Difference between revisions of "2012 Indonesia MO Problems/Problem 5"
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let the collection X be named | let the collection X be named | ||
− | < | + | <cmath>\begin{bmatrix} X_{1,1}&X_{1,2}&\dots&X_{1,n}\\X_{2,1}&X_{2,2}&\dots&X_{2,n}\\\vdots&\vdots&\vdots&\vdots\\X_{m,1}&X_{m,2}&\dots&X_{m,n}\end{bmatrix}</cmath> |
− | + | by (i), for all <math>i</math>, <math>P_{i,1}\geq P_{i,2}\geq \dots \geq P_{i,n}</math>, that means <math>(P_{1,1}+P_{2,1}+\dots+P_{m,1})\geq(P_{1,2}+P_{2,2}+\dots+P_{m,2})\geq\dots\geq(P_{1,n}+P_{2,n}+\dots+P_{m,n})\implies(Q_{1,1}+Q_{2,1}+\dots+Q_{m,1})\geq(Q_{1,2}+Q_{2,2}+\dots+Q_{m,2})\geq\dots\geq(Q_{1,n}+Q_{2,n}+\dots+Q_{m,n})</math> | |
+ | for the collection Q, all the 1's are getting pushed up and the inequality of the sum of columns of Q are pushing it to the left, if we do the same logic to P we also sed it getting pushed up and left. notice how if the sum of some row/column is some number <math>p</math>, then the first p numbers of the row/column is <math>1</math> and the rest <math>0</math>, so we can deduce its equal (sorry i have no idea how to tody the writing, some1 help :() | ||
==See Also== | ==See Also== | ||
{{Indonesia MO box|year=2012|num-b=4|num-a=6}} | {{Indonesia MO box|year=2012|num-b=4|num-a=6}} | ||
[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 21:14, 24 December 2024
Problem
Given positive integers and . Let and be two collections of numbers of and , arranged in rows and columns. An example of such collections for and is Let those two collections satisfy the following properties: (i) On each row of , from left to right, the numbers are non-increasing, (ii) On each column of , from top to bottom, the numbers are non-increasing, (iii) The sum of numbers on the row in equals to the same row in , (iv) The sum of numbers on the column in equals to the same column in . Show that the number on row and column of equals to the number on row and column of for and .
Solution
let the collection X be named by (i), for all , , that means for the collection Q, all the 1's are getting pushed up and the inequality of the sum of columns of Q are pushing it to the left, if we do the same logic to P we also sed it getting pushed up and left. notice how if the sum of some row/column is some number , then the first p numbers of the row/column is and the rest , so we can deduce its equal (sorry i have no idea how to tody the writing, some1 help :()
See Also
2012 Indonesia MO (Problems) | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 6 |
All Indonesia MO Problems and Solutions |