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>). | + | (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>. | + | <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>. | ||
==Solution== | ==Solution== |
Revision as of 18:33, 30 December 2008
Problem
(a) Prove that
where and denotes the greatest integer (e.g., ).
(b) Using (a) or otherwise, prove that
is integral for all positive integral and .
Solution
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See also
1975 USAMO (Problems • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |