Show that if $a$ is an odd integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$ Show that if $a$ is an odd en integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$.
I understand that since $a$ is not divisible by $2$ but $b$ is, the gcd of $a$ and $b$ also can't be divisible by $2$ but I'm getting stuck using this to help me. 
 A: $\,d\mid a\ $ odd $\,\Rightarrow\, d\,$  odd, $ $ so $\ d\mid b\! =\! 2(b/2)\! \iff\! d\mid b/2.\,$ Thus $\ a,b\ $ and $\ a,b/2\ $ have the same set $\,S$ of common divisors $\,d,\,$ hence the same greatest common divisor (= max $S).\ \ $ QED
Or: $ $ Let $\,b = 2c.\,$ $(a,2c) = (a,ac,2c) = (a,(ac,2c)) = (a,\color{#c00}{(a,2)}c) = (a,c)\,$ by $\,\color{#c00}{(a,2)=1}$
Remark $\,\ $ Generally $\,\ (a,bc) = (a,(a,b)c)\ $ follows precisely as in the second proof above.
Therefore we deduce $\ \ (a,bc) = (a,c)\ $ if $\ (a,b) = 1.\, $ Yours is special case: $ $ odd $\,a,\,\ b = 2.$
A: We will show that (i) If $k$ divides $a$ and $b/2$, then $k$ divides $a$ and $b$ and (ii) If $k$ divides $a$ and $b$, then $k$ divides $a$ and $b/2$.
This will sow that the set of common divisors of $a$ and $b$ is the same as the set of common divisors of $a$ and $b/2$. Since the sets are the same, the maximum elements of the sets (the gcds) are the same.
Proof of (i): We need only show that $k$ divides $b$. We have $\frac{b}{2}=kq$ for some integer $q$, and therefore $b=k(2q)$, so $k$ divides $b$.
Proof of (ii): We need to show that $k$ divides $\frac{b}{2}$. Let $\frac{b}{2}=b^\ast$. Then $b=2b^\ast$. 
Because $k$ divides $a$, and $a$ is odd, it follows that $k$ is odd. So the odd integer $k$ divides $2b^\ast$. But $k$ and $2$ are relatively prime, so $k$ divides $b^\ast$. 
A: $$GCD(a,b) = 2^{min(a1,b1)}\cdot3^{min(a2,b2)}\cdot5^{min(a3,b3)}...$$
Now, Here a1 (exponent of 2 in Prime factorization of a) is $0$ (because a is odd integer).
so no matter you divide b by 2, $min(a1,b1-1) = 0$ still, so GCD will remain same, now,to prove,
$$GCD(a,b/2) = 2^{min(a1,b1-1)}\cdot3^{min(a2,b2)}\cdot5^{min(a3,b3)}...$$
since $min(a1,b1)=0, min(a1,b1-1)$ = 0, therefore, 
$$GCD(a,b) =GCD(a,b/2)$$
Hence proved.
(min(x,y) represents the minimum value of either x or y.)
