Difference between revisions of "1967 IMO Problems/Problem 5"
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− | + | ==Problem== | |
− | + | Let <math>a_1,\ldots,a_8</math> be reals, not all equal to zero. Let <cmath>c_n = \sum^8_{k=1} a^n_k</cmath> for <math>n=1,2,3,\ldots</math>. Given that among the numbers of the sequence <math>(c_n)</math>, there are infinitely many equal to zero, determine all the values of <math>n</math> for which <math>c_n = 0.</math> | |
+ | |||
+ | ==Solution== | ||
+ | |||
+ | <math>c_n</math> must be zero for all odd <math>n</math>. | ||
+ | |||
+ | Proof: | ||
+ | WLOG suppose that <math>a_1 \geq a_2 \geq ... \geq a_8</math>. If <math>a_1+a_8 > 0</math> then for sufficiently high odd <math>n</math>, <math>c_n</math> will be dominated by <math>a_1</math> alone i.e. it will always be positive. Similarly if <math>a_1+a_8 < 0</math>; hence <math>a_1=-a_8</math>. Now for odd <math>n</math> these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd <math>n</math>. Since some <math>a_i</math> is nonzero <math>c_n > 0</math> for even <math>n</math>. | ||
+ | |||
+ | The above solution was written by Fiachra and can be found here: [https://aops.com/community/p138611] | ||
+ | |||
+ | |||
+ | ---- | ||
+ | <math>\textbf{Note:}\hspace{4000pt}</math> Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently. | ||
+ | |||
+ | == See Also == | ||
+ | {{IMO box|year=1967|num-b=4|num-a=6}} |
Latest revision as of 19:23, 10 November 2024
Problem
Let be reals, not all equal to zero. Let for . Given that among the numbers of the sequence , there are infinitely many equal to zero, determine all the values of for which
Solution
must be zero for all odd .
Proof: WLOG suppose that . If then for sufficiently high odd , will be dominated by alone i.e. it will always be positive. Similarly if ; hence . Now for odd these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd . Since some is nonzero for even .
The above solution was written by Fiachra and can be found here: [1]
Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently.
See Also
1967 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |