Mock AIME 5 2005-2006 Problems/Problem 11: Difference between revisions
m Created page with "== Problem == == Solution == == See also == {{Mock AIME box|year=2005-2006|n=5|source=76847|num-b=10|num-a=12}} Category:Intermediate Number Theory Problems" |
m →Problem |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
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>) | |||
== Solution == | |||
== Solution == | == Solution == | ||
Latest revision as of 20:19, 8 October 2014
Problem
Let 
be a subset of consecutive elements of 
where 
is a positive integer. Define 
, where 
if 
has an odd number of divisors and 
if has an even number of divisors. For how many 
does there exist an such that 
and 
? (
denotes the cardinality of the set 
, or the number of elements in )
Solution
Solution
See also
| Mock AIME 5 2005-2006 (Problems, Source) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
