Problem:
What is the remainder when 2202+202 is divided by 2101+251+1?
Answer Choices:
A. 100
B. 101
C. 200
D. 201
E. 202
Solution:
2202+202=(2101)2+2⋅2101+1−2⋅2101+201=(2101+1)2−2102+201=(2101−251+1)(2101+251+1)+201
Thus, we see that the remainder is surely (D)201.
OR
Let x=250. We are now looking for the remainder of
2x2+2x+14x4+202
We could proceed with polynomial division, but the numerator looks awfully similar to the Sophie Germain Identity, which states that
a4+4b4=(a2+2b2+2ab)(a2+2b2−2ab)
Let's use the identity, with a=1 and b=x, so we have
1+4x4=(1+2x2+2x)(1+2x2−2x)
Rearranging, we can see that this is exactly what we need:
2x2+2x+14x4+1=2x2−2x+1
So
2x2+2x+14x4+202=2x2+2x+14x4+1+2x2+2x+1201
Since the first half divides cleanly as shown earlier, the remainder must be (D)201.
The problems on this page are the property of the MAA's American Mathematics Competitions