The GCD of given numbers is 1.
Step 1 :
Divide $ 67 $ by $ 35 $ and get the remainder
The remainder is positive ($ 32 > 0 $), so we will continue with division.
Step 2 :
Divide $ 35 $ by $ \color{blue}{ 32 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 3 :
Divide $ 32 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 5 :
Divide $ 2 $ by $ \color{blue}{ 1 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 1 }} $.
We can summarize an algorithm into a following table.
67 | : | 35 | = | 1 | remainder ( 32 ) | ||||||||
35 | : | 32 | = | 1 | remainder ( 3 ) | ||||||||
32 | : | 3 | = | 10 | remainder ( 2 ) | ||||||||
3 | : | 2 | = | 1 | remainder ( 1 ) | ||||||||
2 | : | 1 | = | 2 | remainder ( 0 ) | ||||||||
GCD = 1 |
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.