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 and solutions on this page are the property of the MAA's American Mathematics Competitions