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 $Legendre$ $\text{[math]}$ 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

Remark

Continuing as above, we realize that $L_n < 2^{n^{2}}$ . But this is absurd for all $n \geq 6$ . So the claim holds. Since we have already checked $n = 1, 2$ it remains to check $n = 3, 4, 5$ and we see neither of them work so $(1,1) and (3,2)$ are the unique two solutions.

~ilikemath247365

2019 IMO (Problems) • Resources
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
All IMO Problems and Solutions

Page

Toolbox

Search

2019 IMO Problems/Problem 4

Contents

Problem

Solution 1

Remark

See Also