Problem:
For every m≥2, let Q(m) be the least positive integer with the following property: For every n≥Q(m), there is always a perfect cube k3 in the range n<k3≤m⋅n. Find the remainder when
m=2∑2017Q(m)
is divided by 1000.
Solution:
Each of the sets {2,3,4,5,6,7},{9,10,…,24}, and {32,33,…,62} has the form {n+1,n+2,…,m⋅n} and contains no perfect cubes. Thus Q(7)>1,Q(3)>8, and Q(2)>31.
For a given m let k be the greatest integer such that k3≤Q(m)−1 and m⋅(Q(m)−1)<(k+1)3. It follows that mk3+1≤(k+1)3, so (m−1)k2−3k−3≤0. Solving for k and using the fact that k is an integer yields
k≤⌊2m−23+12m−3⌋
For m=2,3,4, and 8, it follows that k≤3,2,1, and 0, respectively. Because m⋅(Q(m)−1)<(k+1)3, it follows that Q(2)<33,Q(3)<10,Q(4)<3, and Q(8)<89.
By definition Q(m)≥Q(m+1)≥1 for all m. Thus Q(m)=1 for all m≥8. Also, 2≤Q(7)≤Q(6)≤Q(5)≤Q(4)≤2, so Q(4)=Q(5)=Q(6)=Q(7)=2. Finally, Q(3)=9 and Q(2)=32. Therefore