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

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?

• 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? Aug 4, 2020 at 10:16
• The question had $(x - 2)^{200}$ , edited it right now was a typo . Aug 4, 2020 at 10:18
• 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 $x\in\{1,2\}$?). Aug 4, 2020 at 10:19
• @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 $6$. Aug 4, 2020 at 11:19
• Yeah true , I got it before what I had done as a mistake Aug 4, 2020 at 11:20

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 $$b=1$$.

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,

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

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 $$(x-1)(x-2)$$.

• Oh thanks , I got your approach. So I guess my approach was wrong. Aug 4, 2020 at 10:22

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.

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

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

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

$$a=0 , b=1$$

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

• Please, use MathJax to write mathematics in this site. Aug 4, 2020 at 10:33