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Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .

What I tried: In some step I messed up with this problem and so I think I am getting my answer wrong, so please correct me.

We have $x^2 - 3x + 2$ = $(x - 1)(x - 2)$ and I can see $(x - 1)^2 \equiv 1$ $($mod $x - 2)$ . We also have :- $$\frac{(x - 1)^{100}}{(x - 1)(x - 2)} = \frac{(x - 1)^{99}}{(x - 2)}.$$ We have :- $(x - 1)^{98} \equiv 1$ $($mod $x - 2).$ $\rightarrow (x - 1)^{99} \equiv (x - 1)$ $($mod $x - 2)$. Now for the case of $(x - 2)^{200}$ we have :- $$\frac{(x - 2)^{200}}{(x - 1)(x - 2)} = \frac{(x - 2)^{199}}{(x - 1)}.$$

We have :- $(x - 2) \equiv (-1)$ $($mod $x - 1)$ $\rightarrow (x - 2)^{199} \equiv (-1)$ $($mod $x - 1)$.

Adding all these up we have :- $(x - 1)^{100} + (x - 2)^{200} \equiv (x - 2)$ $($mod $x² - 3x + 2)$ .

On checking my answer with wolfram alpha , I found the remainder to be $1$, so I messed up in some step . Can anyone help me?

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  • $\begingroup$ Where did $(x-2)^{200}$ come from? And how did you manage to add "remainder when divided by x-1" with "remainder when divided by x-2? $\endgroup$ Aug 4, 2020 at 10:16
  • $\begingroup$ The question had $(x - 2)^{200}$ , edited it right now was a typo . $\endgroup$
    – Anonymous
    Aug 4, 2020 at 10:18
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    $\begingroup$ You can also try different approach: if $P(x)=(x^2-3x+2)Q(x)+R(x)$, where $\deg R<\deg (x^2-3x+2)=2$, then $R(x)=ax+b$ and you can find $a$ and $b$ by plugging $x=1$ and $x=2$ into the previous equality (question: why we choose $x\in\{1,2\}$?). $\endgroup$
    – richrow
    Aug 4, 2020 at 10:19
  • $\begingroup$ @SouradipDas for example, $6^{20}$ divided by 6 and then $6^{19}$ divided by $5$ will give you a remainder of $1$ but $6^{20}$ divided directly by $30$ will give you a remainder of $6$. $\endgroup$
    – Math Lover
    Aug 4, 2020 at 11:19
  • $\begingroup$ Yeah true , I got it before what I had done as a mistake $\endgroup$
    – Anonymous
    Aug 4, 2020 at 11:20

5 Answers 5

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Write $$(x - 1)^{100} + (x - 2)^{200}=k(x)(x-2)(x-1)+ax+b$$

SInce this is valid for all $x$ it is valid also for $$x=1: \;\;\; 1=a+b$$ and $$x=2: \;\;\; 1=a2+b$$

So $a=0$ and $b=1$.

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Since $(x - 1)^{99} \equiv (x - 1)\;\mod (x - 2)$, we get that

$(x - 1)^{100}\equiv (x-1)^2\;\mod (x-2)(x-1). \quad(*)$

Since $(x-2)^{199}\equiv -1\;\mod (x-1)$, we get that

$(x-2)^{200}\equiv -(x-2)\;\mod (x-1)(x-2). \quad(**)$

So, by adding $(*)$ and $(**)$, it follows that

$(x - 1)^{100} + (x - 2)^{200} \equiv (x-1)^2-(x-2)\\\mod (x-1)(x-2),$

that is

$(x - 1)^{100} + (x - 2)^{200}\equiv x^2-3x+3\mod (x^2-3x+2)$.

Hence,

$(x - 1)^{100} + (x - 2)^{200} \equiv 1\;\mod (x^2-3x+2)$.

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You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies $$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$ More naively, as $$x-1\equiv1\pmod{x-2}$$ then $$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$ Similarly, $$x-2\equiv-1\pmod{x-1}$$ and $$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$ So $(x-1)^{100}+(x-2)^{200}$ is congruent to $1$ modulo both $x-1$ and $x-2$, and so also modulo $(x-1)(x-2)$.

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  • $\begingroup$ Oh thanks , I got your approach. So I guess my approach was wrong. $\endgroup$
    – Anonymous
    Aug 4, 2020 at 10:22
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The remainder of $(x - 1)^{100}$ divided by $(x - 1)(x - 2)$ will be $(x - 1) (2 - 1)^{99} = x - 1$. The remainder of $(x - 2)^{200}$ divided by $(x - 1)(x - 2)$ will be $(x - 2)(1 - 2)^{199} = 2 - x$ Therefore, the total remainder will be 1.

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$P(x)=(x-1)^{100}+(x-2)^{200}=Q(x)×(x^2-3x+2)+ax+b$

$P(1)=(-1)^{200}=a+b, a+b=1$

$P(2)=(1)^{100}=2a+b, 2a+b=1$

$a=0 , b=1$

Hence remainder :- $R(x)=1$

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    $\begingroup$ Please, use MathJax to write mathematics in this site. $\endgroup$
    – DonAntonio
    Aug 4, 2020 at 10:33

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