TITLE: Why $\gcd(b,qb+r)=\gcd(b,r),\,$ so $\,\gcd(b,a) = \gcd(b,a\bmod b)$
QUESTION [42 upvotes]: Given: $a = qb + r$. Then it holds that $\gcd(a,b)=\gcd(b,r)$. That doesn't sound logical to me. Why is this so?

Addendum by LePressentiment on 11/29/2013: (in the interest of http://meta.math.stackexchange.com/a/4110/53259 and averting a duplicate)
What's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally.

REPLY [2 votes]: Let 
$$a = qb + r \implies a\equiv r (\text{ mod }b)\implies r = a\text{ mod }b$$
Bey GCD recursion theorem we have:
$$\gcd\big(a, b\big) = \gcd\big(a, a\text{ mod } b\big)$$
Therefore

$$\gcd\big(a, b\big) = \gcd\big(a, a\text{ mod } b\big) = \gcd\big(a, r\big)$$

See reference

Bibliographic reference:
Thomas H. Cormen (2009). Introduction to Algorithms 3rd, 934.