The GCD of given numbers is 10.
Step 1 :
Divide $ 18190 $ by $ 13190 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 13190 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 3190 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 3190 } $ and get the remainder
The remainder is still positive ($ 1810 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3190 $ by $ \color{blue}{ 1810 } $ and get the remainder
The remainder is still positive ($ 1380 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1810 $ by $ \color{blue}{ 1380 } $ and get the remainder
The remainder is still positive ($ 430 > 0 $), so we will continue with division.
Step 6 :
Divide $ 1380 $ by $ \color{blue}{ 430 } $ and get the remainder
The remainder is still positive ($ 90 > 0 $), so we will continue with division.
Step 7 :
Divide $ 430 $ by $ \color{blue}{ 90 } $ and get the remainder
The remainder is still positive ($ 70 > 0 $), so we will continue with division.
Step 8 :
Divide $ 90 $ by $ \color{blue}{ 70 } $ and get the remainder
The remainder is still positive ($ 20 > 0 $), so we will continue with division.
Step 9 :
Divide $ 70 $ by $ \color{blue}{ 20 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 10 :
Divide $ 20 $ by $ \color{blue}{ 10 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 10 }} $.
We can summarize an algorithm into a following table.
| 18190 | : | 13190 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||
| 13190 | : | 5000 | = | 2 | remainder ( 3190 ) | ||||||||||||||||||
| 5000 | : | 3190 | = | 1 | remainder ( 1810 ) | ||||||||||||||||||
| 3190 | : | 1810 | = | 1 | remainder ( 1380 ) | ||||||||||||||||||
| 1810 | : | 1380 | = | 1 | remainder ( 430 ) | ||||||||||||||||||
| 1380 | : | 430 | = | 3 | remainder ( 90 ) | ||||||||||||||||||
| 430 | : | 90 | = | 4 | remainder ( 70 ) | ||||||||||||||||||
| 90 | : | 70 | = | 1 | remainder ( 20 ) | ||||||||||||||||||
| 70 | : | 20 | = | 3 | remainder ( 10 ) | ||||||||||||||||||
| 20 | : | 10 | = | 2 | remainder ( 0 ) | ||||||||||||||||||
| GCD = 10 | |||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.