Difference between revisions of "1975 USAMO Problems/Problem 1"

(New page: ==Problem== (a) Prove that <center><math>[5x]+[5y]\ge [3x+y]+[3y+x]</math>,</center>where <math>x,y\ge 0</math> and <math>[u]</math> denotes the greatest integer <math>\le u</math> (e.g., ...)
 
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==Problem==
 
==Problem==
(a) Prove that <center><math>[5x]+[5y]\ge [3x+y]+[3y+x]</math>,</center>where <math>x,y\ge 0</math> and <math>[u]</math> denotes the greatest integer <math>\le u</math> (e.g., <math>[\sqrt{2}]=1</math>).
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(a) Prove that
 +
<center><math>[5x]+[5y]\ge [3x+y]+[3y+x]</math>,</center>
 +
where <math>x,y\ge 0</math> and <math>[u]</math> denotes the greatest integer <math>\le u</math> (e.g., <math>[\sqrt{2}]=1</math>).
  
 
+
(b) Using (a) or otherwise, prove that
(b) Using (a) or otherwise, prove that <center><math>\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}</math></center>is integral for all positive integral <math>m</math> and <math>n</math>.
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<center><math>\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}</math></center>
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is integral for all positive integral <math>m</math> and <math>n</math>.
  
 
==Solution==
 
==Solution==

Revision as of 18:33, 30 December 2008

Problem

(a) Prove that

$[5x]+[5y]\ge [3x+y]+[3y+x]$,

where $x,y\ge 0$ and $[u]$ denotes the greatest integer $\le u$ (e.g., $[\sqrt{2}]=1$).

(b) Using (a) or otherwise, prove that

$\frac{(5m)!(5n)!}{m!n!(3m+n)!(3n+m)!}$

is integral for all positive integral $m$ and $n$.

Solution

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See also

1975 USAMO (ProblemsResources)
Preceded by
First Question
Followed by
Problem 2
1 2 3 4 5
All USAMO Problems and Solutions
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