Difference between revisions of "2019 CIME I Problems/Problem 9"
Icematrix2 (talk | contribs) (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>:...") |
Icematrix2 (talk | contribs) |
||
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>( | + | * 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 15:27, 6 October 2020
Let denote the number of strictly increasing sequences of positive integers satisfying the following two rules
- and
- for any if is the number not in the sequence then $(a_ib_{ij(a_jb_{ij<0.$ (Error compiling LaTeX. Unknown error_msg)
Find the largest positive integer such that divides
See also
2019 CIME I (Problems • Answer Key • Resources) | ||
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.