Difference between revisions of "1998 USAMO Problems/Problem 5"
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| + | [[Category:Olympiad Number Theory Problems]] | ||
Latest revision as of 08:44, 20 July 2016
Problem
Prove that for each 
, there is a set 
of 
integers such that 
divides 
for every distinct 
.
Solution
Proof by induction. For n=2, the proof is trivial, since 
satisfies the condition. Assume now that there is such a set S of n elements, 
which satisfy the condition. The key is to note that if 
, then if we define 
for all 
, where k is a positive integer, then 
and 
, and so 
.
Let 
. Consider the set 
. To finish the proof, we simply need to choose a k such that 
for all . Since 
, simply choose k so that 
.
See Also
| 1998 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
