Difference between revisions of "2019 CIME I Problems/Problem 9"

(Created page with "Let <math>\text{N}</math> denote the number of strictly increasing sequences of positive integers <math>a_1,a_2,\cdots, a_{19}</math> satisfying the following two rules<math>:...")
 
Line 1: Line 1:
 
Let <math>\text{N}</math> denote the number of strictly increasing sequences of positive integers <math>a_1,a_2,\cdots, a_{19}</math> satisfying the following two rules<math>:</math>
 
Let <math>\text{N}</math> denote the number of strictly increasing sequences of positive integers <math>a_1,a_2,\cdots, a_{19}</math> satisfying the following two rules<math>:</math>
<math>a_1=1</math> and <math>a_{19}=361,</math>
+
* <math>a_1=1</math> and <math>a_{19}=361,</math>
for any <math>i \neq j,</math> if <math>b_{ij}</math> is the <math>(i \cdot j)^{\text{th}}</math> number not in the sequence<math>,</math> then <math>(a_i-b_{ij})(a_jb_{ij})<0.</math>
+
* for any <math>i \neq j,</math> if <math>b_{ij}</math> is the <math>(i \cdot j)^{\text{th}}</math> number not in the sequence<math>,</math> then <math>(a_ib_{ij(a_jb_{ij<0.</math>
 
Find the largest positive integer <math>k</math> such that <math>2^k</math> divides <math>\text{N}.</math>
 
Find the largest positive integer <math>k</math> such that <math>2^k</math> divides <math>\text{N}.</math>
 +
 +
==See also==
 +
{{CIME box|year=2019|n=I|num-b=8|num-a=10}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Revision as of 16:27, 6 October 2020

Let $\text{N}$ denote the number of strictly increasing sequences of positive integers $a_1,a_2,\cdots, a_{19}$ satisfying the following two rules$:$

  • $a_1=1$ and $a_{19}=361,$
  • for any $i \neq j,$ if $b_{ij}$ is the $(i \cdot j)^{\text{th}}$ number not in the sequence$,$ then $(a_ib_{ij(a_jb_{ij<0.$ (Error compiling LaTeX. Unknown error_msg)

Find the largest positive integer $k$ such that $2^k$ divides $\text{N}.$

See also

2019 CIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All CIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png