Find Greatest Common Divisor of 3158 and 1226, using Euclidean algorithm.
Answer
The GCD of given numbers is 2.
Explanation
Step 1 :
Divide $ 3158 $ by $ 1226 $ and get the remainder
The remainder is positive ($ 706 > 0 $), so we will continue with division.
Step 2 :
Divide $ 1226 $ by $ \color{blue}{ 706 } $ and get the remainder
The remainder is still positive ($ 520 > 0 $), so we will continue with division.
Step 3 :
Divide $ 706 $ by $ \color{blue}{ 520 } $ and get the remainder
The remainder is still positive ($ 186 > 0 $), so we will continue with division.
Step 4 :
Divide $ 520 $ by $ \color{blue}{ 186 } $ and get the remainder
The remainder is still positive ($ 148 > 0 $), so we will continue with division.
Step 5 :
Divide $ 186 $ by $ \color{blue}{ 148 } $ and get the remainder
The remainder is still positive ($ 38 > 0 $), so we will continue with division.
Step 6 :
Divide $ 148 $ by $ \color{blue}{ 38 } $ and get the remainder
The remainder is still positive ($ 34 > 0 $), so we will continue with division.
Step 7 :
Divide $ 38 $ by $ \color{blue}{ 34 } $ and get the remainder
The remainder is still positive ($ 4 > 0 $), so we will continue with division.
Step 8 :
Divide $ 34 $ by $ \color{blue}{ 4 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 9 :
Divide $ 4 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 2 }} $.
We can summarize an algorithm into a following table.
| 3158 | : | 1226 | = | 2 | remainder ( 706 ) | ||||||||||||||||
| 1226 | : | 706 | = | 1 | remainder ( 520 ) | ||||||||||||||||
| 706 | : | 520 | = | 1 | remainder ( 186 ) | ||||||||||||||||
| 520 | : | 186 | = | 2 | remainder ( 148 ) | ||||||||||||||||
| 186 | : | 148 | = | 1 | remainder ( 38 ) | ||||||||||||||||
| 148 | : | 38 | = | 3 | remainder ( 34 ) | ||||||||||||||||
| 38 | : | 34 | = | 1 | remainder ( 4 ) | ||||||||||||||||
| 34 | : | 4 | = | 8 | remainder ( 2 ) | ||||||||||||||||
| 4 | : | 2 | = | 2 | remainder ( 0 ) | ||||||||||||||||
| GCD = 2 | |||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.