Question

Find Greatest Common Divisor of 11607 and 3162, using Euclidean algorithm.

Answer

The GCD of given numbers is 3.

Explanation

Step 1 :
Divide $11607$ by $3162$ and get the remainder

11607 : 3162 = 3 ( remainder is 2121)

The remainder is positive ( $2121 > 0$ ), so we will continue with division.

Step 2 :
Divide $3162$ by $2121$ and get the remainder

3162 : 2121 = 1 ( remainder is 1041)

The remainder is still positive ( $1041 > 0$ ), so we will continue with division.

Step 3 :
Divide $2121$ by $1041$ and get the remainder

2121 : 1041 = 2 ( remainder is 39)

The remainder is still positive ( $39 > 0$ ), so we will continue with division.

Step 4 :
Divide $1041$ by $39$ and get the remainder

1041 : 39 = 26 ( remainder is 27)

The remainder is still positive ( $27 > 0$ ), so we will continue with division.

Step 5 :
Divide $39$ by $27$ and get the remainder

39 : 27 = 1 ( remainder is 12)

The remainder is still positive ( $12 > 0$ ), so we will continue with division.

Step 6 :
Divide $27$ by $12$ and get the remainder

27 : 12 = 2 ( remainder is 3)

The remainder is still positive ( $3 > 0$ ), so we will continue with division.

Step 7 :
Divide $12$ by $3$ and get the remainder

12 : 3 = 4 ( remainder is 0)

The remainder is zero => GCD is the last divisor $3$ .

We can summarize an algorithm into a following table.

This solution can be visualized using a Venn diagram.

0,0

11607

3162

The GCD equals the product of the numbers at the intersection.

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