Try this problem from TOMATO Objective 288, useful for ISI BStat, BMath Entrance Exam based on finding big remainder in a small way.
Problem: Tomato objective 288
The remainder R(x) obtained by dividing the polynomial [latex]x^{100}[/latex] by the polynomial [latex]x^2-3x+2[/latex] is
(A) [latex]2^{100}-1[/latex]
(B) [latex](2^{100}-1)x-(2^{99}-1)[/latex]
(C) [latex]2^{100}x-3(2^{100})[/latex]
(D) [latex](2^{100}-1)x+(2^{99}-1)[/latex]
SOLUTION: (B)
The the divisor is a quadratic term .So, R(x) must be 1 degree less than divisor.
We know , [latex]f(x)=divisor.Q(x)+R(x)[/latex]
[latex]=> x^{100}=(x^2-3x+2).Q(x)+ (ax+b)[/latex]
[latex]=> x^{100}=(x-1)(x-2).Q(x)+ (ax+b)[/latex]
when ,[latex]x=2[/latex]
[latex]=> 2^{100}=(2-1)(2-2).Q(x)+ (2a+b)[/latex]
[latex]=> 2^{100}= (2a+b)..........(i)[/latex]
when ,[latex]x=1[/latex]
[latex]=> 1^{100}=(1-1)(1-2).Q(x)+ (a+b)[/latex]
[latex]=> 1^{100}=(a+b)...........(ii)[/latex]
solving two equation we get, [latex]a=(2^{100}-1)[/latex]
and, [latex]b=-(2^{99}-1)[/latex]
The remainder R(x) is [latex](2^{100}-1)x-(2^{99}-1)[/latex]
