Problem:
Given a real number x, let ⌊x⌋ denote the greatest integer less than or equal to x. For a certain integer k, there are exactly 70 positive integers n1​,n2​,…,n70​ such that k=⌊3n1​​⌋=⌊3n2​​⌋=⋯=⌊3n70​​⌋ and k divides ni​ for all i such that 1≤i≤70. Find the maximum value of kni​​ for 1≤i≤70.
Solution:
Because k≤3ni​​<k+1, it follows that k3≤ni​<(k+1)3=k3+3k2+3k+1. Because k is a divisor of ni​, there are 3k+4 possible values for ni​, namely k3,k3+k,…,k3+3k2+3k. Hence 3k+4=70 and k=22. The desired maximum is kk3+3k2+3k​=k2+3k+3=553​.