PLEASE FANS HELP ME OUT WITH THIS 3x + 11=13mod17
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: .$\displaystyle x \:\equiv\:12\text{ (mod 17)}$
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 (3•12-2)=2•17.
If y is another solution, 3x=3y(mod17) (equivalence) and x=y(mod17), ie, all solutions are “equal” (equivalent) mod17. So all solutions are:
12, 12+17, 12-17, 12+2•17, 12-2•17, ………
* There is no general systematic arithmetical procedure for solving am+bn=c for m and n, all integers.
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 (x_{0},y_{0}) exists, any solution is of the form (x_{0}-tb,y_{0}+ta) where t is an integer. One solution (x_{0},y_{0}) can be found by the Euclidean algorithm.
Thanks for your reply johng. Personally, I really don’t consider the procedure for solving a linear diophantine equation a systematic arithmetic one in the sense of post #3, ie, you don’t 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
“Linear Diophantine equations take the form ax + by = c. If c is the greatest common divisor of a and b then this is Bézout'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.”
Also:
How to Solve a Linear Diophantine Equation: 11 Steps
One notes that Fermat’s last theorem was a diophantine equation. It has been proven, but not by a systematic arithmetic procedure.
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, …, x0+dm1. But x0+dm1=x0modm.
So the incongruent sols of ax=bmodm are [x0], [x1], …[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 wasn’t 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.
Thanks Leibnitz, one can’t have too many blessings.
Without books I would be lost. I don’t understand lectures because I can’t register a stream of information. I have a picture memory for physical things, I can’t 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 don’t 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 “like” 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 don’t understand this and give up, or I’ll 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, won’t 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 father’s will, who called me a lazy bum, “if everybody went to college, who would do the work,” 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 society’s purposes.
My cousin’s 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 wasn’t 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. Can’t remember any of it.
While working as an engineer I became very annoyed that I couldn’t prove that the sqrt2 is irrational, what does that mean, maybe I really am dumb. I think I could do it now. After that it’s a long long story.
But that’s not why I called.
Next Post: Learning Algebra
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- they’re 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 it’s 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 don’t 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.