Difference between revisions of "Mock AIME II 2012 Problems"
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==Problem 1== | ==Problem 1== | ||
Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. | Given that <cmath>\left(\dfrac{6^2-1}{6^2+11}\right)\left(\dfrac{7^2-2}{7^2+12}\right)\left(\dfrac{8^2-3}{8^2+13}\right)\cdots\left(\dfrac{2012^2-2007}{2012^2+2017}\right)=\dfrac{m}{n},</cmath> where <math>m</math> and <math>n</math> are positive relatively prime integers, find the remainder when <math>m+n</math> is divided by <math>1000</math>. |
Revision as of 23:57, 4 April 2012
Problem 1
Given that where and are positive relatively prime integers, find the remainder when is divided by .
Problem 2
Let be a recursion defined such that , and where , and is an integer. If for being a positive integer greater than and being a positive integer greater than 2, find the smallest possible value of .
Problem 3
The of a number is defined as the result obtained by repeatedly adding the digits of the number until a single digit remains. For example, the of is (). Find the of .