Problem:
There is a polynomial P(x) with integer coefficients such that
P(x)=(x105−1)(x70−1)(x42−1)(x30−1)(x2310−1)6
holds for every 0<x<1. Find the coefficient of x2022 in P(x).
Solution:
For 0<x<1, the given rational expression is equal to
(x2310−1)2⋅x105−1x2310−1⋅x70−1x2310−1⋅x42−1x2310−1⋅x30−1x2310−1=(x2310−1)2⋅a=0∑21x105a⋅b=0∑32x70b⋅c=0∑54x42c⋅d=0∑76x30d
The coefficient of x2022 is therefore equal to the number of quadruples of nonnegative integers (a,b,c,d) that satisfy
105a+70b+42c+30d=2022
Considering this equation modulo 2,3,5,7 yields
abcd≡0(mod2)≡0(mod3)≡42−1⋅2022≡1(mod5)≡30−1⋅2022≡3(mod7)
Thus there are nonnegative integers w,x,y, and z satisfying a=2w,b=3x,c=5y+1, and d=7z+3. The given equation becomes
2022=105(2w)+70(3x)+42(5y+1)+30(7z+3)=210(w+x+y+z)+132,
from which w+x+y+z=9. By the sticks-and-stones method, the number of solutions to this equation in nonnegative integers is given by (39+3)=220.
The problems on this page are the property of the MAA's American Mathematics Competitions