The GCD of given numbers is 10.
Step 1 :
Divide $ 16090 $ by $ 11090 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 11090 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 1090 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 1090 } $ and get the remainder
The remainder is still positive ($ 640 > 0 $), so we will continue with division.
Step 4 :
Divide $ 1090 $ by $ \color{blue}{ 640 } $ and get the remainder
The remainder is still positive ($ 450 > 0 $), so we will continue with division.
Step 5 :
Divide $ 640 $ by $ \color{blue}{ 450 } $ and get the remainder
The remainder is still positive ($ 190 > 0 $), so we will continue with division.
Step 6 :
Divide $ 450 $ by $ \color{blue}{ 190 } $ and get the remainder
The remainder is still positive ($ 70 > 0 $), so we will continue with division.
Step 7 :
Divide $ 190 $ by $ \color{blue}{ 70 } $ and get the remainder
The remainder is still positive ($ 50 > 0 $), so we will continue with division.
Step 8 :
Divide $ 70 $ by $ \color{blue}{ 50 } $ and get the remainder
The remainder is still positive ($ 20 > 0 $), so we will continue with division.
Step 9 :
Divide $ 50 $ by $ \color{blue}{ 20 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 10 :
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.
| 16090 | : | 11090 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||
| 11090 | : | 5000 | = | 2 | remainder ( 1090 ) | ||||||||||||||||||
| 5000 | : | 1090 | = | 4 | remainder ( 640 ) | ||||||||||||||||||
| 1090 | : | 640 | = | 1 | remainder ( 450 ) | ||||||||||||||||||
| 640 | : | 450 | = | 1 | remainder ( 190 ) | ||||||||||||||||||
| 450 | : | 190 | = | 2 | remainder ( 70 ) | ||||||||||||||||||
| 190 | : | 70 | = | 2 | 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.