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wednesday09876
FAN AND MEDAL WILL BE REWARDED: Solve this system of congruences 2x=5 (mod 7) 3x=4 (mod 8)
i basically only need to know how to make the 2x and 3x into single x's.
not sure how to explain but I got an answer for the first, x must be equivalent to 6 + 7k, k = 0,1,2,...
would you happen to know how to make the x's into a workable form, i mean into just x??
for the second, x must be equivalent to 4 + 8m, m = 0, 1, 2, ...
lol just learned it online
need explanation?
how is x = 5? 2*5 = 10 = 3 mod 7
where do we get 3 mod 7 though?
jeff, yeah thanks that would be great.
http://www.math.rutgers.edu/~erowland/modulararithmetic.html
dude, 20 is the first x that works... not 5
basically a ≡ b mod m. means you do a-b=km
idk lol I just solved the system with k and x as variables ok maybe you know it better
alright just one sec im gonna try some work real quick
ok 5 doesn't work haha
40 does nice job
so i start with 2x-5 = 7 times k?
because i cant really solve further than that
1 sec asking friend
Think 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)
so then how would i come up with just x though? i meant tkhunny that was great but
x=20mod56 dont ask why
my friend gave a huggge explanation that is like 20 fb messages
lol dude i need the work thats most of the points!! haha thanks tho
do you think you could like copy the messages??
It's so hard...i dont think its worth it..uses chinese remainder theorem
haha nvm then thanks
wed. what class is this for?
wait your in algebra 2? Im in calc 3 and diff eq...
This 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.
ok dude let me copy the messages. Good job in coming up with a problem no1 knows the answer to
how 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 multi-valued. 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 math-competition 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 modular-type questions. If someone asks you for the last 2 digits of some giant product-like thing, think mod 100, or, by the CRT, mods 4 and 25
yup have fun with that I guess.
hey tk, how did you get from 2x=5 (mod 7) to 2*4 = 1 (mod 7)???
There 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)
wow is that that hard, or did i miss come questions in between?
oh i see it is two of them but \(2x\equiv 5(7)\iff 2x\equiv 12(7)\iff x\equiv 6(7)\)
similarly \[3x\equiv4(8)\iff 3x\equiv 12(8)\iff x\equiv 4(8)\]
2*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
it is clear that \(2\times 4=8\) for sure right? and also that \(8\equiv 1(7)\)
@wednesday09876 this i think is too much work, although you may need it for something later
so what do i do after that
as @tkhunny was saying now your job is to find \(x\) so that \[x\equiv 4(\text{ mod }7)\] and also \[x\equiv 4(\text{ mod }8)\]
but how did i legally just get it down to the form where there is one x?
if it was me, i would multiply \(8\times 7\) and then add 4
oh i see what your question is lets go slow
what ive been trying to understand is how the 2 and 3 disappeared from behind the x
ok i got it lets go real slow and get this for sure
you have \[2x\equiv 5( \text{ mod } 7)\] right
ok now forget about this mod for a second suppose you just had \(2x=5\) what would you do to solve for \(x\)?
i 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
so 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
what i really should have written is that \[12\equiv 5(\text { mod } 7)\] but no matter i hope you get the idea
so now we have \[2x\equiv 12 ( \text { mod } 7)\]and now we CAN divide by 2
yeah i got that, but i vaguely recall my teacher saying that when dividing you must divide the mod as well?
divide by \(2\) and we get \[x\equiv 6( \text { mod } 7)\]
alright as long as that is not a rule then yeah i got that part
lets check that this answer is right, so it is not a mystery
suppose \(x=6\) then \(2x=12\) right? and also \(12\equiv 5(\text { mod }7)\) so we know it is right
now lets repeat the process for \(3x\equiv 4(\text { mod }8)\)
you cannot divide 4 by 3, so keep adding 8 until you can
fortunately for you, this requires only one addition sometimes it takes more
since \(8+4=12\) 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
course you still have to solve for \(x\)
yeah i got that part, just needed it simplified so i could do the work
2x = 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.