# Difference between revisions of "Mock AIME 1 2006-2007 Problems/Problem 5"

m |
|||

Line 25: | Line 25: | ||

---- | ---- | ||

− | *[[Mock AIME 1 2006-2007/Problem 4 | Previous Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 4 | Previous Problem]] |

− | *[[Mock AIME 1 2006-2007/Problem 6 | Next Problem]] | + | *[[Mock AIME 1 2006-2007 Problems/Problem 6 | Next Problem]] |

*[[Mock AIME 1 2006-2007]] | *[[Mock AIME 1 2006-2007]] | ||

[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] |

## Latest revision as of 15:52, 3 April 2012

## Modified Problem

For a prime number , define the function as follows: If there exists , , such that

set . Otherwise, set . Compute the sum .

## Original Problem

Let be a prime and satisfy for all integers . is the greatest integer less than or equal to . If for fixed , there exists an integer such that:

then . If there is no such , then . If , find the sum: .

## Solution

The definition of is equivalent to the following: "If has a multiplicative inverse mod , is the member of the set such that . Otherwise, ."

Note that this really gives a well-defined function because that set includes exactly one member from each congruence class modulo , and each invertible element has inverses in only one such class.

From this point onwards, it's clear: as cycles through , also cycles through the same values in some order. We cover those values 11 times. Thus the answer is .