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.