Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution 1)
(→‎See Also)
 
(4 intermediate revisions by 3 users not shown)
Line 8: Line 8:
 
<math>LHS</math>   
 
<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>)
 
<math>RHS = 1</math>(when <math>n = 1</math>), <math>6</math> (when <math>n = 2</math>)
Line 20: Line 20:
 
In all solutions, for any prime <math>p</math> and positive integer <math>N</math>, we will denote by <cmath>v_p(N)</cmath> the exponent of the largest power of <math>p</math> that divides <math>N</math>. The right-hand side of <math>(1)</math> will be denoted by <cmath>L_n;</cmath> that is, <cmath>L_n = (2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})</cmath>
 
In all solutions, for any prime <math>p</math> and positive integer <math>N</math>, we will denote by <cmath>v_p(N)</cmath> the exponent of the largest power of <math>p</math> that divides <math>N</math>. The right-hand side of <math>(1)</math> will be denoted by <cmath>L_n;</cmath> that is, <cmath>L_n = (2^n-1)(2^n-2)(2^n-4)\dots(2^n-2^{n-1})</cmath>
  
−
<cmath>(2^{1+2+3+\dots+(n-1)})(2^n-1)(2^{n-1}-1)(2^{n-2}-1)\dots(2^1-1)</cmath> = <cmath>v_2(L_n) = (2^{\frac{n(n-1)}{2}}) </cmath>  
+
<cmath>(2^{1+2+3+\dots+(n-1)})(2^n-1)(2^{n-1}-1)(2^{n-2}-1)\dots(2^1-1)</cmath> = <cmath>v_2(L_n) = {\frac{n(n-1)}{2}} </cmath>  
  
 
On the other hand, <cmath>v_2(k!)</cmath> is expressed by the <math>Legendre</math> <math>formula</math> as <cmath>v_2(k!) < \sum_{i=1}^{\infty} (\frac{k}{2^i})) = k</cmath>
 
On the other hand, <cmath>v_2(k!)</cmath> is expressed by the <math>Legendre</math> <math>formula</math> as <cmath>v_2(k!) < \sum_{i=1}^{\infty} (\frac{k}{2^i})) = k</cmath>
Line 30: Line 30:
 
For <math>n \geq 6</math> the estimate (3) is true because <cmath>2^{6^2} < (6.9)(10^{10})</cmath>  
 
For <math>n \geq 6</math> the estimate (3) is true because <cmath>2^{6^2} < (6.9)(10^{10})</cmath>  
 
and <cmath>(\frac{n(n-1)}{2})! = 15! > (1.3)(10^{12})</cmath>
 
and <cmath>(\frac{n(n-1)}{2})! = 15! > (1.3)(10^{12})</cmath>
−
~flamewavelight and phoenixfire
+
~flamewavelight and ~phoenixfire
 +
 
 +
==See Also==
 +
 
 +
{{IMO box|year=2019|num-b=3|num-a=5}}

Latest revision as of 00:51, 19 November 2023

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

See Also

2019 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2019_IMO_Problems/Problem_4&oldid=204595"