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anonymous
 2 years ago
FAN AND MEDAL WILL BE REWARDED:
Solve this system of congruences
2x=5 (mod 7)
3x=4 (mod 8)
anonymous
 2 years ago
FAN AND MEDAL WILL BE REWARDED: Solve this system of congruences 2x=5 (mod 7) 3x=4 (mod 8)

This Question is Closed

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i basically only need to know how to make the 2x and 3x into single x's.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0not sure how to explain but I got an answer for the first, x must be equivalent to 6 + 7k, k = 0,1,2,...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0would you happen to know how to make the x's into a workable form, i mean into just x??

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0for the second, x must be equivalent to 4 + 8m, m = 0, 1, 2, ...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0lol just learned it online

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0how is x = 5? 2*5 = 10 = 3 mod 7

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0where do we get 3 mod 7 though?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0jeff, yeah thanks that would be great.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0http://www.math.rutgers.edu/~erowland/modulararithmetic.html

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0dude, 20 is the first x that works... not 5

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0basically a ≡ b mod m. means you do ab=km

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0idk lol I just solved the system with k and x as variables ok maybe you know it better

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0alright just one sec im gonna try some work real quick

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok 5 doesn't work haha

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so i start with 2x5 = 7 times k?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0because i cant really solve further than that

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1Think about equivalence classes 2*0 = 0 (mod 7) 2*1 = 2 (mod 7) 2*2 = 4 (mod 7) 2*3 = 6 (mod 7) 2*4 = 1 (mod 7)  2*5 = 3 (mod 7)  2*6 = 5 (mod 7) 3*0 = 0 (mod 8) 3*1 = 3 (mod 8) 3*2 = 6 (mod 8) 3*3 = 1 (mod 8)  3*4 = 4 (mod 8)  3*5 = 7 (mod 8) 3*6 = 2 (mod 8) 3*7 = 5 (mod 8) 2x=5 (mod 7) 3x=4 (mod 8)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so then how would i come up with just x though? i meant tkhunny that was great but

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0x=20mod56 dont ask why

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0my friend gave a huggge explanation that is like 20 fb messages

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0lol dude i need the work thats most of the points!! haha thanks tho

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0do you think you could like copy the messages??

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0It's so hard...i dont think its worth it..uses chinese remainder theorem

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wed. what class is this for?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wait your in algebra 2? Im in calc 3 and diff eq...

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1This is where the thinking I suggested might have take you... With a little luck, there is a multiplicative inverse for each value presented. 2x=5 (mod 7) 2*4 = 1 (mod 7) (Everyone has an inverse (mod 7)) 4*2*x = 4*5 (mod 7) x = 6 mod 7 3x=4 (mod 8) 3*3 = 1 (mod 8) (Only 1, 3, 5, and 7 have inverses of this type (mod 8)  Think about mutually prime) 3*3*x = 3*4 (mod 8) x = 4 mod 8 And the problem is substantially simplified.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok dude let me copy the messages. Good job in coming up with a problem no1 knows the answer to

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0how would you solve 2x=5 (mod 7) 3x=4 (mod8) Here's the catch: division is unique if the modulus is prime; otherwise, it can be multivalued. So 5 / 2 mod 7 is unique, and equal to... 6, right? what does 5/2 mean? Five divided by two What number, multiplied by 2, yields 5 when operating mod 7? 20? Sure. But 20 = 6 mod 7 Notice that any value you can find which does so is 6 mod 7 That's the easier part. The trickier part is 3x = 4 mod 8. Let's think for the moment: if instead it were 2x = 4 mod 8, then both 2 and 6 would work, right? That's because the 2, from 2x, divides the modulus So division in modulo composite numbers may be multivalued. This led me to a wrong answer on an AIME problem, and I was mad wow Anyway, since 3 does not divide the modulus, 3x = 4 mod 8 has a unique solution, namely 6 what if x were 12? Namely 4*, that was and did you use a ≡ b mod m. a – b = km Here's what we now know (sorry for not using equivalent signs): x = 6 mod 7; x = 4 mod 8 With me? how did you change the 5 to 6 Since 2x = 5 mod 7, x = 5/2 mod 7 = 6 mod 7, as above (5/2 does not mean 2.5, it means "the number, which, upon multiplication by 2, yields 5 in modulo 7") Anyway, we have two different equivalences for x in different moduli. Now comes the key point: the Chinese Remainder Theorem (CRT) It says: if you have a system of equations x = ai mod mi And the mi are relatively prime There there is a unique solution c such that x = c mod (m1*m2*m3*m4*...*mi) It should make a bit of intuitive sense If we know a certain number yields a certain remainder upon division by 3 and 5, say, then we know that it must have a unique remained upon division by 15. right Does this make sense? some thanks though just started mods Okay. So x = c mod (8 * 7), or x = c mod 56 So, can you find a c such that c = 6 mod 7 AND c = 4 mod 8? (and c < 56) Can't find a c List the 6 mod 7 less than 56: 6, 13, 20, 27, 34, 41, 48, 55 List the 4 mod 8 less than 56: 4, 12, 20, 28, 36, 44, 52 Their intersection has exactly 1 element, as predicted by the CRT: 20 nice So c = 20, and our solution is x = 20 mod 56 Note: Although you can solve the original question just by guessing (does ? work, for ? up to 20) The point is, if x = 20 mod 56, then x does not have to be 20 It can be 76 right Or 76 + 56 * z, were z is an integer. All such values will solve your system Sorry to go through this so fast, but that is, in short, how mods work. yea ill have to look into it thanks! Im tired now do u consider mods as easy? Well, you already intuitively know a couple important mods: mod 2 (even/odd) and mod 10 (last digit). And I've been seeing them on mathcompetition type questions since freshman year, so I'd say they are a bit easier than things like cyclic quadrilaterals They can get hard, however... One sec http://www.artofproblemsolving.com/Wiki/index.php/2013_AIME_I_Problems/Problem_11 I thought mods were easy, but this problem was a pure modular arithmetic one which I messed up somewhere The key to my failure was this: part of the problem involves x = 3 mod 9 And trying to divide out the relevant parts (since x is already a multiple of 3, but unknown which one) Anyway You'll always see modulartype questions. If someone asks you for the last 2 digits of some giant productlike thing, think mod 100, or, by the CRT, mods 4 and 25

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yup have fun with that I guess.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0hey tk, how did you get from 2x=5 (mod 7) to 2*4 = 1 (mod 7)???

tkhunny
 2 years ago
Best ResponseYou've already chosen the best response.1There is no "from" in there. 2x = 5 (mod 7) is the given problem statement 2*4 = 1 (mod 7) is the demonstration that 2 has a multiplicative inverse (mod 7). The inverse is 4. I then used the multiplicative inverse with the original problem statement. 2x = 5 (mod 7) 4*2*x = 4*5 (mod 7) Giving x = 6 (mod 7)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0wow is that that hard, or did i miss come questions in between?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh i see it is two of them but \(2x\equiv 5(7)\iff 2x\equiv 12(7)\iff x\equiv 6(7)\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0similarly \[3x\equiv4(8)\iff 3x\equiv 12(8)\iff x\equiv 4(8)\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.02*4 = 1 (mod 7) is the demonstration that 2 has a multiplicative inverse (mod 7). The inverse is 4. How do you get that? sorry about that just dont really get it

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0it is clear that \(2\times 4=8\) for sure right? and also that \(8\equiv 1(7)\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0@wednesday09876 this i think is too much work, although you may need it for something later

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so what do i do after that

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0as @tkhunny was saying now your job is to find \(x\) so that \[x\equiv 4(\text{ mod }7)\] and also \[x\equiv 4(\text{ mod }8)\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0but how did i legally just get it down to the form where there is one x?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0if it was me, i would multiply \(8\times 7\) and then add 4

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0oh i see what your question is lets go slow

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0what ive been trying to understand is how the 2 and 3 disappeared from behind the x

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok i got it lets go real slow and get this for sure

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0you have \[2x\equiv 5( \text{ mod } 7)\] right

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok now forget about this mod for a second suppose you just had \(2x=5\) what would you do to solve for \(x\)?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i assume you mean "divide by 2" and yes, that is what you would do but you cannot divide this by 2 can you? because you have to work with whole numbers

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so here is what we can do to allow us to divide by 2 add \(7\) to \(5\) because \(5( \text{ mod }7)\equiv 12( \text{ mod }7)\) that should be more or less clear both 5 and 12 have a remainder of 5 when divided by 7

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0what i really should have written is that \[12\equiv 5(\text { mod } 7)\] but no matter i hope you get the idea

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0so now we have \[2x\equiv 12 ( \text { mod } 7)\]and now we CAN divide by 2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i got that, but i vaguely recall my teacher saying that when dividing you must divide the mod as well?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0divide by \(2\) and we get \[x\equiv 6( \text { mod } 7)\]

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0alright as long as that is not a rule then yeah i got that part

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0lets check that this answer is right, so it is not a mystery

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0suppose \(x=6\) then \(2x=12\) right? and also \(12\equiv 5(\text { mod }7)\) so we know it is right

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0now lets repeat the process for \(3x\equiv 4(\text { mod }8)\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0you cannot divide 4 by 3, so keep adding 8 until you can

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0fortunately for you, this requires only one addition sometimes it takes more

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0since \(8+4=12\) is should be clear that \[12\equiv 4(\text { mod } 8)\] and so now we have \[3x\equiv 12(\text {mod} 8)\] divide by 4 etc etc

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0course you still have to solve for \(x\)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yeah i got that part, just needed it simplified so i could do the work

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.02x = 7m + 5 , m = 0, 1, 2, ... we want just x so we have to multiply everything by the inverse of 2 mod 7, which is 4. => 8x = 7m + 20 => x = 7m + 6 now we sub that in for x in the other congruence => 3(7m+6) = 21m + 18 = 4 mod 8 => 5m +2 = 4 mod 8 => 5m = 2 mod 8 again we want only x so we multiply by the inverse of 5 mod 8, which is 5. => 25m = 10 mod 6 => m = 2 mod 8 => m = 8n + 2 now we sub that into our original equation for x... => x = 7(8n + 2) + 6 = 56n + 14 + 6 = 56n + 20 so x = 20 mod 56. I couldn't let it go until i figured it out so i apologize if this is any sort of a nuisance.
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