The GCD of given numbers is 10.
Step 1 :
Divide $ 16490 $ by $ 11490 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 11490 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 1490 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 1490 } $ and get the remainder
The remainder is still positive ($ 530 > 0 $), so we will continue with division.
Step 4 :
Divide $ 1490 $ by $ \color{blue}{ 530 } $ and get the remainder
The remainder is still positive ($ 430 > 0 $), so we will continue with division.
Step 5 :
Divide $ 530 $ by $ \color{blue}{ 430 } $ and get the remainder
The remainder is still positive ($ 100 > 0 $), so we will continue with division.
Step 6 :
Divide $ 430 $ by $ \color{blue}{ 100 } $ and get the remainder
The remainder is still positive ($ 30 > 0 $), so we will continue with division.
Step 7 :
Divide $ 100 $ by $ \color{blue}{ 30 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 8 :
Divide $ 30 $ 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.
| 16490 | : | 11490 | = | 1 | remainder ( 5000 ) | ||||||||||||||
| 11490 | : | 5000 | = | 2 | remainder ( 1490 ) | ||||||||||||||
| 5000 | : | 1490 | = | 3 | remainder ( 530 ) | ||||||||||||||
| 1490 | : | 530 | = | 2 | remainder ( 430 ) | ||||||||||||||
| 530 | : | 430 | = | 1 | remainder ( 100 ) | ||||||||||||||
| 430 | : | 100 | = | 4 | remainder ( 30 ) | ||||||||||||||
| 100 | : | 30 | = | 3 | remainder ( 10 ) | ||||||||||||||
| 30 | : | 10 | = | 3 | remainder ( 0 ) | ||||||||||||||
| GCD = 10 | |||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.