The GCD of given numbers is 10.
Step 1 :
Divide $ 17110 $ by $ 12110 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 12110 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 2110 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 2110 } $ and get the remainder
The remainder is still positive ($ 780 > 0 $), so we will continue with division.
Step 4 :
Divide $ 2110 $ by $ \color{blue}{ 780 } $ and get the remainder
The remainder is still positive ($ 550 > 0 $), so we will continue with division.
Step 5 :
Divide $ 780 $ by $ \color{blue}{ 550 } $ and get the remainder
The remainder is still positive ($ 230 > 0 $), so we will continue with division.
Step 6 :
Divide $ 550 $ by $ \color{blue}{ 230 } $ and get the remainder
The remainder is still positive ($ 90 > 0 $), so we will continue with division.
Step 7 :
Divide $ 230 $ by $ \color{blue}{ 90 } $ and get the remainder
The remainder is still positive ($ 50 > 0 $), so we will continue with division.
Step 8 :
Divide $ 90 $ by $ \color{blue}{ 50 } $ and get the remainder
The remainder is still positive ($ 40 > 0 $), so we will continue with division.
Step 9 :
Divide $ 50 $ by $ \color{blue}{ 40 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 10 :
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.
| 17110 | : | 12110 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||
| 12110 | : | 5000 | = | 2 | remainder ( 2110 ) | ||||||||||||||||||
| 5000 | : | 2110 | = | 2 | remainder ( 780 ) | ||||||||||||||||||
| 2110 | : | 780 | = | 2 | remainder ( 550 ) | ||||||||||||||||||
| 780 | : | 550 | = | 1 | remainder ( 230 ) | ||||||||||||||||||
| 550 | : | 230 | = | 2 | remainder ( 90 ) | ||||||||||||||||||
| 230 | : | 90 | = | 2 | remainder ( 50 ) | ||||||||||||||||||
| 90 | : | 50 | = | 1 | remainder ( 40 ) | ||||||||||||||||||
| 50 | : | 40 | = | 1 | 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.