Difference between revisions of "1979 USAMO Problems/Problem 3"
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random. Its value is <math>a</math>. Another member is picked at random, independently of the first. Its value is <math>b</math>. Then a third value, <math>c</math>. Show that the probability that <math>a + b +c</math> is divisible by <math>3</math> is at least <math>\frac14</math>. | random. Its value is <math>a</math>. Another member is picked at random, independently of the first. Its value is <math>b</math>. Then a third value, <math>c</math>. Show that the probability that <math>a + b +c</math> is divisible by <math>3</math> is at least <math>\frac14</math>. | ||
| + | ==Hint== | ||
| + | The given problem is equivalent to proving that <math>4(x^3 + y^3 + z^3 + 6xyz) \ge (x + y + z)^3</math>. | ||
==Solution== | ==Solution== | ||
{{solution}} | {{solution}} | ||
Revision as of 10:03, 19 April 2014
Contents
Problem
is an arbitrary sequence of positive integers. A member of the sequence is picked at
random. Its value is 
. Another member is picked at random, independently of the first. Its value is 
. Then a third value, 
. Show that the probability that 
is divisible by 
is at least 
.
Hint
The given problem is equivalent to proving that 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See Also
| 1979 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
