Find Greatest Common Divisor of 110 and 2511, using Euclidean algorithm.
Answer
The GCD of given numbers is 1.
Explanation
Step 1 :
Divide $ 2511 $ by $ 110 $ and get the remainder
The remainder is positive ($ 91 > 0 $), so we will continue with division.
Step 2 :
Divide $ 110 $ by $ \color{blue}{ 91 } $ and get the remainder
The remainder is still positive ($ 19 > 0 $), so we will continue with division.
Step 3 :
Divide $ 91 $ by $ \color{blue}{ 19 } $ and get the remainder
The remainder is still positive ($ 15 > 0 $), so we will continue with division.
Step 4 :
Divide $ 19 $ by $ \color{blue}{ 15 } $ and get the remainder
The remainder is still positive ($ 4 > 0 $), so we will continue with division.
Step 5 :
Divide $ 15 $ by $ \color{blue}{ 4 } $ and get the remainder
The remainder is still positive ($ 3 > 0 $), so we will continue with division.
Step 6 :
Divide $ 4 $ by $ \color{blue}{ 3 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 7 :
Divide $ 3 $ 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.
| 2511 | : | 110 | = | 22 | remainder ( 91 ) | ||||||||||||
| 110 | : | 91 | = | 1 | remainder ( 19 ) | ||||||||||||
| 91 | : | 19 | = | 4 | remainder ( 15 ) | ||||||||||||
| 19 | : | 15 | = | 1 | remainder ( 4 ) | ||||||||||||
| 15 | : | 4 | = | 3 | remainder ( 3 ) | ||||||||||||
| 4 | : | 3 | = | 1 | remainder ( 1 ) | ||||||||||||
| 3 | : | 1 | = | 3 | remainder ( 0 ) | ||||||||||||
| GCD = 1 | |||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.