Polynomials & Real Numbers: Find Remainder & No. of Divisors

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The discussion centers on two mathematical problems: finding the remainder of x^100 divided by x^2 - 3x + 2 and determining the number of positive divisors of 10^999 that do not divide 10^998. For the polynomial division, participants suggest using polynomial long division and factorization, ultimately concluding that the remainder will be a linear expression. In the second problem, the focus shifts to expressing 10^999 and 10^998 in terms of their prime factors, leading to a deeper exploration of how to count the divisors. The conversation highlights the importance of understanding polynomial division and factorization techniques in solving these mathematical challenges.
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this is a HOT question from polynomials .

1)find the remainder when x^100 is divided by x^2 - 3x + 2

another from real numbers

2) The number of + divisors which divide 10^999 but not 10^998 are ______ i tried them for sometime . but could not solve them .
 
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sambarbarian said:
this is a HOT question from polynomials .

1)find the remainder when x^100 is divided by x^2 - 3x + 2

another from real numbers

2) The number of + divisors which divide 10^999 but not 10^998 are ______


i tried them for sometime . but could not solve them .

What have you tried?
 
well , i dint really know where to begin , so i just tried some factorization of the first equation and then dividing x^100 by them , but that dint work out . I also tried out a load of thinking but dint get anywhere :( . ( as to the second , I am clueless )
 
sambarbarian said:
well , i dint really know where to begin , so i just tried some factorization of the first equation and then dividing x^100 by them , but that dint work out . I also tried out a load of thinking but dint get anywhere :( . ( as to the second , I am clueless )

Let's tackle the first, well, first.

One way to do this sort of problem is to use polynomial long division. While that may be practical for a lower order dividend (numerator), the dividend here is x^{100}, which makes this infeasible. So we have to use another approach.

The divisor here is a quadratic expression. Do you know what form the remainder will take? Would it be a constant (a number) like when you divide a polynomial by a linear divisor? Or something else?

Start by factorising the divisor (x^2 - 3x + 2).
 
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the form of remainder is not mentioned , and i tried the factorization , doesn't work
 
sambarbarian said:
the form of remainder is not mentioned , and i tried the factorization , doesn't work

The remainder will be a polynomial expression. The order of the polynomial is at most one less than the order of the divisor. So if your divisor is a linear expression like x + 5, then you will have a constant remainder (like 3 or -5 or 0). If your divisor is a quadratic expression, your remainder may be a linear expression like x + 2 or 3x - 5, or it may be a constant like 2 or -5. All these are just examples to illustrate the point.

So you should start by assuming the remainder is a linear expression, ax + b, where a and b are constants. If a comes out to be 0, then you know the remainder is simply a constant.

When P(x) is divided by a quadratic expression (x-\alpha)(x-\beta), you can express the result by this equation:

P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b), where the remainder is ax + b.

So your first step is to find out \alpha and \beta, which is why I asked you to factorise the quadratic divisor. Surely, you know how to do that!
 
oh yes , i know that , i told you i tried that , but the equation cannot be solved :/
 
sambarbarian said:
oh yes , i know that , i told you i tried that , but the equation cannot be solved :/

So WHAT is the factorisation of the divisor?
 
You say you know to factorise the quadratic divisor, and this is all I want to see. Once you do that, I can walk you through the rest of the solution. But you have to factorise the quadratic, which is a very basic step, which you say you know how to do.

So: just factorise x^2 - 3x + 2.

It's late where I am, so I'm going to have to turn in now. I'll check in again in the morning my time (about 8 hours' time) and help you, provided no one else has guided you to a solution by then.
 
  • #10
sambarbarian said:
2) The number of + divisors which divide 10^999 but not 10^998 are ______

OP, start by writing the two numbers as products of a power of 2 and a power of 5. In other words,
10^{999} = 2^{(something)} \cdot 5^{(something)}
 
  • #11
Curious3141 said:
Start by factorising the divisor (x^2 - 3x + 2).

sambarbarian said:
... and i tried the factorization , doesn't work

If you can't factor that, why would you be expect to be able to solve this problem? Perhaps you should take a look in a high school algebra book and do a review.
 
  • #12
LCKurtz said:
If you can't factor that, why would you be expect to be able to solve this problem? Perhaps you should take a look in a high school algebra book and do a review.

Yes. Either he knows how to factor a quadratic, in which case he should show his work, or he doesn't, in which case this problem is not suitable for him at this stage.
 
  • #13
guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/
 
  • #14
eumyang said:
OP, start by writing the two numbers as products of a power of 2 and a power of 5. In other words,
10^{999} = 2^{(something)} \cdot 5^{(something)}

u mean 2^999 * 3^999 ... ( why did you write 'something' ? )
 
  • #15
sambarbarian said:
u mean 2^999 * 3^999 ... ( why did you write 'something' ? )

Because I wanted you to fill in (something) with a number. And by the way, the 2nd base should be 5, not 3. So,
10^{999} = 2^{999} \cdot 5^{999} and
10^{998} = 2^{998} \cdot 5^{998}

You'll notice that 2999 is not a factor of 10998. In fact, any factor of 10999 that is divisible by 2999 cannot be a factor of 10998. So how many of these factors of 10999 would that make altogether? (And no, the answer to this question is not the final answer of the problem.)
 
  • #16
sambarbarian said:
guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

First find the remainder when you divide x100 by (x-1).(hint: is xn-1 divisible by (x-1)?)

ehild
 
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  • #17
sambarbarian said:
guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

Take another look at Curious3141's post (#6). What do you get when you look at x=\alpha and x=\beta? Apply that to your polynomial.
 
  • #18
sambarbarian said:
guys , i obviously know how to factor the equation ( (x-2)(x-1) ) , i wanted to say that i tried this approach but dint get anywhere :/

OK, finally, you've factored it (and shown it). (BTW, that's not an "equation", merely an expression).

You must have tried the wrong approach, because it's very easy from here on.

Use this equation I previously quoted:

P(x) = (x-\alpha)(x-\beta)Q(x) + (ax + b), where the remainder is ax + b.

for the current problem as:

x^{100} = (x-1)(x-2)Q(x) + (ax + b)

and successively put x = 1 and x = 2 to get two simple, linear simultaneous equations in a and b. Solve those, and just write down the remainder as ax + b, using what you've found.
 
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  • #19
case 1 . 1^100 = (1)a + b => 1 = a+b
case 2 . 2^100 = (2)a = b => 2^100 = 2a+b

if i solve these for a and b , which value of x will i take for the answer ( 1 or 2 ) , as the remainder is ax+b.

this is the part where i failed.
 
  • #20
eumyang said:
because i wanted you to fill in (something) with a number. And by the way, the 2nd base should be 5, not 3. So,
10^{999} = 2^{999} \cdot 5^{999} and
10^{998} = 2^{998} \cdot 5^{998}

you'll notice that 2999 is not a factor of 10998. In fact, any factor of 10999 that is divisible by 2999 cannot be a factor of 10998. So how many of these factors of 10999 would that make altogether? (and no, the answer to this question is not the final answer of the problem.)

2^999 , 5^999 ?
 
  • #21
sambarbarian said:
case 1 . 1^100 = (1)a + b => 1 = a+b
case 2 . 2^100 = (2)a = b => 2^100 = 2a+b

Correct so far.

if i solve these for a and b , which value of x will i take for the answer ( 1 or 2 ) , as the remainder is ax+b.

this is the part where i failed.

No, the remainder IS (ax + b). The remainder when dividing by a quadratic can be a linear expression like that. You're probably getting confused by division by linear divisors, when the remainder is just a number (a constant).

I already explained this in post #6.
 
  • #22
by substitution , b=1-a => 2^100 = 1+a => a= 2^100 - 1
a=1-b =>2^100 = 2-b => b = 2 - 2^100
 
  • #23
sambarbarian said:
by substitution , b=1-a => 2^100 = 1+a => a= 2^100 - 1
a=1-b =>2^100 = 2-b => b = 2 - 2^100

Yes, now you can just write down the answer.
 
  • #24
x(2^100 - 1 ) + 2 - 2^100

2^100x -x +2 - 2^100 ? is this the ans ?
 
  • #25
sambarbarian said:
2^999 , 5^999 ?

No, I believe Eumyang was thinking about numbers like 2^{999}.5, 2^{999}.5^4 and 2^{999}.5^{999} (just examples). Have you considered those?

There are many more such numbers to consider.
 
  • #26
sambarbarian said:
x(2^100 - 1 ) + 2 - 2^100

2^100x -x +2 - 2^100 ? is this the ans ?

I would just leave it as (2^{100} - 1 )x + 2 - 2^{100}.
 
  • #27
any chance this is equal to 2 - x ,,, that's the answer at the back of the textbook
 
  • #28
sambarbarian said:
any chance this is equal to 2 - x ,,, that's the answer at the back of the textbook

No. How can it be? If 2 - x is the answer in your book, it's wrong.
 
  • #29
begin {offtopic}

Sambarbarian, I am curious to know if you grew up in or near northeastern South Dakota. If your answer is yes, I will tell you why I thought that. Otherwise nevermind.

end{offtopic}
 
  • #30
nopes , I am an indian ... but please tell me y u thought that
 
  • #31
LCKurtz said:
begin {offtopic}

Sambarbarian, I am curious to know if you grew up in or near northeastern South Dakota. If your answer is yes, I will tell you why I thought that. Otherwise nevermind.

end{offtopic}

sambarbarian said:
nopes , I am an indian ... but please tell me y u thought that

Because I did grow up in South Dakota, and I have met people from the eastern part of the state that use the word "dint" when they mean "didn't", which is a contraction for "did not". The word "dint" means something else.
 
  • #32
eumyang said:
You'll notice that 2999 is not a factor of 10998. In fact, any factor of 10999 that is divisible by 2999 cannot be a factor of 10998. So how many of these factors of 10999 would that make altogether? (And no, the answer to this question is not the final answer of the problem.)

Curious3141 said:
No, I believe Eumyang was thinking about numbers like 2^{999} \cdot 5, 2^{999} \cdot 5^4 and 2^{999} \cdot 5^{999} (just examples). Have you considered those?

There are many more such numbers to consider.

Right. 2999 (or 2999 x 50) is not a factor of 10998.
2999 x 5 (or 2999 x 51) is not a factor of 10998.
2999 x 52 is not a factor of 10998.
These are some of the factors of 10999, divisible by 2999, but are not factors of 10998. So how many factors would this make altogether?
 
  • #33
sambarbarian said:
nopes , I am an indian ... but please tell me y u thought that

Well, hello, I'm (ethnically) Indian too, although I'm a Singaporean by birth and citizenship.

And as LCKurtz points out, "dint" has its own meaning. It means by virtue of (something) when used in a phrase like "by dint of his dogged hard work". It's usually used emphatically, to suggest force, power or perseverance. It can also be an alternative spelling for "dent".

Back to the math now! :biggrin:
 
  • #34
eumyang said:
right. 2999 (or 2999 x 50) is not a factor of 10998.
2999 x 5 (or 2999 x 51) is not a factor of 10998.
2999 x 52 is not a factor of 10998.
These are some of the factors of 10999, divisible by 2999, but are not factors of 10998. So how many factors would this make altogether?

999 + 999 = 1998 ?
 
  • #35
sambarbarian said:
999 + 999 = 1998 ?

I have one more than that. How did you arrive at your figure (exactly)?
 
  • #36
Now you are just guessing. The critical point is that 10^{998}= (5^{998})(2^{998}). That means that 5^n is a factor for every n from 0 to 998. But it also means that 2^n, for n= 0 to 99 is also a factor. That's 2(999) right there, and and we haven't even started combining "2"s and "5"s.

One way to look at it is this: there are 999 ways to find factors using only "5"s and there are 999 ways using only "2"s. And the "fundamental theorem of counting" says that the if there are m ways of doing one thing and n ways of doing another (independently of the first) then there are mn ways of doing both.
 
  • #37
HallsofIvy said:
Now you are just guessing. The critical point is that 10^{998}= (5^{998})(2^{998}). That means that 5^n is a factor for every n from 0 to 998. But it also means that 2^n, for n= 0 to 99 is also a factor. That's 2(999) right there, and and we haven't even started combining "2"s and "5"s.

Yes, but everything from 5^0 to 5^{998} are also factors of 10^{998} and are therefore excluded from consideration. Ditto for powers of 2. Every number that should be considered should have as a factor, either 2^{999} or 5^{999} or both (the only one that has both is 10^{999}).
 
  • #38
i don't quite understand
 
  • #39
sambarbarian said:
i don't quite understand

OK, start by considering the numbers with 2^{999} as a factor. Now, everything from 2^{999}.5^0 to 2^{999}.5^{998} divides 10^{999} but not 10^{998}. Agreed? So how many is that?

Now consider numbers with 5^{999} as a factor in a similar way. How many do we have here?

Finally, don't forget 2^{999}.5^{999} = 10^{999} itself, which I avoided including till the last step so that we don't double-count it.
 
  • #40
Curious3141 said:
OK, start by considering the numbers with 2^{999} as a factor. Now, everything from 2^{999}.5^0 to 2^{999}.5^{998} divides 10^{999} but not 10^{998}. Agreed? So how many is that?

2^999 . 5^0 is equal to 2^999 . 5^1 ,,, (998 - 1) = 997 ? i am not sure
 
  • #41
sambarbarian said:
2^999 . 5^0 is equal to 2^999 . 5^1 ,,, (998 - 1) = 997 ? i am not sure

You're not thinking clearly, and what you wrote makes no sense.

2^{999}.5^0 = 2^{999} because any nonzero number raised to the power zero is 1.

2^{999} is still a number you have to include in the count because it divides 10^{999} but not 10^{998}.

I asked you to count the numbers from 2^{999}.5^0 to 2^{999}.5^{998} going in consecutive exponents of 5. This is equivalent to counting the integers from 0 to 998 inclusive. You replied with '997'. Sure of this?

How many integers are there from 0 to 1 inclusive? 0 to 10? 0 to 998?
 
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