Difference between revisions of "Mock AIME II 2012 Problems"
(Created page with "==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}...") |
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[[Mock AIME II 2012 Problems/Problem 1 |Solution]] | [[Mock AIME II 2012 Problems/Problem 1 |Solution]] | ||
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+ | ==Problem 2== | ||
+ | Let <math>\{a_n\}</math> be a recursion defined such that <math>a_1=1, a_2=20</math>, and <math>a_n=\sqrt{\left| a_{n-1}^2-a_{n-2}^2 \right|}</math> where <math>n\ge 3</math>, and <math>n</math> is an integer. If <math>a_m=k</math> for <math>k</math> being a positive integer greater than <math>1</math> and <math>m</math> being a positive integer greater than 2, find the smallest possible value of <math>m+k</math>. | ||
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+ | [[Mock AIME II 2012 Problems/Problem 2| Solution]] | ||
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+ | ==Problem 3== | ||
+ | The <math>\textit{digital root}</math> 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 <math>\textit{digital root}</math> of <math>237</math> is <math>3</math> (<math>2+3+7=12, 1+2=3</math>). Find the <math>\textit{digital root}</math> of <math>2012^{2012^{2012}}</math>. | ||
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+ | [[Mock AIME II 2012 Problems/Problem 3| Solution]] |
Revision as of 23:56, 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 .