Difference between revisions of "2012 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
For <math>n</math> equal to any odd prime <math>p</math>, the sequence <math>\left\{a_i = \left(\frac{1-n}{2}\right)^{m_p\left(i\right)}\right\}</math>, where <math>p^{m_p\left(i\right)}</math> is the greatest power of <math>p</math> that divides <math>i</math>, gives a valid sequence. Therefore, the set of possible values for <math>n</math> is at least the set of odd primes. | For <math>n</math> equal to any odd prime <math>p</math>, the sequence <math>\left\{a_i = \left(\frac{1-n}{2}\right)^{m_p\left(i\right)}\right\}</math>, where <math>p^{m_p\left(i\right)}</math> is the greatest power of <math>p</math> that divides <math>i</math>, gives a valid sequence. Therefore, the set of possible values for <math>n</math> is at least the set of odd primes. | ||
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==See Also== | ==See Also== | ||
Revision as of 08:53, 25 April 2012
Problem
Determine which integers 
have the property that there exists an infinite sequence 
, 
, 
, 
of nonzero integers such that the equality
![\[a_k + 2a_{2k} + \dots + na_{nk} = 0\]](http://latex.artofproblemsolving.com/b/b/5/bb59afa27abb3e054ab7f2cb301e50ffcca5ad62.png)
holds for every positive integer 
.
Solution
For 
equal to any odd prime 
, the sequence 
, where 
is the greatest power of that divides 
, gives a valid sequence. Therefore, the set of possible values for is at least the set of odd primes.
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See Also
| 2012 USAMO (Problems • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 1 • 2 • 3 • 4 • 5 • 6 | ||
| All USAMO Problems and Solutions | ||
