Find Greatest Common Divisor of 377 and 610, using Euclidean algorithm.
Answer
The GCD of given numbers is 1.
Explanation
Step 1 :
Divide $ 610 $ by $ 377 $ and get the remainder
The remainder is positive ($ 233 > 0 $), so we will continue with division.
Step 2 :
Divide $ 377 $ by $ \color{blue}{ 233 } $ and get the remainder
The remainder is still positive ($ 144 > 0 $), so we will continue with division.
Step 3 :
Divide $ 233 $ by $ \color{blue}{ 144 } $ and get the remainder
The remainder is still positive ($ 89 > 0 $), so we will continue with division.
Step 4 :
Divide $ 144 $ by $ \color{blue}{ 89 } $ and get the remainder
The remainder is still positive ($ 55 > 0 $), so we will continue with division.
Step 5 :
Divide $ 89 $ by $ \color{blue}{ 55 } $ and get the remainder
The remainder is still positive ($ 34 > 0 $), so we will continue with division.
Step 6 :
Divide $ 55 $ by $ \color{blue}{ 34 } $ and get the remainder
The remainder is still positive ($ 21 > 0 $), so we will continue with division.
Step 7 :
Divide $ 34 $ by $ \color{blue}{ 21 } $ and get the remainder
The remainder is still positive ($ 13 > 0 $), so we will continue with division.
Step 8 :
Divide $ 21 $ by $ \color{blue}{ 13 } $ and get the remainder
The remainder is still positive ($ 8 > 0 $), so we will continue with division.
Step 9 :
Divide $ 13 $ by $ \color{blue}{ 8 } $ and get the remainder
The remainder is still positive ($ 5 > 0 $), so we will continue with division.
Step 10 :
Divide $ 8 $ by $ \color{blue}{ 5 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 11 :
Divide $ 5 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 12 :
Divide $ 3 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 13 :
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.
| 610 | : | 377 | = | 1 | remainder ( 233 ) | ||||||||||||||||||||||||
| 377 | : | 233 | = | 1 | remainder ( 144 ) | ||||||||||||||||||||||||
| 233 | : | 144 | = | 1 | remainder ( 89 ) | ||||||||||||||||||||||||
| 144 | : | 89 | = | 1 | remainder ( 55 ) | ||||||||||||||||||||||||
| 89 | : | 55 | = | 1 | remainder ( 34 ) | ||||||||||||||||||||||||
| 55 | : | 34 | = | 1 | remainder ( 21 ) | ||||||||||||||||||||||||
| 34 | : | 21 | = | 1 | remainder ( 13 ) | ||||||||||||||||||||||||
| 21 | : | 13 | = | 1 | remainder ( 8 ) | ||||||||||||||||||||||||
| 13 | : | 8 | = | 1 | remainder ( 5 ) | ||||||||||||||||||||||||
| 8 | : | 5 | = | 1 | remainder ( 3 ) | ||||||||||||||||||||||||
| 5 | : | 3 | = | 1 | 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.