+0  
 
-1
70
3
avatar+195 

(deleted)

 Apr 29, 2020
edited by somehelpplease  May 8, 2020

Best Answer 

 #1
avatar+656 
+1

 

We can use Euclid's algorithm for a few steps:

gcd(2π‘Ž2+29π‘Ž+65,π‘Ž+13)=gcd(2π‘Ž2+29π‘Ž+65βˆ’2π‘Ž(π‘Ž+13),π‘Ž+13)=gcd(3π‘Ž+65,π‘Ž+13)=gcd(3π‘Ž+65βˆ’3(π‘Ž+13),π‘Ž+13)=gcd(26,π‘Ž+13)gcd(2a2+29a+65,a+13)=gcd(2a2+29a+65βˆ’2a(a+13),a+13)=gcd(3a+65,a+13)=gcd(3a+65βˆ’3(a+13),a+13)=gcd(26,a+13)

which is necessarily 1,2,131,2,13 or 2626, just by looking at the first term. By factoring 11831183, and considering that π‘Ža is an odd multiple, you should be able to conclude the rest

 

I'm sorry for the bad spacing I wrote this in apple notes and copy-pasted

 

Here to help

:D

 Apr 29, 2020
 #1
avatar+656 
+1
Best Answer

 

We can use Euclid's algorithm for a few steps:

gcd(2π‘Ž2+29π‘Ž+65,π‘Ž+13)=gcd(2π‘Ž2+29π‘Ž+65βˆ’2π‘Ž(π‘Ž+13),π‘Ž+13)=gcd(3π‘Ž+65,π‘Ž+13)=gcd(3π‘Ž+65βˆ’3(π‘Ž+13),π‘Ž+13)=gcd(26,π‘Ž+13)gcd(2a2+29a+65,a+13)=gcd(2a2+29a+65βˆ’2a(a+13),a+13)=gcd(3a+65,a+13)=gcd(3a+65βˆ’3(a+13),a+13)=gcd(26,a+13)

which is necessarily 1,2,131,2,13 or 2626, just by looking at the first term. By factoring 11831183, and considering that π‘Ža is an odd multiple, you should be able to conclude the rest

 

I'm sorry for the bad spacing I wrote this in apple notes and copy-pasted

 

Here to help

:D

LuckyDucky Apr 29, 2020
 #2
avatar+195 
+1

Haha, nah it's fine. Thanks for helping :D

somehelpplease  Apr 29, 2020
 #3
avatar+368 
+2

Ooh, I've seen this one, it was on stack exchange! I don't want to just copy it from there though, so here's the link (very good hints):

 

https://math.stackexchange.com/questions/2513736/given-that-a-is-an-odd-multiple-of-1183-find-the-greatest-common-divisor-of

CentsLord  Apr 29, 2020

9 Online Users

avatar