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8. Show that if (1 02 03, < 04 C5 06 are natural numbers; then 9 | (aj Ci _ Or 9 | (aj + 4i) , for some pair G; # aj- Hint: Pigeon hole principle: Approach 1: co...

Question

8. Show that if (1 02 03, < 04 C5 06 are natural numbers; then 9 | (aj Ci _ Or 9 | (aj + 4i) , for some pair G; # aj- Hint: Pigeon hole principle: Approach 1: consider the remainder of &; when divided by 9. I.e., for each i, letCi9qi + Ti,where 0 < ri < 8 If r; = Tj for some pair G; # 4j, then aj li 9(4; Yi) so 9 | (aj Wi). If not, then T: T6 are all distinct elements of the set {0,1,2,3,4,5,6,7, 8}, s0 at least 5 of these remainders are in the set {1,2,3,4,5,6,7,8}. These are your

8. Show that if (1 02 03, < 04 C5 06 are natural numbers; then 9 | (aj Ci _ Or 9 | (aj + 4i) , for some pair G; # aj- Hint: Pigeon hole principle: Approach 1: consider the remainder of &; when divided by 9. I.e., for each i, let Ci 9qi + Ti, where 0 < ri < 8 If r; = Tj for some pair G; # 4j, then aj li 9(4; Yi) so 9 | (aj Wi). If not, then T: T6 are all distinct elements of the set {0,1,2,3,4,5,6,7, 8}, s0 at least 5 of these remainders are in the set {1,2,3,4,5,6,7,8}. These are your "pigeons' DOW consider the pigeon holes {1,8}, {2,7}, {3,6}, {4,5}. Approach 2: consider the 10 natural numbers (a6 (5 ) (a6 (4) < (a6 a1) < (a6- + (1) < (a6 + (5) . Use the PHP to show that there are two distinct numbers on this list whose difference is divisible by 9. Now what?



Answers

Find the set of possible remainders when an integer is divided by the given integer.
Seven

So we want the remainder when um is divided by B. And the remainder is uh has to be less than or equal to zero but strictly less than B. Okay, so I was set contains the remainder Can be zero. It can be one. I guess. You can say that this is less than or equal to or less than BS five. Okay, so the remainder, they can be 01 or two, 34. I cannot be five. So these are all the possible remainders.

Okay, so um if m and n are relatively prime, we first need to show by example that um the equation um plus VN equals one um can be satisfied by multiple values. So let's say like M equals three and um and equals five um sure I can come up with two different pairs, let's say two times three plus minus one times five that equals one. And what about two times 5 -3 times 3? So we'll write it this way to keep it in the same order. That also equals one. Um So these are two different pairs. You and be here, you comma b Is two, comma -1 and here you can movie is -3 02. Okay, so now we're going to restrict um what if we restrict you to be between zero and absolute value of N but not including in um Okay. And you is this first coefficient? Um philosophy? And it's all right. Yes, equals one. Okay, well um let's suppose there were uh this was not unique. So let's say there was another pair. U one, M plus B. One, N equal one. You come a the Not the same as you want to have a view one. Then what would happen? We have these two equations. What if we subtract them then? We have u minus U one M Plus V -3 1 and Equals 1 -1, which is zero. So we have -11 M equals V one minus V times N. But M and n are relatively prime and um the right hand side is divisible by n. Um because well, first of all if if the if the V's are the same, then they use have to be the same because you can solve for the use in terms of the V's So the Vs are different. So the right side is not zero and it's divisible by N. That means the left side has to be divisible by N. And no part of them is going to help because M and n are relatively prime. So that means that -11 has to be divisible by Ed but we said these were both true and then these were both true because that was the constraint that we added and we're trying to show that it's unique. So we're supposing that it's not in hoping for a contradiction. Um Okay, but how can that be true? You and you won are both between zero and n absolute value and not including n And and divides the difference. The only way that's true is U equals U- one equals zero. Uh In which case the equals V one as well um Actually probably not even possible because if those are about zero then what are we going to put their some Manager Times The equals one unless and it was one. Um Okay. And then in this case uh is the absolute value of v less than the absolute value of em um Well let's see, we have um V n um equals one minus. Um Right so V equals one minus um over n. And then we said the uh you was greater than equal to zero, less than the absolute value of N. So that means the absolute value of um is less than the absolute value of N. M. Um And so the absolute value of one minus um you can only get oh this is tough use and ends look pretty similar. Okay let's to make sure using ends using ends look different. Okay um now one minus the absolute value of um it can actually be one bigger than the absolute value of um But so we're going to write less than or equal to the absolute value of an M. Okay uh so that actually gives now then the absolute value of V is less than or equal to the absolute value of N. Um absolute value of em because we're dividing here by N. And we got to absolute value of N. M. Okay, can it be equal though to the absolute value of em? Say the absolute value of the equals absolute value of em? Um That's not going to work because now I am then m is going to divide um plus VN the end so um plus VN is not one Unless and More one. Um But they said relatively prime managers differing from plus or -1 so we're good. Okay so we've done it

Okay. So when the remainder when some interior A. Is uh divided by B. So A mad B. Uh huh. So it's A Mad B. Is too so a matter, so the set of possible commanders uh remainder can either be zero? Oh what? Uh huh. Mhm. So these are all of the possible remainders because our has to be Less than or equal to zero and it's strictly less than B. Okay, so all of the possible things here B is too right and our has to be less than or equal to zero. So it can be zero, it can be one but it cannot be too So that's all of the possible remainders we have.


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