Difference between revisions of "2021 AMC 12A Problems/Problem 8"
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==Solution== | ==Solution== | ||
Making a small chart, we have | Making a small chart, we have | ||
| − | < | + | <cmath>\begin{tabular}{c|c|c|c|c|c|c|c|c|c} |
| − | + | D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline | |
0&0&1&1&1&2&3&4&6&9\\\hline | 0&0&1&1&1&2&3&4&6&9\\\hline | ||
E&E&O&O&O&E&O&E&E&O | E&E&O&O&O&E&O&E&E&O | ||
| − | \end{tabular | + | \end{tabular}</cmath> |
This starts repeating every 7 terms, so <math>D_{2021}=D_5=</math> E, <math>D_{2022}=D_6=</math> O, and <math>D_{2023}=D_7=</math> E. Thus, the answer is \boxed{C) \text{(E, O, E)}} | This starts repeating every 7 terms, so <math>D_{2021}=D_5=</math> E, <math>D_{2022}=D_6=</math> O, and <math>D_{2023}=D_7=</math> E. Thus, the answer is \boxed{C) \text{(E, O, E)}} | ||
~JHawk0224 | ~JHawk0224 | ||
Revision as of 16:31, 11 February 2021
Problem
A sequence of numbers is defined by 
and 
for 
. What are the parities (evenness or oddness) of the triple of numbers 
, where 
denotes even and 
denotes odd?
Solution
Making a small chart, we have
\[\begin{tabular}{c|c|c|c|c|c|c|c|c|c}
D_0&D_1&D_2&D_3&D_4&D_5&D_6&D_7&D_8&D_9\\\hline
0&0&1&1&1&2&3&4&6&9\\\hline
E&E&O&O&O&E&O&E&E&O
\end{tabular}\] (Error compiling LaTeX. Unknown error_msg)
This starts repeating every 7 terms, so 
E, 
O, and 
E. Thus, the answer is \boxed{C) \text{(E, O, E)}}
~JHawk0224
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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. 
