Apply the algorithm for the GCD in Section 10.1 to 15 and 46 , and then use the results to deter

Apply the algorithm for the GCD in Section 10.1 to 15 and 46...

Key Concepts

Extended Euclidean Algorithm

The Extended Euclidean algorithm not only computes the GCD of two integers but also finds a pair of coefficients (often referred to as Bezout coefficients) that express the GCD as a linear combination of the two original numbers. This is essential for finding multiplicative inverses in modular arithmetic.

Euclidean Algorithm

The Euclidean algorithm is a process used to determine the greatest common divisor (GCD) of two integers. It works by repeatedly applying the division algorithm, replacing the larger number with the remainder until the remainder is zero, at which point the last nonzero remainder is the GCD.

Multiplicative Inverse in Modular Arithmetic

In modular arithmetic, the multiplicative inverse of an integer a modulo n is an integer b such that the product ab is congruent to 1 modulo n. The Extended Euclidean algorithm is typically used to compute this inverse when the GCD of a and n is 1, ensuring that a has an inverse modulo n.

Transcript

00:01 In this video we'll be using the euclidean algorithm to find the gcd or the greatest common divisor of six different pairs of integers.

00:09 So we'll start with part a which is asking us to find the gcd of 1 and 5.

00:17 Okay so we can start with the big number on the left equals 1 times 5 and we can see that there's no remainder here.

00:28 So in this case 1 and 5 don't have a...

00:36 In this case the gcd is just 1.

00:40 So 1 and 5 are said to be relatively prime or they're called co -prime because their gcd is 1.

00:50 Okay so now we continue on to part b which is asking us to find the gcd of 100 and 101.

01:02 Okay we write the big number on the left so 101 equals 100 times 1 plus the remainder of 1.

01:13 So then we write 100 equals 1 times 100 and then there's no remainder.

01:21 So at this point we recognize that we have a value of 0 as a remainder and we look up one row above to find that the gcd in this case is also 1.

01:33 So again you can say that 100 and 101 are co -prime because the gcd of those numbers ends up being 1.

01:44 So for problem c we have the gcd of 123 and 277.

01:52 So we have...

01:52 So we write 277 equals 123...

01:57 Oops...

01:59 Equals 123 times 2 and what does that leave us with as a remainder? that leaves us with a remainder of 31.

02:11 So now i write 123 equals 31 times 3 or 4.

02:18 Let's see 31 times 3...

02:21 93...

02:24 I want to say...

02:25 Can't go 4.

02:28 Right okay times 3 and then the remainder would be...

02:33 If i can do math...

02:36 30 of course right? so 31 equals 1 times 30 and then you have a remainder of 1 and then you write 30 equals 30 times 1.

02:50 That's the remainder of 0.

02:52 Oops 0.

02:53 And at this point we can stop, look one above above and you can see that again the gcd is equal to 1.

03:02 Whatever that squiggle was.

03:05 Okay so that's part c.

03:07 Now we can go down here and write part d which is asking us to find the gcd of 1529 and 14039.

03:23 Okay so 14039 equals 1529 times...

03:35 So using a calculator i found that multiplying these or 1529 by 9 and adding a remainder of 278 gives you the number of 14039 on the left.

03:49 Okay so now we continue on.

03:52 1529 equals 278 times 5 and then you're left with a remainder of 139...

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