Problem:
For each positive integer n, let f(n)=∑k=1100⌊log10(kn)⌋. Find the largest value of n for which f(n)≤300.
Note: ⌊x⌋ is the greatest integer less than or equal to x.
Solution:
Note that f(1)=9⋅0+90⋅1+2=92, and f(10n)=100+f(n), so f(100)=292 and f(1000)=392. For 0≤j<900,log10(k(100+j))≥2. Furthermore log10(k(100+j))≥3 if and only if
k≥100+j1000=10−100+j10j, that is, k≥10−⌊100+j10j⌋
Therefore the number of terms in the sequence having a value of at least 3 is 91+⌊100+j10j⌋. Similarly, the number of terms having a value of 4 is 1+⌊100+j100j⌋, which implies
f(100+j)=200+91+1+⌊100+j10j⌋+⌊100+j100j⌋.
Thus the required value of n=100+j must satisfy 100j<9(100+j), and therefore j≤9. It can be verified that f(109)=300, so the answer is 109.
The problems on this page are the property of the MAA's American Mathematics Competitions