Difference between revisions of "2005 AMC 12B Problems/Problem 10"
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| − | Performing this operation several times yields the results of | + | Performing this operation several times yields the results of '''133''' 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>. |
== See also == | == See also == | ||
Revision as of 19:39, 14 July 2020
- 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
The first term of a sequence is 
. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 
term of the sequence?
Solution
Performing this operation several times yields the results of 133 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 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
