Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 11"

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== Problem ==
 
== Problem ==
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Let <math>A</math> be a subset of consecutive elements of <math>S = \{n, n+1, \ldots, n+999\}</math> where <math>n</math> is a positive integer. Define <math>\mu(A) = \sum_{k \in A} \tau(k)</math>, where <math>\tau(k) = 1</math> if <math>k</math> has an odd number of divisors and <math>\tau(k) = 0</math> if <math>k</math> has an even number of divisors. For how many <math>n \le 1000</math> does there exist an <math>A</math> such that <math>|A| = 620</math> and <math>\mu(A) = 11</math>? (<math>|X|</math> denotes the cardinality of the set <math>X</math>, or the number of elements in <math>X</math>)
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== Solution ==
  
 
== Solution ==
 
== Solution ==

Latest revision as of 21:19, 8 October 2014

Problem

Let $A$ be a subset of consecutive elements of $S = \{n, n+1, \ldots, n+999\}$ where $n$ is a positive integer. Define $\mu(A) = \sum_{k \in A} \tau(k)$, where $\tau(k) = 1$ if $k$ has an odd number of divisors and $\tau(k) = 0$ if $k$ has an even number of divisors. For how many $n \le 1000$ does there exist an $A$ such that $|A| = 620$ and $\mu(A) = 11$? ($|X|$ denotes the cardinality of the set $X$, or the number of elements in $X$)

Solution

Solution

See also

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by
Problem 10
Followed by
Problem 12
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