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Difference between revisions of "2005 AMC 12B Problems/Problem 10"
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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #10]] and [[2005 AMC 10B Problems|2005 AMC 10B #11]]}}
 
== 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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The first term of a sequence is <math>2005</math>. Each succeeding term is the sum of the cubes of the digits of the previous terms. What is the <math>2005^\text{th}</math> term of the sequence?
  
 
<math>\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}</math>
 
<math>\mathrm{(A)}\ {{{29}}} \qquad \mathrm{(B)}\ {{{55}}} \qquad \mathrm{(C)}\ {{{85}}} \qquad \mathrm{(D)}\ {{{133}}} \qquad \mathrm{(E)}\ {{{250}}}</math>
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== See also ==
 
== See also ==
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{{AMC10 box|year=2005|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2005|ab=B|num-b=9|num-a=11}}
 
{{AMC12 box|year=2005|ab=B|num-b=9|num-a=11}}

Revision as of 14:10, 16 July 2011

The following problem is from both the 2005 AMC 12B #10 and 2005 AMC 10B #11, so both problems redirect to this page.

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 $2005^\text{th}$ term of the sequence?

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

Solution

Performing this operation several times yields the results of $133$ for the second term, $55$ for the third term, and $250$ for the fourth term. The sum of the cubes of the digits of $250$ equal $133$, a complete cycle. The cycle is... excluding the first term, the $2^{\text{nd}}$, $3^{\text{rd}}$, and $4^{\text{th}}$ terms will equal $133$, $55$, and $250$, following the fourth term. Any term number that is equivalent to $1\ (\text{mod}\ 3)$ will produce a result of $250$. It just so happens that $2005\equiv 1\ (\text{mod}\ 3)$, which leads us to the answer of $\boxed{\mathrm{(E)}\ 250}$.

See also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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