# 2019 IMO Problems/Problem 4

## Problem

Find all pairs $(k,n)$ of positive integers such that

$$k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}).$$

## Solution 1

$LHS$

$k! = 1$ (when $k = 1$), $2$ (when $k = 2$), $6$ (when $k = 3$)

$RHS = 1$(when $n = 1$), $6$ (when $n = 2$)

Hence, $(1,1)$, $(3,2)$ satisfy

For $k = 2: RHS$ is strictly increasing, and will never satisfy $k$ = 2 for integer n since $RHS = 6$ when $n = 2$.

$$k!=(2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1}) ... (1)$$

In all solutions, for any prime $p$ and positive integer $N$, we will denote by $$v_p(N)$$ the exponent of the largest power of $p$ that divides $N$. The right-hand side of $(1)$ will be denoted by $$L_n;$$ that is, $$L_n = (2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})$$

$$(2^{1+2+3+\dots+(n-1)})(2^n-1)(2^{n-1}-1)(2^{n-2}-1)\dots(2^1-1)$$ = $$v_2(L_n) = {\frac{n(n-1)}{2}}$$

On the other hand, $$v_2(k!)$$ is expressed by the $Legendre$ $formula$ as $$v_2(k!) < \sum_{i=1}^{\infty} (\frac{k}{2^i})) = k$$

Thus, $k! = L_n$ implies the inequality $$\frac{n(n-1)}{2} < k ... (2)$$ In order to obtain an opposite estimate, observe that $$L_n = (2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})$$ We claim that $$2^{n^2} < (\frac{n(n-1)}{2})! ... (3)$$ for all $n \geq 6$

For $n \geq 6$ the estimate (3) is true because $$2^{6^2} < (6.9)(10^{10})$$ and $$(\frac{n(n-1)}{2})! = 15! > (1.3)(10^{12})$$ ~flamewavelight and ~phoenixfire