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wednesday09876

  • one year ago

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

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  1. wednesday09876
    • one year ago
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    i basically only need to know how to make the 2x and 3x into single x's.

  2. pgpilot326
    • one year ago
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    not sure how to explain but I got an answer for the first, x must be equivalent to 6 + 7k, k = 0,1,2,...

  3. wednesday09876
    • one year ago
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    would you happen to know how to make the x's into a workable form, i mean into just x??

  4. pgpilot326
    • one year ago
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    for the second, x must be equivalent to 4 + 8m, m = 0, 1, 2, ...

  5. jefftheloveableguy
    • one year ago
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    x is 5

  6. jefftheloveableguy
    • one year ago
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    lol just learned it online

  7. jefftheloveableguy
    • one year ago
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    need explanation?

  8. pgpilot326
    • one year ago
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    how is x = 5? 2*5 = 10 = 3 mod 7

  9. wednesday09876
    • one year ago
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    where do we get 3 mod 7 though?

  10. wednesday09876
    • one year ago
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    jeff, yeah thanks that would be great.

  11. jefftheloveableguy
    • one year ago
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    http://www.math.rutgers.edu/~erowland/modulararithmetic.html

  12. pgpilot326
    • one year ago
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    dude, 20 is the first x that works... not 5

  13. jefftheloveableguy
    • one year ago
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    basically a ≡ b mod m. means you do a-b=km

  14. jefftheloveableguy
    • one year ago
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    idk lol I just solved the system with k and x as variables ok maybe you know it better

  15. wednesday09876
    • one year ago
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    alright just one sec im gonna try some work real quick

  16. jefftheloveableguy
    • one year ago
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    ok 5 doesn't work haha

  17. jefftheloveableguy
    • one year ago
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    40 does nice job

  18. wednesday09876
    • one year ago
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    so i start with 2x-5 = 7 times k?

  19. wednesday09876
    • one year ago
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    because i cant really solve further than that

  20. jefftheloveableguy
    • one year ago
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    1 sec asking friend

  21. wednesday09876
    • one year ago
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    alright thanks

  22. tkhunny
    • one year ago
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    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)

  23. pgpilot326
    • one year ago
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    have a look

  24. wednesday09876
    • one year ago
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    so then how would i come up with just x though? i meant tkhunny that was great but

  25. jefftheloveableguy
    • one year ago
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    x=20mod56 dont ask why

  26. jefftheloveableguy
    • one year ago
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    my friend gave a huggge explanation that is like 20 fb messages

  27. wednesday09876
    • one year ago
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    lol dude i need the work thats most of the points!! haha thanks tho

  28. wednesday09876
    • one year ago
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    do you think you could like copy the messages??

  29. jefftheloveableguy
    • one year ago
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    It's so hard...i dont think its worth it..uses chinese remainder theorem

  30. wednesday09876
    • one year ago
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    haha nvm then thanks

  31. jefftheloveableguy
    • one year ago
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    wed. what class is this for?

  32. wednesday09876
    • one year ago
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    algebra 2 trig

  33. jefftheloveableguy
    • one year ago
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    wait your in algebra 2? Im in calc 3 and diff eq...

  34. tkhunny
    • one year ago
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    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.

  35. jefftheloveableguy
    • one year ago
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    ok dude let me copy the messages. Good job in coming up with a problem no1 knows the answer to

  36. jefftheloveableguy
    • one year ago
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    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

  37. jefftheloveableguy
    • one year ago
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    yup have fun with that I guess.

  38. wednesday09876
    • one year ago
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    hey tk, how did you get from 2x=5 (mod 7) to 2*4 = 1 (mod 7)???

  39. tkhunny
    • one year ago
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    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)

  40. satellite73
    • one year ago
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    wow is that that hard, or did i miss come questions in between?

  41. satellite73
    • one year ago
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    oh i see it is two of them but \(2x\equiv 5(7)\iff 2x\equiv 12(7)\iff x\equiv 6(7)\)

  42. satellite73
    • one year ago
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    similarly \[3x\equiv4(8)\iff 3x\equiv 12(8)\iff x\equiv 4(8)\]

  43. wednesday09876
    • one year ago
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    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

  44. satellite73
    • one year ago
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    it is clear that \(2\times 4=8\) for sure right? and also that \(8\equiv 1(7)\)

  45. satellite73
    • one year ago
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    @wednesday09876 this i think is too much work, although you may need it for something later

  46. wednesday09876
    • one year ago
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    yeah i got it

  47. wednesday09876
    • one year ago
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    so what do i do after that

  48. satellite73
    • one year ago
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    as @tkhunny was saying now your job is to find \(x\) so that \[x\equiv 4(\text{ mod }7)\] and also \[x\equiv 4(\text{ mod }8)\]

  49. wednesday09876
    • one year ago
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    but how did i legally just get it down to the form where there is one x?

  50. satellite73
    • one year ago
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    if it was me, i would multiply \(8\times 7\) and then add 4

  51. satellite73
    • one year ago
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    oh i see what your question is lets go slow

  52. wednesday09876
    • one year ago
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    what ive been trying to understand is how the 2 and 3 disappeared from behind the x

  53. wednesday09876
    • one year ago
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    yeah sorry

  54. satellite73
    • one year ago
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    ok i got it lets go real slow and get this for sure

  55. satellite73
    • one year ago
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    you have \[2x\equiv 5( \text{ mod } 7)\] right

  56. wednesday09876
    • one year ago
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    yeah

  57. satellite73
    • one year ago
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    ok now forget about this mod for a second suppose you just had \(2x=5\) what would you do to solve for \(x\)?

  58. wednesday09876
    • one year ago
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    divide two over

  59. satellite73
    • one year ago
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    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

  60. wednesday09876
    • one year ago
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    yeah

  61. satellite73
    • one year ago
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    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

  62. wednesday09876
    • one year ago
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    got it

  63. satellite73
    • one year ago
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    what i really should have written is that \[12\equiv 5(\text { mod } 7)\] but no matter i hope you get the idea

  64. satellite73
    • one year ago
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    so now we have \[2x\equiv 12 ( \text { mod } 7)\]and now we CAN divide by 2

  65. satellite73
    • one year ago
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    k?

  66. wednesday09876
    • one year ago
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    yeah i got that, but i vaguely recall my teacher saying that when dividing you must divide the mod as well?

  67. satellite73
    • one year ago
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    divide by \(2\) and we get \[x\equiv 6( \text { mod } 7)\]

  68. satellite73
    • one year ago
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    no

  69. wednesday09876
    • one year ago
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    alright as long as that is not a rule then yeah i got that part

  70. satellite73
    • one year ago
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    lets check that this answer is right, so it is not a mystery

  71. satellite73
    • one year ago
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    suppose \(x=6\) then \(2x=12\) right? and also \(12\equiv 5(\text { mod }7)\) so we know it is right

  72. wednesday09876
    • one year ago
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    aha got it

  73. satellite73
    • one year ago
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    now lets repeat the process for \(3x\equiv 4(\text { mod }8)\)

  74. satellite73
    • one year ago
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    you cannot divide 4 by 3, so keep adding 8 until you can

  75. satellite73
    • one year ago
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    fortunately for you, this requires only one addition sometimes it takes more

  76. wednesday09876
    • one year ago
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    yeah i got it

  77. satellite73
    • one year ago
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    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

  78. wednesday09876
    • one year ago
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    ah thank you

  79. satellite73
    • one year ago
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    yw

  80. satellite73
    • one year ago
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    course you still have to solve for \(x\)

  81. wednesday09876
    • one year ago
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    yeah i got that part, just needed it simplified so i could do the work

  82. satellite73
    • one year ago
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    k good

  83. pgpilot326
    • one year ago
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    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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