Difference between revisions of "2005 AMC 12B Problems/Problem 10"

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== Problem ==
 
== Problem ==
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The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the 2005th term of the sequence?
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<math>\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}</math>
  
 
== Solution ==
 
== Solution ==

Revision as of 22:10, 29 July 2009

Problem

The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the 2005th term of the sequence?

$\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}$

Solution

See also