The GCD of given numbers is 10.
Step 1 :
Divide $ 18130 $ by $ 13130 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 13130 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 3130 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 3130 } $ and get the remainder
The remainder is still positive ($ 1870 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3130 $ by $ \color{blue}{ 1870 } $ and get the remainder
The remainder is still positive ($ 1260 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1870 $ by $ \color{blue}{ 1260 } $ and get the remainder
The remainder is still positive ($ 610 > 0 $), so we will continue with division.
Step 6 :
Divide $ 1260 $ by $ \color{blue}{ 610 } $ and get the remainder
The remainder is still positive ($ 40 > 0 $), so we will continue with division.
Step 7 :
Divide $ 610 $ by $ \color{blue}{ 40 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 8 :
Divide $ 40 $ 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.
| 18130 | : | 13130 | = | 1 | remainder ( 5000 ) | ||||||||||||||
| 13130 | : | 5000 | = | 2 | remainder ( 3130 ) | ||||||||||||||
| 5000 | : | 3130 | = | 1 | remainder ( 1870 ) | ||||||||||||||
| 3130 | : | 1870 | = | 1 | remainder ( 1260 ) | ||||||||||||||
| 1870 | : | 1260 | = | 1 | remainder ( 610 ) | ||||||||||||||
| 1260 | : | 610 | = | 2 | remainder ( 40 ) | ||||||||||||||
| 610 | : | 40 | = | 15 | remainder ( 10 ) | ||||||||||||||
| 40 | : | 10 | = | 4 | remainder ( 0 ) | ||||||||||||||
| GCD = 10 | |||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.