Question

Given that $\frac{((3!)!)!}{3!} = k \cdot n!,$ where $k$ and $n$ are positive integers and $n$ is as large as possible, find $k + n.$

Answer

Answer: 839

Concepts:factorial bounding to limit cases

Difficulty rating: 1670

Small Hint:

Compute the inner factorials: $3! = 6,$ so the expression is $\frac{720!}{6}$

Big Hint:

Write $720! = 720 \cdot 719!$ and note that $n = 720$ would make $k \cdot n!$ too large

Solution:

Since $3! = 6$ and $6! = 720,$ the expression is $\begin{aligned} \frac{((3!)!)!}{3!} &= \frac{720!}{6} \\ &= \frac{720 \cdot 719!}{6} \\ &= 120 \cdot 719!. \end{aligned}$

If $n$ were $720$ or more, then $k \cdot n! \ge 720!,$ which exceeds $\frac{720!}{6}.$ So the largest possible value of $n$ is $719,$ achieved with $k = 120,$ and $k + n = 120 + 719 = 839.$

2003 AIME I Problem 1

Answer: 839

Difficulty rating: 1670

Small Hint:

Big Hint:

Solution:

Problem 1 in Other Years