Arithmetic/Divisibility homework problem

In summary: After you make that correction, there is still another one you need to make.In summary, Scurty said that you should use the euclidean division to solve the homework equations. Then he said to use the quadratic formula to find n.
  • #1
mtayab1994
584
0

Homework Statement



Solve for n in the following equations:

[tex]n-2\mid n^{2}-n+3[/tex]

Homework Equations


The Attempt at a Solution



Should i use the euclidean division to solve it? Thank you before hand.

Edit: When i used euclidean division i got n+1 and a remainder of 5 but i don't know how to get n.
 
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  • #2


Start by using the definition of divides to produce an equation. This equation will introduce a new variable, say, k. Solve for n using the quadratic formula (which will involve k) and then choose values of k so that n is an integer (I assume n is an integer in this problem).

See what you come up with!
 
  • #3


So that means you should get: n^2-n+3=k(n-2) am i right?
 
  • #4


mtayab1994 said:
So that means you should get: n^2-n+3=k(n-2) am i right?
Right.
 
  • #5


Mark44 said:
Right.

Yes now how can i continue from there?
 
  • #6


Ever heard of quadratic equations?
 
  • #7


Whovian said:
Ever heard of quadratic equations?

Yes, but n^2-n+3 doesn't have any real roots, if that's what you're talking about.
 
  • #8


No, we group everything on one side of the equation so we have coefficients in terms of k and solve THAT.
 
  • #9


Whovian said:
No, we group everything on one side of the equation so we have coefficients in terms of k and solve THAT.

So we get: n^2-k(n-2)-n+3=0, but where are we supposed to get b^2-4ac. Is b=k(n-2)?
 
  • #10


Not quite. We get n^2-(k+1)n+(2k+3)=0 after multiplying out and grouping like terms, and it should be obvious how to proceed from here.
 
  • #11


So for Δ i got k^2-6k-11 and then i counted the discriminant again and i got 80. Is what I've done so far correct?
 
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  • #12


Reread what scurty wrote in post #2. He laid out pretty much what you need to do.
 
  • #13


Mark44 said:
Reread what scurty wrote in post #2. He laid out pretty much what you need to do.

So there is no need to find the discriminant of the equation : n^2-(k+1)n+(2k+3)=0?
 
  • #14


I don't see any need for finding the discriminant in this problem.
 
  • #15


Mark44 said:
I don't see any need for finding the discriminant in this problem.

But the quadratic formula involves finding the discriminant, or do i just use random values of k?
 
  • #16


Of course you find the discriminant in the quadratic formula. What I meant was that there's no reason to find it and stop there.

Again, take a look at what scurty said in post #2. You are asking questions that suggest you haven't read it or didn't understand it.
 
  • #17


Mark44 said:
Of course you find the discriminant in the quadratic formula. What I meant was that there's no reason to find it and stop there.

Again, take a look at what scurty said in post #2. You are asking questions that suggest you haven't read it or didn't understand it.

Well, he said to produce a new equation, which I did and then he said to find n using the quadratic formula, which will involve k, and that's what I did. I'm confused because we have just started this lesson, and we haven't done any problem like this so far.
 
  • #18


Did you solve the quadratic equation for n? If so, what did you get?
 
  • #19


Mark44 said:
Did you solve the quadratic equation for n? If so, what did you get?

For n^2-(k+1)n+(2k+3)=0 I got a discriminant of k^2-6k-11 then i solved that equation and I got Δ=80 so i got k1=3+2√5 and k2=3-2√5.
 
  • #21


Mark44 said:
Solve n^2-(k+1)n+(2k+3)=0 for n.

Ok when solving for n i got : [tex]n=\frac{k+\sqrt{k^{2}-8k+4}}{2},\frac{k-\sqrt{k^{2}-8k+4}}{2}[/tex]
 
  • #22


mtayab1994 said:
Ok when solving for n i got : [tex]n=\frac{k+\sqrt{k^{2}-8k+4}}{2},\frac{k-\sqrt{k^{2}-8k+4}}{2}[/tex]

No.

Do you know how to use the quadratic formula? You said before that the discriminant was k2 - 6k -11, but that's not what you show above.
 
  • #23


Mark44 said:
No.

Do you know how to use the quadratic formula? You said before that the discriminant was k2 - 6k -11, but that's not what you show above.

yes instead of k^2-8k+4 should be k^2-6k-11
 
  • #24


I think i got it for k^2-6k-11=0 we get n =(k+1)/2 right?
 
  • #25


After you make that correction, there is still another one you need to make.

Then READ POST #2!
 
  • #26


For what it's worth, when I solved this, I got 4 answers for n. I'm not sure if there is a theorem that's states a maximum number for n based on certain criteria, but I tested out about 30 values for k and only two of them worked. Maybe someone else can chime in.
 
  • #27


mtayab1994 said:
I think i got it for k^2-6k-11=0 we get n =(k+1)/2 right?
Why do you think that k^2-6k-11=0?
 
  • #28


Well after all you really don't need to form and equation and give values for k all you have to do is:
[tex]n-2\mid n^{2}-n+3\Longleftrightarrow\frac{(n-2)(n+1)+5}{n-2}\Longleftrightarrow n+1+\frac{5}{n-2}[/tex]

and since: [tex](n+1)\in Z[/tex] therefore: [tex]\frac{5}{n-2}\in Z[/tex]

thus: [tex]n-2\mid5[/tex]Then you form a table with 2 rows one row contains the divisors of '5' and the other row contains n the number that we have to plug into n-2 to get that divisor of 5. Like the following.

(n-2) 1 -1 5 -5

(n) 3 1 7 -3

And then the solutions are: [tex]S=\{-3,1,3,7\}[/tex]

So how's that? I think my way is a lot easier then solving all of those equations.
 
  • #29


Those are the answers I got. There are often more ways than one to solve a problem, in this case, that method would probably be the easiest way to solve the problem. However, you seemed to struggle with simple algebra involving solving for a quadratic equation.. You might want to touch up on that a bit for the future. Algebra techniques are common in number theory, I remember using the quadratic formula numerous times when I took the course, it would be wise to understand why the other method works as well.

Edit: I just noticed that you don't have k anywhere in your equation. The definition of divides introduces a new variable, you can't just assume k=1, you have to explain why certain values of k work and others don't. It works in this case but not all cases.
 
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  • #30


Sorry for butting in, but I like to think of myself as a decent mathematician who has terrible difficulty with number theory.

I couldn't solve this at all. I had to write a program to solve for all cases for 1 to 1000 and I got the same answers: 3 & 7

Could someone please tell me where I'm going wrong analytically?

PS I've read post #2.
 
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  • #31


oay said:
Sorry for butting in, but I like to think of myself as a decent mathematician who has terrible difficulty with number theory.

I couldn't solve this at all. I had to write a program to solve for all cases for 1 to 1000 and I got the same answers: 3 & 7

Could someone please tell me where I'm going wrong analytically?

PS I've read post #2.

But you're missing -3 and 1 they are also answers in Z.
 
  • #32


mtayab1994 said:
But you're missing -3 and 1 they are also answers in Z.
I got those too, I assumed n was positive, but I admit n=1 is a fair solution but is trivial.

How come Z comes suddenly into it?
 
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  • #33


oay said:
I got those too, I assumed n was positive, but I admit n=1 is a fair solution.

Oh alright it's pretty easy solving it my way just look at my steps and see what I did. By the way that 5 I got in the numerator is the remainder you get when you do the Euclidean Division.
 
  • #34


mtayab1994 said:
Oh alright it's pretty easy solving it my way just look at my steps and see what I did.
I actually appreciated your method; I didn't say anything against it.
 
  • #35


oay said:
I actually appreciated your method; I didn't say anything against it.

What method do you use to solve them and by the way i have another one for you to solve if you want. Look at the other thread I'm going to post so you can see it.
 
<h2>1. What is arithmetic?</h2><p>Arithmetic is a branch of mathematics that deals with the study of numbers and their basic operations such as addition, subtraction, multiplication, and division.</p><h2>2. What is divisibility?</h2><p>Divisibility is the ability of one number to be divided by another number without leaving a remainder. For example, 12 is divisible by 3 because 12 divided by 3 equals 4 with no remainder.</p><h2>3. How do I determine if a number is divisible by another number?</h2><p>To determine if a number is divisible by another number, you can use the divisibility rules. For example, a number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. A number is divisible by 3 if the sum of its digits is divisible by 3.</p><h2>4. What is the importance of divisibility in mathematics?</h2><p>Divisibility is important in mathematics because it helps us identify patterns and relationships between numbers. It also allows us to simplify calculations and solve more complex problems.</p><h2>5. How can I improve my skills in arithmetic and divisibility?</h2><p>The best way to improve your skills in arithmetic and divisibility is to practice regularly. You can also use online resources, such as tutorials and practice problems, to help you understand and master these concepts.</p>

1. What is arithmetic?

Arithmetic is a branch of mathematics that deals with the study of numbers and their basic operations such as addition, subtraction, multiplication, and division.

2. What is divisibility?

Divisibility is the ability of one number to be divided by another number without leaving a remainder. For example, 12 is divisible by 3 because 12 divided by 3 equals 4 with no remainder.

3. How do I determine if a number is divisible by another number?

To determine if a number is divisible by another number, you can use the divisibility rules. For example, a number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. A number is divisible by 3 if the sum of its digits is divisible by 3.

4. What is the importance of divisibility in mathematics?

Divisibility is important in mathematics because it helps us identify patterns and relationships between numbers. It also allows us to simplify calculations and solve more complex problems.

5. How can I improve my skills in arithmetic and divisibility?

The best way to improve your skills in arithmetic and divisibility is to practice regularly. You can also use online resources, such as tutorials and practice problems, to help you understand and master these concepts.

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