The GCD of given numbers is 10.
Step 1 :
Divide $ 17190 $ by $ 12190 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 12190 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 2190 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 2190 } $ and get the remainder
The remainder is still positive ($ 620 > 0 $), so we will continue with division.
Step 4 :
Divide $ 2190 $ by $ \color{blue}{ 620 } $ and get the remainder
The remainder is still positive ($ 330 > 0 $), so we will continue with division.
Step 5 :
Divide $ 620 $ by $ \color{blue}{ 330 } $ and get the remainder
The remainder is still positive ($ 290 > 0 $), so we will continue with division.
Step 6 :
Divide $ 330 $ by $ \color{blue}{ 290 } $ and get the remainder
The remainder is still positive ($ 40 > 0 $), so we will continue with division.
Step 7 :
Divide $ 290 $ 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.
| 17190 | : | 12190 | = | 1 | remainder ( 5000 ) | ||||||||||||||
| 12190 | : | 5000 | = | 2 | remainder ( 2190 ) | ||||||||||||||
| 5000 | : | 2190 | = | 2 | remainder ( 620 ) | ||||||||||||||
| 2190 | : | 620 | = | 3 | remainder ( 330 ) | ||||||||||||||
| 620 | : | 330 | = | 1 | remainder ( 290 ) | ||||||||||||||
| 330 | : | 290 | = | 1 | remainder ( 40 ) | ||||||||||||||
| 290 | : | 40 | = | 7 | 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.