Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 5"
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Using those three recursive rules, and that <math>a_2 = 4, b_2 = 2, c_2=2</math>, we can make a table: | Using those three recursive rules, and that <math>a_2 = 4, b_2 = 2, c_2=2</math>, we can make a table: | ||
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{array}{|r||r|r|r|} |
\hline | \hline | ||
&a_n&b_n&c_n \\ | &a_n&b_n&c_n \\ | ||
Line 25: | Line 25: | ||
10 & 472 & 648 & 816 \\ | 10 & 472 & 648 & 816 \\ | ||
\hline | \hline | ||
− | \end{ | + | \end{array} |
</cmath> | </cmath> | ||
For simplicity, we used <math>\mod 1000</math>. Thus, the answer is <math>a_{10} + b_{10} + c_{10} \equiv \boxed{936} \pmod{1000}</math>. (the real answer is <math>15936</math>.) | For simplicity, we used <math>\mod 1000</math>. Thus, the answer is <math>a_{10} + b_{10} + c_{10} \equiv \boxed{936} \pmod{1000}</math>. (the real answer is <math>15936</math>.) |
Revision as of 19:24, 10 March 2015
Problem
In Zuminglish, all words consist only of the letters and . As in English, is said to be a vowel and and are consonants. A string of and is a word in Zuminglish if and only if between any two there appear at least two consonants. Let denote the number of -letter Zuminglish words. Determine the remainder obtained when is divided by .
Solution
Solution 1 (recursive)
Let denote the number of -letter words ending in two constants (CC), denote the number of -letter words ending in a constant followed by a vowel (CV), and let denote the number of -letter words ending in a vowel followed by a constant (VC - the only other combination, two vowels, is impossible due to the problem statement). Then, note that:
- We can only form a word of length with CC at the end by appending a constant () to the end of a word of length that ends in a constant. Thus, we have the recursion , as there are two possible constants we can append.
- We can only form a word of length with a CV by appending to the end of a word of length that ends with CC. This is because we cannot append a vowel to VC, otherwise we'd have two vowels within characters of each other. Thus, .
- We can only form a word of length with a VC by appending a constant to the end of a word of length that ends with CV. Thus, .
Using those three recursive rules, and that , we can make a table: For simplicity, we used . Thus, the answer is . (the real answer is .)
Solution 2 (combinatorics)
See solutions pdf.
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |