Results 1 to 11 of 11
Like Tree2Thanks
  • 1 Post By romsek
  • 1 Post By johng

Thread: Modulus

  1. #1
    Newbie leibnitz's Avatar
    Joined
    Feb 2013
    From
    Onitsha
    Posts
    12
    Thanks
    2

    Post Modulus

    PLEASE FANS HELP ME OUT WITH THIS 3x + 11=13mod17
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    6,154
    Thanks
    2614

    Re: Modulus

    do you understand what equalling some number mod another means? How have you attacked it so far?
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    849

    Re: Modulus

    Hello, leibnitz!

    Even if you don't know Modulo Arithmetic,
    you can work out the answer with Algebra.


    $\displaystyle 3x + 11\:\equiv\:13\text{ (mod17)}$

    We have: .$\displaystyle 3x \:\equiv\:2\text{ (mod 17)}$
    Then: .$\displaystyle 3x \:=\:17a + 2 \quad\Rightarrow\quad x \:=\:\frac{17a+2}{3}$
    That is: .$\displaystyle x \:=\:5a + \frac{2a+2}{3}\;\text{ for some integer }a.\;\;[1]$

    Since $\displaystyle x$ is an integer, $\displaystyle 2a+2$ must be a multiple of 3.

    Then: .$\displaystyle 2a + 2 \:=\:3b\;\text{ for some integer }b.$

    Then: .$\displaystyle a \:=\:\frac{3b-2}{2} \:=\:b + \frac{b-2}{2}\;\;[2]$

    Since $\displaystyle a$ is an integer, $\displaystyle b-2$ must be a multiple of 2.

    The first time this happens is: $\displaystyle b = 2.$

    Substitute into [2]: .$\displaystyle a \:=\:2 + \frac{2-2}{2} \quad\Rightarrow\quad a \:=\:2$

    Substitute into [1]: .$\displaystyle x \:=\:5(2) + \frac{2(2)+2}{3} \:=\:10 + 2$

    Therefore: .$\displaystyle x \:\equiv\:12\text{ (mod 17)}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Modulus

    Def: a=b(modm) if m divides (a-b), ie, (a-b) = km for some k.

    It follows that you can add (subtract) to, and multiply, both sides of a congruence. If (c,m)=1, you can cancel c in ca=cb(modm).

    3x+11=13(mod17)
    3x=2(mod17)
    By inspection*, 12 is a solution because (3ē12-2)=2ē17.

    If y is another solution, 3x=3y(mod17) (equivalence) and x=y(mod17), ie, all solutions are ďequalĒ (equivalent) mod17. So all solutions are:
    12, 12+17, 12-17, 12+2ē17, 12-2ē17, ÖÖÖ

    * There is no general systematic arithmetical procedure for solving am+bn=c for m and n, all integers.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie leibnitz's Avatar
    Joined
    Feb 2013
    From
    Onitsha
    Posts
    12
    Thanks
    2

    Re: Modulus

    Thanks fans,from the option given,12 is the answer to it
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2012
    From
    Athens, OH, USA
    Posts
    1,143
    Thanks
    476

    Re: Modulus

    Hartlw,
    I beg to differ with you. Given the Diophantine equation ax + by = c, a solution exists if and only if gcd(a,b) divides c. Assuming a solution (x0,y0) exists, any solution is of the form (x0-tb,y0+ta) where t is an integer. One solution (x0,y0) can be found by the Euclidean algorithm.
    Thanks from topsquark
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Modulus

    Thanks for your reply johng. Personally, I really donít consider the procedure for solving a linear diophantine equation a systematic arithmetic one in the sense of post #3, ie, you donít start off by trying to solve it using standard artithemetic. And you have to think about the worked-out procedure, uninteresting for me personally.

    But let leibnitz judge for himself:

    From:
    Diophantine equation - Wikipedia, the free encyclopedia

    ďLinear Diophantine equations take the form ax + by = c. If c is the greatest common divisor of a and b then this is Bťzout's identity, and the equation has an infinite number of solutions. These can be found by applying the extended Euclidean algorithm. It follows that there are also infinitely many solutions if c is a multiple of the greatest common divisor of a and b. If c is not a multiple of the greatest common divisor of a and b, then the Diophantine equation ax + by = c has no solutions.Ē

    Also:
    How to Solve a Linear Diophantine Equation: 11 Steps

    One notes that Fermatís last theorem was a diophantine equation. It has been proven, but not by a systematic arithmetic procedure.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Sols of ax=b(modm)

    SOLUTIONS OF ax=bmodm

    Let (a,b)=d=gcd

    I) Euclidean algorithm (a=bq+r) can be used to find (a,b) as a linear combination of a and b. For example, find (57,21)
    57=2(21)+15
    21=1(15)+6
    15=2(6)+3
    6=2(3)
    (57,21)=(21,15)=(15,6)=(6,3)=3
    3=15-2(6)=15-2[21-15]
    3=-2(21)+3(15)
    3=-2(21)+3[57-2(21)]
    3=-8(21)=3(57) ck it with a calculator.

    II) ax=bmodm has a unique solution if (a,m)=1
    From I), 1=sa+tm, b=(bs)a+(bt)m and x=bs is a solution.
    If x1 & x2 are sols, then ax1=ax2modm and x1=x2modm because (a,m)=1.

    III) ax=bmodm has a sol iff (a,m)|b.
    Let d=(a,m), a=a1d, b=b1d and m=m1d.

    If ax=bmodm has a sol, ax-b=qm ̶> (a1d)x-b=q(m1d) ̶> d|b

    If d|b, then (a1,m1)=1 and a1x=b1modm1 has a sol (by II) ̶> a1x=b1+km1 ̶> da1x=db1+dkm1 ̶> ax-b=km and x is sol of ax=bmodm.

    IV) If (a,m)|b, a1x=b1modm1 has a unique sol x0modm1 and x=x0, x0+m1, x0+2m1, Ö, x0+dm1. But x0+dm1=x0modm.
    So the incongruent sols of ax=bmodm are [x0], [x1], Ö[xd-1]modm


    So you see what you did johng? All leibnitz wanted to know is x=12 and all I wanted to point out was that it wasnít simply a matter of elementary algebra. Now leibnitz will probably become a plumber and make more money than you do- serves you right.

    In much wisdom is much grief, and he who increases knowledge increases sorrow. Ecclesiastes.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie leibnitz's Avatar
    Joined
    Feb 2013
    From
    Onitsha
    Posts
    12
    Thanks
    2

    Thumbs up Re: Modulus

    Thanks for your great help Hartlw,may the creator reward you in all your good help...just that I need text book that will help me in mathematiccs
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Modulus

    Thanks Leibnitz, one canít have too many blessings.

    Without books I would be lost. I donít understand lectures because I canít register a stream of information. I have a picture memory for physical things, I canít remember a formula. My modus operandi is to try and remember where things are- leverage.

    How does one intentionally learn something, ie, other than what life imposes on us ?

    Spirit and time (money, opportunity):

    Spirit: Motivation and confidence.

    Motivation: Money, Success, Power, Curiousity, Interest, Egotism, Prejudice (I donít like mathematicians), and Sex- not necessarily in that order. An inner drive (voice) related to a Natural Talent.

    Confidence: 99% cultural, If your parents, or people ďlikeĒ you can do it, you can do it. What happens after you read the first few pages of a math book depends on confidence: I donít understand this and give up, or Iíll get it sooner or later and keep going. Once everybody understands this, society collapses. The only counter-measure is to pay uneducated people more than educated people, which would happen in a purely capitalistic society, but educated people, who rule, wonít tolerate it.

    But you have to be careful- if you doggedly keep going you can destroy yourself.

    Natural Talent: I reluctantly concede this. It goes against my fundamental belief that anybody can do anything. I do believe that natural talent is equally distributed among everybody, that say 25% of a random population are Mozarts and then its a matter of chance and circumstance, ie, time and money, exposure and the financial ability to experiment.

    Confidence has to be kept from people in order to rule over them. That is the primary function of formal education. See how smart we are, we deserve to rule over you. Confidence is a fixed quantity. You increase confidence by taking it away from someone else.

    Personal Experience:
    Working Class background. No money. Assumed only smart people could go to college. Took mechanical course (machinist) in HS. After working a while applied to and got into free (at that time) City College. Much against my fatherís will, who called me a lazy bum, ďif everybody went to college, who would do the work,Ē I decided to go. Floundered for first two years. Then dropped out (auto accident) for one semester, working part time. Then went back, and really tried hard, and amazed to discover I could do it. After that did very well.
    Today this experience would be impossible to duplicate without money. You borrow a fortune and then get one shot at it, after which you are up to your ears in debt. Quite vicious, but serves societyís purposes.

    My cousinís son was doing verry poorly in grade school. Dropped back one semester. I decided to help him with his math homework. I quickly and easily solved a few problems for him which totally demoralized him. That wasnít working. So I pretended not to know the answer to some simple question, which he answered and explained to me, and then suddenly lit up. After a little bit of this his confidence began to grow. He did very well in math and became an accountant.

    After first week of an electronics course totally lost. Stopped doing HW and taking exams till end of semester. Instead went over first chapter of text over and over and over again till it finally registered. At the end of semester handed in all the HW and told Prof what I had done. Took final and got an A. Canít remember any of it.

    While working as an engineer I became very annoyed that I couldnít prove that the sqrt2 is irrational, what does that mean, maybe I really am dumb. I think I could do it now. After that itís a long long story.

    But thatís not why I called.

    Next Post: Learning Algebra
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Banned
    Joined
    Aug 2010
    Posts
    961
    Thanks
    98

    Re: Modulus

    Learning Algebra:
    You need a selection of books, which you can get real cheap used (Amazon and AbeBooks Official Site - New & Used Books, Textbooks, & Rare Books). The older ones are better, because they are less abstracted.

    Math developed from the intuitive and rational to the abstract. What makes it difficult is that only the abstract is taught, which favors the people with a better memory and is deadly for the engineering mentality, which is why so many good engineers get knocked out in the first few semesters.

    My personal goal was to develop a simple, logical structure for a math subject which I could remember and that would carry the whole subject. An exercise in futility. In the final analysis, my memory fails me.

    I try to master the fundamentals, first two chapters of a good book, plus other books to fill in the gaps. Write things down and create a notebook. Avoid the problems- theyíre a confidence destroyer. Take a shot at fundamental questions on MHF.

    For algebra beyond elementary college algebra, you can put a lot of things together from:

    Birkhoff and McLean. First 2 chapters are a good summary. After a few years on the first 2 chapters you might start to feel a little comfortable with it. Same holds for first 2 chapters of Rudin. You start to feel comfortable with it when you begin to realize itís basically useless but fun for math games and putting down BS.

    Ayres, Schaums Outline of College Algebra. Terrrible typos but good for looking up things you donít understand, which is a good general principle- always try to get another explanation.

    Meserve, Fundamentals of Algebra. Much, much too wordy, but good for elabration on a particular topic.

    Lutfiya, Modern Algebra. Interesting crisp summary of principles.

    Between Birkhoff, Meserve, Ayres, and Lutfiya, you might be able to get a feel for congruence arithmetic, and condense it to a structure.

    Personally, vector geometry and analysis suits me, as does anything that can be represented on an x,y,z axis sketch.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Modulus
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 22nd 2011, 06:11 AM
  2. Modulus
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Nov 4th 2009, 10:09 AM
  3. Modulus x^n
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 30th 2009, 11:15 AM
  4. Use of modulus..
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jun 30th 2009, 10:27 AM
  5. modulus
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: Sep 19th 2008, 08:45 AM

Search Tags


/mathhelpforum @mathhelpforum