Difference between revisions of "2019 IMO Problems/Problem 4"

Revision as of 01:29, 9 February 2020

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, $L_n = (2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})$

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

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

@@ Line 8: / Line 8: @@
 <math>LHS</math>
-<math>k</math>! = 1(when <math>k = 1</math>), 2 (when <math>k = 2</math>), 6(when <math>k = 3</math>)
+<math>k! = 1</math> (when <math>k = 1</math>), <math>2</math> (when <math>k = 2</math>), <math>6</math> (when <math>k = 3</math>)
 <math>RHS = 1</math>(when <math>n = 1</math>), <math>6</math> (when <math>n = 2</math>)

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Difference between revisions of "2019 IMO Problems/Problem 4"

Revision as of 01:29, 9 February 2020

Problem

Solution 1