The GCD of given numbers is 3.
Step 1 :
Divide $ 202851 $ by $ 22773 $ and get the remainder
The remainder is positive ($ 20667 > 0 $), so we will continue with division.
Step 2 :
Divide $ 22773 $ by $ \color{blue}{ 20667 } $ and get the remainder
The remainder is still positive ($ 2106 > 0 $), so we will continue with division.
Step 3 :
Divide $ 20667 $ by $ \color{blue}{ 2106 } $ and get the remainder
The remainder is still positive ($ 1713 > 0 $), so we will continue with division.
Step 4 :
Divide $ 2106 $ by $ \color{blue}{ 1713 } $ and get the remainder
The remainder is still positive ($ 393 > 0 $), so we will continue with division.
Step 5 :
Divide $ 1713 $ by $ \color{blue}{ 393 } $ and get the remainder
The remainder is still positive ($ 141 > 0 $), so we will continue with division.
Step 6 :
Divide $ 393 $ by $ \color{blue}{ 141 } $ and get the remainder
The remainder is still positive ($ 111 > 0 $), so we will continue with division.
Step 7 :
Divide $ 141 $ by $ \color{blue}{ 111 } $ and get the remainder
The remainder is still positive ($ 30 > 0 $), so we will continue with division.
Step 8 :
Divide $ 111 $ by $ \color{blue}{ 30 } $ and get the remainder
The remainder is still positive ($ 21 > 0 $), so we will continue with division.
Step 9 :
Divide $ 30 $ by $ \color{blue}{ 21 } $ and get the remainder
The remainder is still positive ($ 9 > 0 $), so we will continue with division.
Step 10 :
Divide $ 21 $ by $ \color{blue}{ 9 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 11 :
Divide $ 9 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 3 }} $.
We can summarize an algorithm into a following table.
| 202851 | : | 22773 | = | 8 | remainder ( 20667 ) | ||||||||||||||||||||
| 22773 | : | 20667 | = | 1 | remainder ( 2106 ) | ||||||||||||||||||||
| 20667 | : | 2106 | = | 9 | remainder ( 1713 ) | ||||||||||||||||||||
| 2106 | : | 1713 | = | 1 | remainder ( 393 ) | ||||||||||||||||||||
| 1713 | : | 393 | = | 4 | remainder ( 141 ) | ||||||||||||||||||||
| 393 | : | 141 | = | 2 | remainder ( 111 ) | ||||||||||||||||||||
| 141 | : | 111 | = | 1 | remainder ( 30 ) | ||||||||||||||||||||
| 111 | : | 30 | = | 3 | remainder ( 21 ) | ||||||||||||||||||||
| 30 | : | 21 | = | 1 | remainder ( 9 ) | ||||||||||||||||||||
| 21 | : | 9 | = | 2 | remainder ( 3 ) | ||||||||||||||||||||
| 9 | : | 3 | = | 3 | remainder ( 0 ) | ||||||||||||||||||||
| GCD = 3 | |||||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.