The GCD of given numbers is 10.
Step 1 :
Divide $ 18470 $ by $ 13470 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 13470 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 3470 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 3470 } $ and get the remainder
The remainder is still positive ($ 1530 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3470 $ by $ \color{blue}{ 1530 } $ and get the remainder
The remainder is still positive ($ 410 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1530 $ by $ \color{blue}{ 410 } $ and get the remainder
The remainder is still positive ($ 300 > 0 $), so we will continue with division.
Step 6 :
Divide $ 410 $ by $ \color{blue}{ 300 } $ and get the remainder
The remainder is still positive ($ 110 > 0 $), so we will continue with division.
Step 7 :
Divide $ 300 $ by $ \color{blue}{ 110 } $ and get the remainder
The remainder is still positive ($ 80 > 0 $), so we will continue with division.
Step 8 :
Divide $ 110 $ by $ \color{blue}{ 80 } $ and get the remainder
The remainder is still positive ($ 30 > 0 $), so we will continue with division.
Step 9 :
Divide $ 80 $ by $ \color{blue}{ 30 } $ and get the remainder
The remainder is still positive ($ 20 > 0 $), so we will continue with division.
Step 10 :
Divide $ 30 $ by $ \color{blue}{ 20 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 11 :
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.
| 18470 | : | 13470 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||||
| 13470 | : | 5000 | = | 2 | remainder ( 3470 ) | ||||||||||||||||||||
| 5000 | : | 3470 | = | 1 | remainder ( 1530 ) | ||||||||||||||||||||
| 3470 | : | 1530 | = | 2 | remainder ( 410 ) | ||||||||||||||||||||
| 1530 | : | 410 | = | 3 | remainder ( 300 ) | ||||||||||||||||||||
| 410 | : | 300 | = | 1 | remainder ( 110 ) | ||||||||||||||||||||
| 300 | : | 110 | = | 2 | remainder ( 80 ) | ||||||||||||||||||||
| 110 | : | 80 | = | 1 | remainder ( 30 ) | ||||||||||||||||||||
| 80 | : | 30 | = | 2 | remainder ( 20 ) | ||||||||||||||||||||
| 30 | : | 20 | = | 1 | 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.