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Mastering Greatest Common Divisor(GCD) using Euclidean Method ||Simple Method Explained with Example

Chirag Lecture Tutorials · YouTube · 4:50

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The greatest common divisor is the largest positive divisor shared by the given positive integers. The Euclidean algorithm finds it by repeatedly replacing a pair with the divisor and the remainder.

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Chapters

0:19Compare common divisors1:30Keep the GCD while replacing the pair3:15Trace the example to the base case

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For 1010 and 1515, the common positive divisors are 11 and 55, so their greatest common divisor is 55. Listing divisors works for small inputs but is not the efficient general method.

For nonnegative aa and positive bb, the Euclidean step is gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b). The remainder is smaller than bb, which makes repeated steps terminate.

The sequence is gcd⁡(10,15)=gcd⁡(15,10)=gcd⁡(10,5)=gcd⁡(5,0)=5\gcd(10,15)=\gcd(15,10)=\gcd(10,5)=\gcd(5,0)=5. Stop at the zero second argument rather than attempting a remainder with a zero divisor.

Knowledge cards

01

Compare common divisors

For 1010 and 1515, the common positive divisors are 11 and 55, so their greatest common divisor is 55. Listing divisors works for small inputs but is not the efficient general method.

02

Keep the GCD while replacing the pair

For nonnegative aa and positive bb, the Euclidean step is gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b). The remainder is smaller than bb, which makes repeated steps terminate.

03

Trace the example to the base case

The sequence is gcd⁡(10,15)=gcd⁡(15,10)=gcd⁡(10,5)=gcd⁡(5,0)=5\gcd(10,15)=\gcd(15,10)=\gcd(10,5)=\gcd(5,0)=5. Stop at the zero second argument rather than attempting a remainder with a zero divisor.

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  • Greatest common divisor ExplanationAt 0:19
    Why this connection?

    For 1010 and 1515, the common positive divisors are 11 and 55, so their greatest common divisor is 55. Listing divisors works for small inputs but is not the efficient general method. For nonnegative aa and positive bb, the Euclidean step is gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b). The remainder is smaller than bb, which makes repeated steps terminate. The sequence is gcd⁡(10,15)=gcd⁡(15,10)=gcd⁡(10,5)=gcd⁡(5,0)=5\gcd(10,15)=\gcd(15,10)=\gcd(10,5)=\gcd(5,0)=5. Stop at the zero second argument rather than attempting a remainder with a zero divisor.

  • Greatest common divisor ExplanationAt 0:19
    Why this connection?

    For 1010 and 1515, the common positive divisors are 11 and 55, so their greatest common divisor is 55. Listing divisors works for small inputs but is not the efficient general method. For nonnegative aa and positive bb, the Euclidean step is gcd⁡(a,b)=gcd⁡(b,a mod b)\gcd(a,b)=\gcd(b,a\bmod b). The remainder is smaller than bb, which makes repeated steps terminate. The sequence is gcd⁡(10,15)=gcd⁡(15,10)=gcd⁡(10,5)=gcd⁡(5,0)=5\gcd(10,15)=\gcd(15,10)=\gcd(10,5)=\gcd(5,0)=5. Stop at the zero second argument rather than attempting a remainder with a zero divisor.

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