The GCD of given numbers is 10.
Step 1 :
Divide $ 18110 $ by $ 13110 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 13110 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 3110 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 3110 } $ and get the remainder
The remainder is still positive ($ 1890 > 0 $), so we will continue with division.
Step 4 :
Divide $ 3110 $ by $ \color{blue}{ 1890 } $ and get the remainder
The remainder is still positive ($ 1220 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1890 $ by $ \color{blue}{ 1220 } $ and get the remainder
The remainder is still positive ($ 670 > 0 $), so we will continue with division.
Step 6 :
Divide $ 1220 $ by $ \color{blue}{ 670 } $ and get the remainder
The remainder is still positive ($ 550 > 0 $), so we will continue with division.
Step 7 :
Divide $ 670 $ by $ \color{blue}{ 550 } $ and get the remainder
The remainder is still positive ($ 120 > 0 $), so we will continue with division.
Step 8 :
Divide $ 550 $ by $ \color{blue}{ 120 } $ and get the remainder
The remainder is still positive ($ 70 > 0 $), so we will continue with division.
Step 9 :
Divide $ 120 $ by $ \color{blue}{ 70 } $ and get the remainder
The remainder is still positive ($ 50 > 0 $), so we will continue with division.
Step 10 :
Divide $ 70 $ by $ \color{blue}{ 50 } $ and get the remainder
The remainder is still positive ($ 20 > 0 $), so we will continue with division.
Step 11 :
Divide $ 50 $ by $ \color{blue}{ 20 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 12 :
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.
| 18110 | : | 13110 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||||||
| 13110 | : | 5000 | = | 2 | remainder ( 3110 ) | ||||||||||||||||||||||
| 5000 | : | 3110 | = | 1 | remainder ( 1890 ) | ||||||||||||||||||||||
| 3110 | : | 1890 | = | 1 | remainder ( 1220 ) | ||||||||||||||||||||||
| 1890 | : | 1220 | = | 1 | remainder ( 670 ) | ||||||||||||||||||||||
| 1220 | : | 670 | = | 1 | remainder ( 550 ) | ||||||||||||||||||||||
| 670 | : | 550 | = | 1 | remainder ( 120 ) | ||||||||||||||||||||||
| 550 | : | 120 | = | 4 | remainder ( 70 ) | ||||||||||||||||||||||
| 120 | : | 70 | = | 1 | remainder ( 50 ) | ||||||||||||||||||||||
| 70 | : | 50 | = | 1 | remainder ( 20 ) | ||||||||||||||||||||||
| 50 | : | 20 | = | 2 | 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.