1995 USAMO Problems/Problem 4: Difference between revisions
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==Problem== | |||
Suppose <math>q_1,q_2,...</math> is an infinite sequence of integers satisfying the following two conditions: | Suppose <math>q_1,q_2,...</math> is an infinite sequence of integers satisfying the following two conditions: | ||
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Prove that there is a polynomial <math>Q</math> such that <math>q_n = Q(n)</math> for each <math>n</math>. | Prove that there is a polynomial <math>Q</math> such that <math>q_n = Q(n)</math> for each <math>n</math>. | ||
==Solution== | |||
{{solution}} | |||
==See Also== | |||
{{USAMO box|year=1995|num-b=3|num-a=5}} | |||
{{MAA Notice}} | |||
[[Category:Olympiad Algebra Problems]] | |||
Revision as of 07:07, 19 July 2016
Problem
Suppose 
is an infinite sequence of integers satisfying the following two conditions:
(a) 
divides 
for 
(b) There is a polynomial 
such that 
for all 
.
Prove that there is a polynomial 
such that 
for each .
Solution
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See Also
| 1995 USAMO (Problems • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
