Find Greatest Common Divisor of 1369 and 2597, using Euclidean algorithm.
Answer
The GCD of given numbers is 1.
Explanation
Step 1 :
Divide $ 2597 $ by $ 1369 $ and get the remainder
The remainder is positive ($ 1228 > 0 $), so we will continue with division.
Step 2 :
Divide $ 1369 $ by $ \color{blue}{ 1228 } $ and get the remainder
The remainder is still positive ($ 141 > 0 $), so we will continue with division.
Step 3 :
Divide $ 1228 $ by $ \color{blue}{ 141 } $ and get the remainder
The remainder is still positive ($ 100 > 0 $), so we will continue with division.
Step 4 :
Divide $ 141 $ by $ \color{blue}{ 100 } $ and get the remainder
The remainder is still positive ($ 41 > 0 $), so we will continue with division.
Step 5 :
Divide $ 100 $ by $ \color{blue}{ 41 } $ and get the remainder
The remainder is still positive ($ 18 > 0 $), so we will continue with division.
Step 6 :
Divide $ 41 $ by $ \color{blue}{ 18 } $ and get the remainder
The remainder is still positive ($ 5 > 0 $), so we will continue with division.
Step 7 :
Divide $ 18 $ by $ \color{blue}{ 5 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 8 :
Divide $ 5 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 9 :
Divide $ 3 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 10 :
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.
| 2597 | : | 1369 | = | 1 | remainder ( 1228 ) | ||||||||||||||||||
| 1369 | : | 1228 | = | 1 | remainder ( 141 ) | ||||||||||||||||||
| 1228 | : | 141 | = | 8 | remainder ( 100 ) | ||||||||||||||||||
| 141 | : | 100 | = | 1 | remainder ( 41 ) | ||||||||||||||||||
| 100 | : | 41 | = | 2 | remainder ( 18 ) | ||||||||||||||||||
| 41 | : | 18 | = | 2 | remainder ( 5 ) | ||||||||||||||||||
| 18 | : | 5 | = | 3 | 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.