Question

@KingGeorge Prove or disprove that if a = bm + r then gcd(a,m) = gcd(a,r) @hartnn @jdoe0001 @aaronq @lalaly @jhonyy9 @agent0smith @tkhunny @e.mccormick

Answer 1

For this one, the most important part will be using the fact that if \(d\) divides \(a,b\), then \(d\) divides \(ax+by\) for any choice of integers \(x,y\).

Answer 2

I see right away that r = a-bm

Answer 3

Excellent. So that means that \(\gcd(a,m)\le\gcd(a,r)\). Then when you notice that \(bm=a-r\), you get that \(\gcd(a,r)\le\gcd(a,m)\). Combine those two facts, and you get equality.

Answer 4

Lmao, with a little computation I proved that r=m, therefore the gcd's are the same!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 5

\(r\neq m\). If you get that, something went wrong.

Answer 6

Set a-bm = (a-r)/b and you get r=m

Answer 7

How do you know that \(a-bm=(a-r)/b\)?

Answer 8

\[\large a - bm = r\] \[\large \frac{ a-r }{b } = m\] why would you set them equal to each other?

Answer 9

how did you even get r=m from here...?\[\large a-bm = (a-r)/b\] \[\large ab - b^2m = a-r\] \[\large -ab + b^2m + a = r\] that last step sure doesn't look like r=m.

Answer 10

a-b(a-r/b) = (a-r)/b

a-a+r = a/b - r/b

r=a-r/b

rb=a-r

rb+r=a

rb+r=bm+r

rb=mb

r=m

Answer 11

I have no idea what you're doing there, but it doesn't appear to be reasonable algebra.

Answer 12

a = bm + r

From this single equation you can in no way prove that r=m.

For example:
a = bm + r
24 = 2*6 + 12
clearly, 6 is not equal to 12.