We have, $(x-1)^{100}+(x-2)^{200}=(x-1)(x-2)q(x)+r(x)$

Since divisor is quadratic, the remainder must be linear.

So, $r(x)=ax+b$

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

Putting $x=1$ above yields $a+b=1$

Substituting $x=2$, yields $2a+b=1$

Clearly, a=0 & b=1

So, remainder $=ax+b=1$.