The GCD of given numbers is 1.
Step 1 :
Divide $ 53540 $ by $ 11553 $ and get the remainder
The remainder is positive ($ 7328 > 0 $), so we will continue with division.
Step 2 :
Divide $ 11553 $ by $ \color{blue}{ 7328 } $ and get the remainder
The remainder is still positive ($ 4225 > 0 $), so we will continue with division.
Step 3 :
Divide $ 7328 $ by $ \color{blue}{ 4225 } $ and get the remainder
The remainder is still positive ($ 3103 > 0 $), so we will continue with division.
Step 4 :
Divide $ 4225 $ by $ \color{blue}{ 3103 } $ and get the remainder
The remainder is still positive ($ 1122 > 0 $), so we will continue with division.
Step 5 :
Divide $ 3103 $ by $ \color{blue}{ 1122 } $ and get the remainder
The remainder is still positive ($ 859 > 0 $), so we will continue with division.
Step 6 :
Divide $ 1122 $ by $ \color{blue}{ 859 } $ and get the remainder
The remainder is still positive ($ 263 > 0 $), so we will continue with division.
Step 7 :
Divide $ 859 $ by $ \color{blue}{ 263 } $ and get the remainder
The remainder is still positive ($ 70 > 0 $), so we will continue with division.
Step 8 :
Divide $ 263 $ by $ \color{blue}{ 70 } $ and get the remainder
The remainder is still positive ($ 53 > 0 $), so we will continue with division.
Step 9 :
Divide $ 70 $ by $ \color{blue}{ 53 } $ and get the remainder
The remainder is still positive ($ 17 > 0 $), so we will continue with division.
Step 10 :
Divide $ 53 $ by $ \color{blue}{ 17 } $ and get the remainder
The remainder is still positive ($ 2 > 0 $), so we will continue with division.
Step 11 :
Divide $ 17 $ by $ \color{blue}{ 2 } $ and get the remainder
The remainder is still positive ($ 1 > 0 $), so we will continue with division.
Step 12 :
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.
| 53540 | : | 11553 | = | 4 | remainder ( 7328 ) | ||||||||||||||||||||||
| 11553 | : | 7328 | = | 1 | remainder ( 4225 ) | ||||||||||||||||||||||
| 7328 | : | 4225 | = | 1 | remainder ( 3103 ) | ||||||||||||||||||||||
| 4225 | : | 3103 | = | 1 | remainder ( 1122 ) | ||||||||||||||||||||||
| 3103 | : | 1122 | = | 2 | remainder ( 859 ) | ||||||||||||||||||||||
| 1122 | : | 859 | = | 1 | remainder ( 263 ) | ||||||||||||||||||||||
| 859 | : | 263 | = | 3 | remainder ( 70 ) | ||||||||||||||||||||||
| 263 | : | 70 | = | 3 | remainder ( 53 ) | ||||||||||||||||||||||
| 70 | : | 53 | = | 1 | remainder ( 17 ) | ||||||||||||||||||||||
| 53 | : | 17 | = | 3 | remainder ( 2 ) | ||||||||||||||||||||||
| 17 | : | 2 | = | 8 | 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.