, num

bsf        ecx, [num]
push      ecx
shr        [num], cl        ;now we have num = num/(2^ecx), which is odd (but not strange)
.
.
.
your code
.
.
.
pop      ebx
retn     4