The GCD of given numbers is 10.
Step 1 :
Divide $ 17910 $ by $ 12910 $ and get the remainder
The remainder is positive ($ 5000 > 0 $), so we will continue with division.
Step 2 :
Divide $ 12910 $ by $ \color{blue}{ 5000 } $ and get the remainder
The remainder is still positive ($ 2910 > 0 $), so we will continue with division.
Step 3 :
Divide $ 5000 $ by $ \color{blue}{ 2910 } $ and get the remainder
The remainder is still positive ($ 2090 > 0 $), so we will continue with division.
Step 4 :
Divide $ 2910 $ by $ \color{blue}{ 2090 } $ and get the remainder
The remainder is still positive ($ 820 > 0 $), so we will continue with division.
Step 5 :
Divide $ 2090 $ by $ \color{blue}{ 820 } $ and get the remainder
The remainder is still positive ($ 450 > 0 $), so we will continue with division.
Step 6 :
Divide $ 820 $ by $ \color{blue}{ 450 } $ and get the remainder
The remainder is still positive ($ 370 > 0 $), so we will continue with division.
Step 7 :
Divide $ 450 $ by $ \color{blue}{ 370 } $ and get the remainder
The remainder is still positive ($ 80 > 0 $), so we will continue with division.
Step 8 :
Divide $ 370 $ by $ \color{blue}{ 80 } $ and get the remainder
The remainder is still positive ($ 50 > 0 $), so we will continue with division.
Step 9 :
Divide $ 80 $ by $ \color{blue}{ 50 } $ and get the remainder
The remainder is still positive ($ 30 > 0 $), so we will continue with division.
Step 10 :
Divide $ 50 $ by $ \color{blue}{ 30 } $ and get the remainder
The remainder is still positive ($ 20 > 0 $), so we will continue with division.
Step 11 :
Divide $ 30 $ by $ \color{blue}{ 20 } $ and get the remainder
The remainder is still positive ($ 10 > 0 $), so we will continue with division.
Step 12 :
Divide $ 20 $ by $ \color{blue}{ 10 } $ and get the remainder
The remainder is zero => GCD is the last divisor $ \color{blue}{ \boxed { 10 }} $.
We can summarize an algorithm into a following table.
| 17910 | : | 12910 | = | 1 | remainder ( 5000 ) | ||||||||||||||||||||||
| 12910 | : | 5000 | = | 2 | remainder ( 2910 ) | ||||||||||||||||||||||
| 5000 | : | 2910 | = | 1 | remainder ( 2090 ) | ||||||||||||||||||||||
| 2910 | : | 2090 | = | 1 | remainder ( 820 ) | ||||||||||||||||||||||
| 2090 | : | 820 | = | 2 | remainder ( 450 ) | ||||||||||||||||||||||
| 820 | : | 450 | = | 1 | remainder ( 370 ) | ||||||||||||||||||||||
| 450 | : | 370 | = | 1 | remainder ( 80 ) | ||||||||||||||||||||||
| 370 | : | 80 | = | 4 | remainder ( 50 ) | ||||||||||||||||||||||
| 80 | : | 50 | = | 1 | remainder ( 30 ) | ||||||||||||||||||||||
| 50 | : | 30 | = | 1 | remainder ( 20 ) | ||||||||||||||||||||||
| 30 | : | 20 | = | 1 | remainder ( 10 ) | ||||||||||||||||||||||
| 20 | : | 10 | = | 2 | remainder ( 0 ) | ||||||||||||||||||||||
| GCD = 10 | |||||||||||||||||||||||||||
This solution can be visualized using a Venn diagram.
The GCD equals the product of the numbers at the intersection.