20170917, 06:35  #1 
Sep 2017
2^{3}·3 Posts 
math gcd problem need help
Dear all im new in this Greatest forum i have problem in mathematica i need calculate gcd from this string:
Code:
Z1:=(7580104959982914123117804044806461820154572036369549529405552162648856956231245030405896783337631871321017085933334306430501343909365379890910193903441(7601611438277968963930435339966375537762723788929138241505092335032216337024111944145650540579951639515460945444623092013993572866002842083492323809287290^17592186175489)); Z2:= (7580104959982914123117804044806461820154572036369549301773204651044654156328611994019678602790933052049997074737942147009504600337458372628655015658163(5472021051232020294637156058031796107019454396525172260491306567922450488302345750910648307191204110931670031994886144725045972546892458995859985546716414^17592186175489)); Z3:= (7580104959982914050609440727680305970398447676033232174897745114621599628756001756455762578505630805317234747401759098214730033873908508468766892578356(2716401471494295676405217606243925186946305783849093640906079739928627393562677984831598246598185728284424287135181309232467496805930321500988187816347957^17592186175489)); ToUpperCase[IntegerString[(GCD[Z1,Z2,Z3]),16]] 
20170918, 06:50  #2 
∂^{2}ω=0
Sep 2002
República de California
5×2,351 Posts 
Your problem is not 'in mathematica', your problem is that you are blindly attempting to get mathematica to do a computation without having thought through how large those exponentials you are inputting are. So tell us  and you don't need any fancy maths package for this, just some simple digitcounting and basic algebra  roughly how many decimal digits will your GCD inputs have once the exponentiations are done?

20170918, 11:00  #3 
Sep 2017
11000_{2} Posts 
Dear sir,
without mathe matica how its possibel its 64bit gcd computation my laptop dell i7 with 8GB RAM... 
20170918, 11:34  #4 
Jun 2003
2×2,719 Posts 
The problem is the size of the calculation. What you want to do is impossible

20170918, 11:37  #5 
"Rashid Naimi"
Oct 2015
Remote to Here/There
928_{16} Posts 
Those numbers are too large for the available memory. Why don't you try the GCD function without the exponents and then raise it to the power of the smallest exponent?
The GCD (8^a,6^b,18^c)= 2^( Smallest of a,b ,c) You can do all that programmatically if you wish. 
20170918, 11:38  #6 
Sep 2017
2^{3}×3 Posts 
i need 512 rsakey from this string please help me..

20170918, 14:22  #7 
"Curtis"
Feb 2005
Riverside, CA
5643_{10} Posts 

20170918, 14:24  #8 
"Curtis"
Feb 2005
Riverside, CA
3^{3}×11×19 Posts 

20170918, 15:22  #9 
Aug 2006
5,987 Posts 

20170918, 15:48  #10 
Feb 2017
Nowhere
3·31·67 Posts 
Hmm. I'm not familiar with Mathematica syntax. The three integers all seem to be of the form
a  b^17592186175489. The three values of a are 7580104959982914123117804044806461820154572036369549529405552162648856956231245030405896783337631871321017085933334306430501343909365379890910193903441 7580104959982914123117804044806461820154572036369549301773204651044654156328611994019678602790933052049997074737942147009504600337458372628655015658163 7580104959982914050609440727680305970398447676033232174897745114621599628756001756455762578505630805317234747401759098214730033873908508468766892578356 which have a suspiciously long block of MSD's in common. Beyond that, I don't see anything particularly useful. 
20170918, 16:26  #11 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4450_{8} Posts 

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