The GCD of given numbers is 10.
Step 1 :
Divide $ 18170 $ by $ 13170 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 13170 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 3170 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 3170 } $ and get the remainder
The remainder is still positive ($ 1830 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3170 $ by $ \color{blue}{ 1830 } $ and get the remainder
The remainder is still positive ($ 1340 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1830 $ by $ \color{blue}{ 1340 } $ and get the remainder
The remainder is still positive ($ 490 > 0 $), so we will continue with division.
Step 6 :
Divide $ 1340 $ by $ \color{blue}{ 490 } $ and get the remainder
The remainder is still positive ($ 360 > 0 $), so we will continue with division.
Step 7 :
Divide $ 490 $ by $ \color{blue}{ 360 } $ and get the remainder
The remainder is still positive ($ 130 > 0 $), so we will continue with division.
Step 8 :
Divide $ 360 $ by $ \color{blue}{ 130 } $ and get the remainder
The remainder is still positive ($ 100 > 0 $), so we will continue with division.
Step 9 :
Divide $ 130 $ by $ \color{blue}{ 100 } $ and get the remainder
The remainder is still positive ($ 30 > 0 $), so we will continue with division.
Step 10 :
Divide $ 100 $ by $ \color{blue}{ 30 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 11 :
Divide $ 30 $ 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.
| 18170 | : | 13170 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||||
| 13170 | : | 5000 | = | 2 | remainder ( 3170 ) | ||||||||||||||||||||
| 5000 | : | 3170 | = | 1 | remainder ( 1830 ) | ||||||||||||||||||||
| 3170 | : | 1830 | = | 1 | remainder ( 1340 ) | ||||||||||||||||||||
| 1830 | : | 1340 | = | 1 | remainder ( 490 ) | ||||||||||||||||||||
| 1340 | : | 490 | = | 2 | remainder ( 360 ) | ||||||||||||||||||||
| 490 | : | 360 | = | 1 | remainder ( 130 ) | ||||||||||||||||||||
| 360 | : | 130 | = | 2 | remainder ( 100 ) | ||||||||||||||||||||
| 130 | : | 100 | = | 1 | remainder ( 30 ) | ||||||||||||||||||||
| 100 | : | 30 | = | 3 | remainder ( 10 ) | ||||||||||||||||||||
| 30 | : | 10 | = | 3 | remainder ( 0 ) | ||||||||||||||||||||
| GCD = 10 | |||||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.