Problem:
Let f(x)=(x2+3x+2)cos(Οx). Find the sum of all positive integers n for which
β£β£β£β£β£β£βk=1βnβlog10βf(k)β£β£β£β£β£β£β=1
Solution:
Note that
f(k)=[(k+1)(k+2)](β1)k={(k+1)(k+2)(k+1)(k+2)1ββ if k is even if k is odd ββ
Therefore
k=1βnβlog10βf(k)=log10β(k=1βnβf(k))=β©βͺβ¨βͺβ§βlog10β(2β
3β
4β―(n+1)3β
4β
5β―(n+2)β)=log10β(2n+2β)log10β(2β
3β
4β―(n+2)3β
4β
5β―(n+1)β)=log10β(2(n+2)1β)=βlog10β(2n+4)β if n is even if n is odd. ββ
For β£βk=1nβlog10βf(k)β£ to be 1, either 2n+2β=10 with n even or 2n+4=10 with n odd, so n=18 or n=3. Thus the requested sum is 18+3=21β.
The problems on this page are the property of the MAA's American Mathematics Competitions