Is gcd(a,b)%m=gcd(a%m,b%m)? [on hold]

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In one of my assignments i was thinking as if gcd(a,b)%m=gcd(a%m,b%m).If yes then please provide a proof and if no then can you provide a method for doing gcd(a,b)%m for large numbers




greatest-common-divisor




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put on hold as off-topic by Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u 21 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Welcome to MSE. What do you mean by %?
    – José Carlos Santos
    yesterday






  • 1




    What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
    – Patrick Stevens
    yesterday






  • 1




    @JoséCarlosSantos In programming, a % b is $a pmodb$.
    – Patrick Stevens
    yesterday







  • 1




    Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
    – rtybase
    yesterday






  • 1




    By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
    – rtybase
    yesterday














up vote
-4
down vote

favorite












In one of my assignments i was thinking as if gcd(a,b)%m=gcd(a%m,b%m).If yes then please provide a proof and if no then can you provide a method for doing gcd(a,b)%m for large numbers




greatest-common-divisor




share|cite|improve this question











put on hold as off-topic by Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u 21 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 2




    Welcome to MSE. What do you mean by %?
    – José Carlos Santos
    yesterday






  • 1




    What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
    – Patrick Stevens
    yesterday






  • 1




    @JoséCarlosSantos In programming, a % b is $a pmodb$.
    – Patrick Stevens
    yesterday







  • 1




    Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
    – rtybase
    yesterday






  • 1




    By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
    – rtybase
    yesterday












up vote
-4
down vote

favorite









up vote
-4
down vote

favorite











In one of my assignments i was thinking as if gcd(a,b)%m=gcd(a%m,b%m).If yes then please provide a proof and if no then can you provide a method for doing gcd(a,b)%m for large numbers




greatest-common-divisor




share|cite|improve this question











In one of my assignments i was thinking as if gcd(a,b)%m=gcd(a%m,b%m).If yes then please provide a proof and if no then can you provide a method for doing gcd(a,b)%m for large numbers





greatest-common-divisor





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked yesterday









Punit Jain

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put on hold as off-topic by Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u 21 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u 21 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Taroccoesbrocco, amWhy, Shailesh, Henrik, Delta-u
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 2




    Welcome to MSE. What do you mean by %?
    – José Carlos Santos
    yesterday






  • 1




    What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
    – Patrick Stevens
    yesterday






  • 1




    @JoséCarlosSantos In programming, a % b is $a pmodb$.
    – Patrick Stevens
    yesterday







  • 1




    Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
    – rtybase
    yesterday






  • 1




    By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
    – rtybase
    yesterday












  • 2




    Welcome to MSE. What do you mean by %?
    – José Carlos Santos
    yesterday






  • 1




    What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
    – Patrick Stevens
    yesterday






  • 1




    @JoséCarlosSantos In programming, a % b is $a pmodb$.
    – Patrick Stevens
    yesterday







  • 1




    Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
    – rtybase
    yesterday






  • 1




    By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
    – rtybase
    yesterday







2




2




Welcome to MSE. What do you mean by %?
– José Carlos Santos
yesterday




Welcome to MSE. What do you mean by %?
– José Carlos Santos
yesterday




1




1




What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
– Patrick Stevens
yesterday




What's $mathrmgcd(0,0)$? If you don't have a good answer to that one, then you might be in trouble.
– Patrick Stevens
yesterday




1




1




@JoséCarlosSantos In programming, a % b is $a pmodb$.
– Patrick Stevens
yesterday





@JoséCarlosSantos In programming, a % b is $a pmodb$.
– Patrick Stevens
yesterday





1




1




Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
– rtybase
yesterday




Also, the answer to the main question is no. $gcd(3cdot 3 + 1, 3cdot 5 + 1)=2$ and $gcd(3cdot 3 + 1, 3cdot 5 + 1) equiv 2 pmod3$ but $$3cdot 3 + 1 equiv 1 pmod3$$ $$3cdot 5 + 1 equiv 1 pmod3$$ and $$gcd(3cdot 3 + 1 pmod3, 3cdot 5 + 1 pmod3) = 1$$
– rtybase
yesterday




1




1




By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
– rtybase
yesterday




By the way, is this question related to codechef.com/AUG18B/problems/GCDMOD?
– rtybase
yesterday















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