Find Greatest Common Divisor of 1970 and 1066, using Euclidean algorithm.
Answer
The GCD of given numbers is 2.
Explanation
Step 1 :
Divide $ 1970 $ by $ 1066 $ and get the remainder
The remainder is positive ($ 904 > 0 $), so we will continue with division.
Step 2 :
Divide $ 1066 $ by $ \color{blue}{ 904 } $ and get the remainder
The remainder is still positive ($ 162 > 0 $), so we will continue with division.
Step 3 :
Divide $ 904 $ by $ \color{blue}{ 162 } $ and get the remainder
The remainder is still positive ($ 94 > 0 $), so we will continue with division.
Step 4 :
Divide $ 162 $ by $ \color{blue}{ 94 } $ and get the remainder
The remainder is still positive ($ 68 > 0 $), so we will continue with division.
Step 5 :
Divide $ 94 $ by $ \color{blue}{ 68 } $ and get the remainder
The remainder is still positive ($ 26 > 0 $), so we will continue with division.
Step 6 :
Divide $ 68 $ by $ \color{blue}{ 26 } $ and get the remainder
The remainder is still positive ($ 16 > 0 $), so we will continue with division.
Step 7 :
Divide $ 26 $ by $ \color{blue}{ 16 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 8 :
Divide $ 16 $ by $ \color{blue}{ 10 } $ and get the remainder
The remainder is still positive ($ 6 > 0 $), so we will continue with division.
Step 9 :
Divide $ 10 $ by $ \color{blue}{ 6 } $ and get the remainder
The remainder is still positive ($ 4 > 0 $), so we will continue with division.
Step 10 :
Divide $ 6 $ by $ \color{blue}{ 4 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 11 :
Divide $ 4 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 2 }} $.
We can summarize an algorithm into a following table.
| 1970 | : | 1066 | = | 1 | remainder ( 904 ) | ||||||||||||||||||||
| 1066 | : | 904 | = | 1 | remainder ( 162 ) | ||||||||||||||||||||
| 904 | : | 162 | = | 5 | remainder ( 94 ) | ||||||||||||||||||||
| 162 | : | 94 | = | 1 | remainder ( 68 ) | ||||||||||||||||||||
| 94 | : | 68 | = | 1 | remainder ( 26 ) | ||||||||||||||||||||
| 68 | : | 26 | = | 2 | remainder ( 16 ) | ||||||||||||||||||||
| 26 | : | 16 | = | 1 | remainder ( 10 ) | ||||||||||||||||||||
| 16 | : | 10 | = | 1 | remainder ( 6 ) | ||||||||||||||||||||
| 10 | : | 6 | = | 1 | remainder ( 4 ) | ||||||||||||||||||||
| 6 | : | 4 | = | 1 | remainder ( 2 ) | ||||||||||||||||||||
| 4 | : | 2 | = | 2 | remainder ( 0 ) | ||||||||||||||||||||
| GCD = 2 | |||||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.