Find Greatest Common Divisor of 31415 and 14142, using Euclidean algorithm.
Answer
The GCD of given numbers is 1.
Explanation
Step 1 :
Divide $ 31415 $ by $ 14142 $ and get the remainder
The remainder is positive ($ 3131 > 0 $), so we will continue with division.
Step 2 :
Divide $ 14142 $ by $ \color{blue}{ 3131 } $ and get the remainder
The remainder is still positive ($ 1618 > 0 $), so we will continue with division.
Step 3 :
Divide $ 3131 $ by $ \color{blue}{ 1618 } $ and get the remainder
The remainder is still positive ($ 1513 > 0 $), so we will continue with division.
Step 4 :
Divide $ 1618 $ by $ \color{blue}{ 1513 } $ and get the remainder
The remainder is still positive ($ 105 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1513 $ by $ \color{blue}{ 105 } $ and get the remainder
The remainder is still positive ($ 43 > 0 $), so we will continue with division.
Step 6 :
Divide $ 105 $ by $ \color{blue}{ 43 } $ and get the remainder
The remainder is still positive ($ 19 > 0 $), so we will continue with division.
Step 7 :
Divide $ 43 $ by $ \color{blue}{ 19 } $ and get the remainder
The remainder is still positive ($ 5 > 0 $), so we will continue with division.
Step 8 :
Divide $ 19 $ by $ \color{blue}{ 5 } $ and get the remainder
The remainder is still positive ($ 4 > 0 $), so we will continue with division.
Step 9 :
Divide $ 5 $ by $ \color{blue}{ 4 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 10 :
Divide $ 4 $ 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.
| 31415 | : | 14142 | = | 2 | remainder ( 3131 ) | ||||||||||||||||||
| 14142 | : | 3131 | = | 4 | remainder ( 1618 ) | ||||||||||||||||||
| 3131 | : | 1618 | = | 1 | remainder ( 1513 ) | ||||||||||||||||||
| 1618 | : | 1513 | = | 1 | remainder ( 105 ) | ||||||||||||||||||
| 1513 | : | 105 | = | 14 | remainder ( 43 ) | ||||||||||||||||||
| 105 | : | 43 | = | 2 | remainder ( 19 ) | ||||||||||||||||||
| 43 | : | 19 | = | 2 | remainder ( 5 ) | ||||||||||||||||||
| 19 | : | 5 | = | 3 | remainder ( 4 ) | ||||||||||||||||||
| 5 | : | 4 | = | 1 | remainder ( 1 ) | ||||||||||||||||||
| 4 | : | 1 | = | 4 | remainder ( 0 ) | ||||||||||||||||||
| GCD = 1 | |||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.