The GCD of given numbers is 10.
Step 1 :
Divide $ 15310 $ by $ 10130 $ and get the remainder
The remainder is positive ($ 5180 > 0 $), so we will continue with division.
Step 2 :
Divide $ 10130 $ by $ \color{blue}{ 5180 } $ and get the remainder
The remainder is still positive ($ 4950 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5180 $ by $ \color{blue}{ 4950 } $ and get the remainder
The remainder is still positive ($ 230 > 0 $), so we will continue with division.
Step 4 :
Divide $ 4950 $ by $ \color{blue}{ 230 } $ and get the remainder
The remainder is still positive ($ 120 > 0 $), so we will continue with division.
Step 5 :
Divide $ 230 $ by $ \color{blue}{ 120 } $ and get the remainder
The remainder is still positive ($ 110 > 0 $), so we will continue with division.
Step 6 :
Divide $ 120 $ by $ \color{blue}{ 110 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 7 :
Divide $ 110 $ 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.
| 15310 | : | 10130 | = | 1 | remainder ( 5180 ) | ||||||||||||
| 10130 | : | 5180 | = | 1 | remainder ( 4950 ) | ||||||||||||
| 5180 | : | 4950 | = | 1 | remainder ( 230 ) | ||||||||||||
| 4950 | : | 230 | = | 21 | remainder ( 120 ) | ||||||||||||
| 230 | : | 120 | = | 1 | remainder ( 110 ) | ||||||||||||
| 120 | : | 110 | = | 1 | remainder ( 10 ) | ||||||||||||
| 110 | : | 10 | = | 11 | remainder ( 0 ) | ||||||||||||
| GCD = 10 | |||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.