The GCD of given numbers is 1.
Step 1 :
Divide $ 4711 $ by $ 1024 $ and get the remainder
The remainder is positive ($ 615 > 0 $), so we will continue with division.
Step 2 :
Divide $ 1024 $ by $ \color{blue}{ 615 } $ and get the remainder
The remainder is still positive ($ 409 > 0 $), so we will continue with division.
Step 3 :
Divide $ 615 $ by $ \color{blue}{ 409 } $ and get the remainder
The remainder is still positive ($ 206 > 0 $), so we will continue with division.
Step 4 :
Divide $ 409 $ by $ \color{blue}{ 206 } $ and get the remainder
The remainder is still positive ($ 203 > 0 $), so we will continue with division.
Step 5 :
Divide $ 206 $ by $ \color{blue}{ 203 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 6 :
Divide $ 203 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 7 :
Divide $ 3 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 8 :
Divide $ 2 $ by $ \color{blue}{ 1 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 1 }} $.
We can summarize an algorithm into a following table.
4711 | : | 1024 | = | 4 | remainder ( 615 ) | ||||||||||||||
1024 | : | 615 | = | 1 | remainder ( 409 ) | ||||||||||||||
615 | : | 409 | = | 1 | remainder ( 206 ) | ||||||||||||||
409 | : | 206 | = | 1 | remainder ( 203 ) | ||||||||||||||
206 | : | 203 | = | 1 | remainder ( 3 ) | ||||||||||||||
203 | : | 3 | = | 67 | remainder ( 2 ) | ||||||||||||||
3 | : | 2 | = | 1 | remainder ( 1 ) | ||||||||||||||
2 | : | 1 | = | 2 | remainder ( 0 ) | ||||||||||||||
GCD = 1 |
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.