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Use the Euclidean algorithm to hand-calculate the greatest common dlvisor for the Integers glven below:2,808 and 455Step 1: Flnd 91 and { so that2,808 455 91 + r1&#...

Question

Use the Euclidean algorithm to hand-calculate the greatest common dlvisor for the Integers glven below:2,808 and 455Step 1: Flnd 91 and { so that2,808 455 91 + r1' where 0 < 01 455.Then2,808 455 91'Step 2: Flnd 92 and rz So that455 r1 92 + r2' where 0 < r2 ` <T1'Then45592Step 3: Flnd 93 and r3 so that=r2 ` 93 + r3' where 0 < r3 r2'Then93Step 4: Find 94 and r4 so thatr2 = r3 94 + r4' where 0 < r4 < 13'ThenStep 5: Conclude that gcd(2,808, 4

Use the Euclidean algorithm to hand-calculate the greatest common dlvisor for the Integers glven below: 2,808 and 455 Step 1: Flnd 91 and { so that 2,808 455 91 + r1' where 0 < 01 455. Then 2,808 455 91' Step 2: Flnd 92 and rz So that 455 r1 92 + r2' where 0 < r2 ` <T1' Then 455 92 Step 3: Flnd 93 and r3 so that =r2 ` 93 + r3' where 0 < r3 r2' Then 93 Step 4: Find 94 and r4 so that r2 = r3 94 + r4' where 0 < r4 < 13' Then Step 5: Conclude that gcd(2,808, 455) 94"



Answers

Use Euclid's division algorithm to find the HCF of : (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

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.


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