Find Greatest Common Divisor of 101 and 3267, using Euclidean algorithm.
Answer
The GCD of given numbers is 1.
Explanation
Step 1 :
Divide $ 3267 $ by $ 101 $ and get the remainder
The remainder is positive ($ 35 > 0 $), so we will continue with division.
Step 2 :
Divide $ 101 $ by $ \color{blue}{ 35 } $ and get the remainder
The remainder is still positive ($ 31 > 0 $), so we will continue with division.
Step 3 :
Divide $ 35 $ by $ \color{blue}{ 31 } $ and get the remainder
The remainder is still positive ($ 4 > 0 $), so we will continue with division.
Step 4 :
Divide $ 31 $ by $ \color{blue}{ 4 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 5 :
Divide $ 4 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 6 :
Divide $ 3 $ 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.
| 3267 | : | 101 | = | 32 | remainder ( 35 ) | ||||||||||
| 101 | : | 35 | = | 2 | remainder ( 31 ) | ||||||||||
| 35 | : | 31 | = | 1 | remainder ( 4 ) | ||||||||||
| 31 | : | 4 | = | 7 | remainder ( 3 ) | ||||||||||
| 4 | : | 3 | = | 1 | remainder ( 1 ) | ||||||||||
| 3 | : | 1 | = | 3 | remainder ( 0 ) | ||||||||||
| GCD = 1 | |||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.