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Hi the Euclidean Algorithm You combination [ that "obuceuwods 1 integers. 28) express by other using words. the find grbetEsuctodemon nonzero H andy # #5_ the ...

Question

Hi the Euclidean Algorithm You combination [ that "obuceuwods 1 integers. 28) express by other using words. the find grbetEsuctodemon nonzero H andy # #5_ the iNcgets 3

Hi the Euclidean Algorithm You combination [ that "obuceuwods 1 integers. 28) express by other using words. the find grbetEsuctodemon nonzero H andy # #5_ the iNcgets 3



Answers

Use the extended Euclidean algorithm to express $\operatorname{gcd}(26,91)$ as a linear combination of 26 and $91 .$

Right. So when you use the extended Euclidean algorithm to take the G c d of 26 91 s, So first we get that 26 is equal to zero times many one plus 26 right? Can you get that? 91 is equal to three times 26 plus 13 and that 26 is equal to two times 13. This means our cue one is equal. Zero Q two is equal to three and cue wth reasonable to two. Our second remainder are two's equal to 26 our first remainder is equal to 13. Missile minute. That's here. If you know any, equals three rounds and we get a few initialization czar algorithms, so we know that that's not tickle toe One s one is equal to zero. T not is equal to zero and t one is equal to one. So now it just plugged the stuff into the algorithm. We see that s two's equal. Thio s naw, minus Q one s one. It has to be one minus zero is equal the one and t to use equal to t not minus Q one t one. Okay, this is gonna be zero minus zero Rex, the Q and A zero. Okay, So because it's zero now, as three, it's gonna be equal to s one minus Q two U S, too. Okay, so that's one we notice. Zero, uh, que two times as too. So it's gonna be three times one. It's gonna be three, because it's a negative. Three and t three then is going to be t one minus Q Thio tea too. Case the T one is one. Subtract that by a cue to which was three times T too, which was zero. Okay, says just equal to one. So then what we have we have s three times a plus t three times b is gonna give us our g c d of 26 91. Okay, so we get negative three times 26 plus one times 91 which is indeed equal to 13

Were asked to use the Euclidean algorithm to find the greatest common divisor of a pair of numbers. In part, they were asked to find the greatest common divisor of one in five. Well, we see that five is the larger of the two numbers, so five is equal to one times five, and therefore it follows that the greatest common divisor of one in five is equal to one now in part B for us to find the greatest common divisor of 101 on one. So 11 is the greater of these two numbers. One of one is equal to 100 times one plus one, and we have that 100 is the greater of 101 and so this is equal to one times 100. So we see the last non zero remainder is one. So it follows that the greatest common divisor of 100 in 101 is one in part C were asked to find the greatest common divisor of 123 and 277. The greater of these two numbers is 277 so dividing to 77 by 1 23 we get to send seven is equal to 1 23 times two. This is 246 plus 11 and sorry plus 31. And then we'll divide 1 23 by 31 since 31 is a non zero remainder. So 1 23 is equal to 31 times three. This is 93 plus 30. This is another non zero remainder, so we'll divide 31 by 30. When we get 31 is equal to 30 times one plus one, and one is a none zero remainder. So finally will divide 30 by one. 30 is equal to 30 times one, but one times 30 I guess to be consistent So it follows. The last non zero remainder is one, and therefore that the greatest common divisor of 277 and 123 is one. In Part D were asked to find the greatest common divisor of 1529 and 14,039. Now we had that 14,000 29 is the greater of these. Two will divide it by 1529. This is equal to 1529 and then multiply this by nine, so we get 9000 plus 4500. So that's 13 1005 100 13,500 plus 180 is 13,600 80 plus 81 13,000 761 and then we have the 14,038 minus 13,000. Whatever I said is 277. Now we see that this is a non zero. Remainders will divide 1529 by 277. So this is equal to 2 77 times five. That gives us 200 times five is 1000 plus 350 is 13 50 plus 35 is 13 85 then subtracting that from 15 29 we get 1 44. Now, this is another non zero remainder. So will divide to 77 by 144. This is equal to 1 44 times one plus, then to 77 minus 1 44 which is 1 33. This is another non zero remainder. Soul divide 1 44 by 1. 33. So this is 1 33 times one plus 11. This is another non zero remainder. So we'll divide 1 33 by 11. This is 11 times. Yeah, 12. So this gives us 120. Plus 12 is 1 32. So we've remainder of one. And this is still a non zero remainder. So we'll divide 11 by one. So we get 11 is equal to one times 11 and finally we have a zero remainder. So the last non zero remainder is one and therefore follows that the greatest common divisor of the two numbers 15 29 and 14,000. Excuse me? 38 is one in part. Uh huh. Mhm. Actually made a mistake here. So first of all, this should be 14,039 for Part D. And then this is equal to 15 129 times nine plus 2. 77. Yeah, but minus one. So this is going to be to 78 actually to account for the extra print, the nine. And so now we divide 15 29 by 2. 78. So 15. 29. Divided by 2. 78. Well, this is gonna be to 78 times five as before. Plus now, 1. 39 since we have to reduce by five and then we divide to 78 by 1. 39. Yeah, and we get to 78 is actually equal toe 1 39 times, two with no remainder. So this last step could be the last few steps here could be admitted as they're incorrect. And therefore the answer actually is also incorrect. The answer is the last none zero remainder, which is 31 39. So the greatest common divisor of 1529 and 14,000 39 is equal to 1. 39. Now, with this in mind in parts E were asked to find the greatest common divisor of 1529 and 14,000 38 so very similar to the previous problem. Except now we're dividing 14,038 by 1529 and this is pretty much the same way that before it really mixed up the different parts. The problem. This is going to be 15 29 times nine plus 2 77 and then we have dividing 15 29 by 2 77. This is going to be 2 77 times five plus 1 44 and because when 44 is a non zero remainder, we divide to 77 by 1 44 to get 1 44 times one plus 1 33 because 133 is non zero, we divide 1 44 by 1 33 to get 1 33 times one plus 11. And because 11 90 we divide 1 33 by 11 to get 11 times 12 plus one. And because one's non zero, we divide 11 by one to get one times 11. And because the last non zero remainder is one, the greatest common divisor of 1529 and 14,000 38 is one. Finally, in part, F were asked to find the greatest common divisor of the numbers. 11,111 and 111,111. So we had. The larger of the two is 111,000. 111. We'll divide this by 11,111 we get this. Is this number times 10. The shifts it over. Yeah, and then we need to add a one bit to the end so we'll add a one changes the zero to a one sort of the way I went about this instead of actually dividing these numbers and then because one is a non zero remainder, we divide 11,111 by one to get one times 11,111. So they remainder of zero in the algorithm terminates now because the last non zero remainder was a one. It follows that the greatest common divisor of 111,000, 111 and 11,111 is one.

Were asked to use the extended Euclidean algorithm to express the greatest common denominator of 252 and 356 as a linear combination of 252 and 356. So first, let's just apply the Euclidean algorithm we have that the larger of the two numbers to 50 52 3 56 is 3 56. So we'll divide 3 56 by 2 52. So we have 3. 56 is equal to one times 2 52 plus 104. This is a non zero remainder, so we'll divide 252 by 104 I get to 52 is equal to two times 104 plus 44. This is a non zero remainder, so I'll divide 104 by 44. I get one of four is equal to two times 44 plus 16. This is a non zero remainder, so I'll divide 44 by 16. I'll get 44 equals two times 16 plus 12. 12 is a non zero. Remainder saw Divide 16 by 12. I get 16 is equal to one times 12 plus four, which is a non zero remainder. So divide 12 by four. I get 12 is equal to three times four and end up with zero remainder. So the Euclidean algorithm terminates. And from the Euclidean algorithm, I get the list of questions. Q one equals one. Cute, too, from the second step is to Q three from the third step is to Q four from the fourth step is to Q five from the fifth step is one and Q six. We're not including because there's there's a zero remainder at that step. Now, with the extended Euclidean algorithm we have, that s zero equals one s one equals zero and we have that t zero equals zero and t one equals one, and we'll execute the extended algorithm to this algorithm. Tells us that as to is equal to s one minus Q one, we're sorry s zero minus Q one times s one which is equal to as Syria was one minus one times zero, which is one and likewise we have that t two is equal to t zero minus Q one t one. This is going to be zero minus one times one, which is a negative one. S three is by the Rikers in formula s one minus Que To s to this is going to be s 10 So zero minus Q two, which is to times s to which we calculated to be one. So this is negative. Two Likewise, t three is t one minus cute too. T two t one were given is one que two is two and t two. We calculated to be negative one. So we get one plus two, which is three, and we're going to continue until we get to s six and t six. Since we had six steps from the previous part, so s four is equal to s two minus Q three s three. We calculated us too. To be one. Q three is too and s to We're sorry s three. We calculated to be negative too. So this is one plus four, which is five. Likewise, t four is t two minus Q three t three. We calculated t to to be negative one Q three is to and t three be calculated to be three. So this is negative one minus six which is negative seven next as five is equal to s three minus Q four s four. We calculated as three to be negative, too. Q four is to and we calculated s four to be five. So this is negative. Two minus 10 or negative 12. Likewise, t five is t three minus Q four t four. We calculated t three to be three. Q four is too, and we calculated t four to be negative seven. So we get three plus 14 which is 17 and now s six is s four minus Q five s five. We calculated s four to be five Q five. We calculated to be one and s five is calculated to be negative. 12. This is five plus 12 which is 17 and t six. Likewise is T four minus Q five t five. We calculated t four to be negative. Seven. T five is one our Sorry Q five is one in t five. We calculated to be 17 so we get negative seven minus 17 which is negative. 24. And here the extended Euclidean algorithm terminates and we have that the greatest common divisor of 252 and 356 which from our Euclidean algorithm was the last non zero remainder, which is four. And from the extended Euclidean algorithm. This is going to be s six. I'm 252 plus t six. Uh huh. Sorry. We have a is greater than be so really think about this the other way. The greatest common divisor of 356 and 252. This is four. And from the extended Euclidean algorithm, this is equal to S six, which is 17 times 3. 56 plus t six, which is negative. 24 times 252.

Very so here we have that A and B are positive integers I'm just gonna go ahead and soon That is greater than or equal to be. And one of them should be gooder than or equal to the other. We want to prove that the grace common divisor of to a two out of a minus one and two to the power of B minus one is equal to 22 Very So here we have that A and B are positive integers I'm just gonna go ahead and soon that is greater than or equal to be. And one of them should be gooder than or equal to the other. We want to prove that the gray's common divisor that are it's gonna be equal to a might be okay. So that means that the g c d of A and B being able to obesity of being ours also equal to the G c d a b A such that a is equal to be times que plus r okay and by the Oiler uh, division methods. So we we get that the Ukrainian method apologize. The G C D of A and B is equal to the G C d, uh, B and R. Okay. And I also know that our is gonna be equal to might just be right. So it's very ready that there most we can use that in our equation. Right? So for looking at the G c D of two to the power of a minus one and two to the power of B minus one, there's gonna be equal to the common divisor of two to the part of B minus one A my baby. Okay, so that means that the g c d of A and B being able to obesity of being ours, also equal to the G C D a b A might just be right. So it's finally ready that there most. We can use that in our equation. Right? So for looking at the G c D of two to the power of a minus one and two to the power of B minus one, I was gonna be to part of a minus one mod to use the power of B minus one. Okay, so I did. There was used basically this expression to rewrite what we were given. Pulled to the is common advisor of To to kind of be minus one, 22 part of a minus one mod to use the power of B minus one. Okay, so what I did there was used basically this expression to rewrite what we were given. That too, to a minus one mod to be minus one is equal to two to power a mod be well, we know that too. To a minus one mod to be minus one is equal to two to power a mod be, but is equal to two out of B minus one two. However a mod be so I'm gonna use that and put it in our equation above. To get that, the G C D is equal to two out of B minus one two. However a mod be just one. I just won. Okay, well, so then that means that this is equal to the G c d. There's this is equal to if we follow the okay, well, so then that means that this is equal to the G c d. Be the algorithm right where we continuously take the smaller number and okay, the G e g c d ah. Using the module Oh, that repeated calculation we end up with is two to the power of the G C D B minus. So is this is equal to if we follow the algorithm right where we continuously take the smaller number and okay, the G e g c d ah, using the module. Oh, that repeated calculation we end up with is two to the power of the G C D B minus one.


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