Difference between revisions of "2005 AMC 12B Problems/Problem 10"
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== 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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== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2005|ab=B|num-b=9|num-a=11}} | |
Revision as of 14:23, 4 July 2011
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?
Solution
Performing this operation several times yields the results of 
for the second term, 
for the third term, and 
for the fourth term. The sum of the cubes of the digits of equal , a complete cycle. The cycle is... excluding the first term, the 
, 
, and 
terms will equal , , and , following the fourth term. Any term number that is equivalent to 
will produce a result of . It just so happens that 
, which leads us to the answer of 
.
See also
| 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 | |
