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]]}}
 
{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #10]] and [[2005 AMC 10B Problems|2005 AMC 10B #11]]}}
 
== Problem 10 ==
 
== Problem 10 ==
 
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436445
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 term. 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>
 
  
 
== Solution ==
 
== Solution ==
  
Performing this operation several times yields the results of <math>133</math> for the second term, <math>55</math> for the third term, and <math>250</math> for the fourth term. The sum of the cubes of the digits of <math>250</math> equal <math>133</math>, a complete cycle. The cycle is, excluding the first term, the <math>2^{\text{nd}}</math>, <math>3^{\text{rd}}</math>, and <math>4^{\text{th}}</math> terms will equal <math>133</math>, <math>55</math>, and <math>250</math>, following the fourth term. Any term number that is equivalent to <math>1\ (\text{mod}\ 3)</math> will produce a result of <math>250</math>. It just so happens that <math>2005\equiv 1\ (\text{mod}\ 3)</math>, which leads us to the answer of <math>\boxed{\mathrm{(E)}\ 250}</math>.
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Performing this operation several times yields the results of <math>133</math> for the second term, <math>55</math> for the third term, and <math>250</math> for the fourth term. The sum of the cubes of the digits of <math>250</math> equal <math>133</math>, a complete cycle. The cycle is, excluding the first term, the <math>2^{\text{nd}}</math>, <math>3^{\text{rd}}</math>, and <math>4^{\text{th}}</math> terms will equal <math>133</math>, <math>55</math>, and <math>250</math>, following the fourth term. Any term number that is equivalent to <math>1\ (\text{mod}\ 3)</math> will produce a result of <math>250</math>. It just so happens that <math>2005\equiv 1\ (\text{mod}\ 3)</math>, which leads us to the answer of <math>\boxed{\textbf{(E) } 250}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 19:49, 24 October 2024

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

Problem 10

436445

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 $\boxed{\textbf{(E) } 250}$.

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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