How to prove $\gcd(a_1,\dots,ka_i,ka_{i+1},\dots,a_n) \mid k\cdot\gcd(a_1,\dots,a_i,a_{i+1},\dots,a_n)$

Asked Apr 22 '17 at 08:22

Active Apr 22 '17 at 12:47

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How to prove: $$\gcd(a_1,\dots,k*a_i,k*a_{i+1},\dots,a_n) \mid k\cdot\gcd(a_1,\dots,a_i,a_{i+1},\dots,a_n)$$

maybe i have to change this question? I'm just reading a contest solution Codeforces Round #410 (Div. 2) Editorial

In problem C he use that $$\gcd(a_1,\dots,2*a_i,2*a_{i+1},\dots,a_n)\mid2\cdot\gcd(a_1,\dots,a_i,a_{i+1},\dots,a_n)$$

edited Apr 22 '17 at 12:08

Martin Sleziak

asked Apr 22 '17 at 08:22

ENE KaIku

I just don't know how to prove it with mathematical words. – ENE KaIku Apr 22 '17 at 08:24
What about if k=2? – ENE KaIku Apr 22 '17 at 08:26
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Please share your thoughts so far :) – Shaun Apr 22 '17 at 08:26
Just two of the $a_j$'s multiplied by $k$? Or all of them? – Magdiragdag Apr 22 '17 at 08:29
@Magdiragdag just two – ENE KaIku Apr 22 '17 at 08:36
This follows from stronger claim that $\gcd(ka_1,\dots,ka_n) = k \gcd(a_1,\dots,a_n)$. Combined with the fact that $\gcd(a_1,\dots,ka_i,ka_{i+1},\dots,a_n) \mid \gcd(ka_1,\dots,ka_i,ka_{i+1},\dots,ka_n)$ you get the claim you are asking about. Maybe you can start by thinking about two variables rather than $n$, i.e., proof that $\gcd(ka,kb)=k\gcd(a,b)$. – Martin Sleziak Apr 22 '17 at 12:04
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For the case of two variables, you should be able to find several posts on this site. For example this one: How to prove that $z\gcd(a,b)=\gcd(za,zb)$. And also other posts linked there. You might find some other similar posts using Approach0. – Martin Sleziak Apr 22 '17 at 12:06

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