Question

Given a real number $x,$ let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x.$ For a certain integer $k,$ there are exactly $70$ positive integers $n_1,$ $n_2,$ $\ldots,$ $n_{70}$ such that $\begin{aligned} k &= \lfloor\sqrt[3]{n_1}\rfloor = \lfloor\sqrt[3]{n_2}\rfloor \\ &= \cdots = \lfloor\sqrt[3]{n_{70}}\rfloor \end{aligned}$ and $k$ divides $n_i$ for all $i$ such that $1 \le i \le 70.$ Find the maximum value of $\frac{n_i}{k}$ for $1 \le i \le 70.$

Answer

Answer: 553

Concepts:floor and ceiling functions multiple counting integers in a range

Difficulty rating: 2510

Small Hint:

$\lfloor\sqrt[3]{n}\rfloor = k$ means $k^3 \le n \lt k^3 + 3k^2 + 3k + 1;$ count the multiples of $k$ in that range

Big Hint:

There are $3k + 4$ such multiples, so solve $3k + 4 = 70,$ then take the largest multiple

Solution:

The condition $\lfloor\sqrt[3]{n}\rfloor = k$ means $k^3 \le n \lt (k+1)^3$ $= k^3 + 3k^2 + 3k + 1.$ The multiples of $k$ in this range are $k \cdot k^2,$ $k(k^2 + 1),$ $\ldots,$ $k(k^2 + 3k + 3),$ so there are exactly $3k + 4$ of them.

Setting $3k + 4 = 70$ gives $k = 22.$ The maximum of $\frac{n_i}{k}$ is $k^2 + 3k + 3 = 484 + 66 + 3$ $= 553.$

2007 AIME II Problem 7

Answer: 553

Difficulty rating: 2510

Small Hint:

Big Hint:

Solution:

Problem 7 in Other Years