anonymous
  • anonymous
FAN AND MEDAL WILL BE REWARDED: Solve this system of congruences 2x=5 (mod 7) 3x=4 (mod 8)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
i basically only need to know how to make the 2x and 3x into single x's.
anonymous
  • anonymous
not sure how to explain but I got an answer for the first, x must be equivalent to 6 + 7k, k = 0,1,2,...
anonymous
  • anonymous
would you happen to know how to make the x's into a workable form, i mean into just x??

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

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