Difference between revisions of "2008 Mock ARML 1 Problems/Problem 2"
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Mathgeek2006 (talk | contribs) m (→Solution) |
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Setting up a table, | Setting up a table, | ||
<cmath> | <cmath> | ||
− | \begin{ | + | \begin{array}{|r||r|r|r|} |
\hline | \hline | ||
n & a_n & b_n & c_n \\ | n & a_n & b_n & c_n \\ | ||
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8 & 353 & 283 & 157 \\ | 8 & 353 & 283 & 157 \\ | ||
\hline | \hline | ||
− | \end{ | + | \end{array} |
</cmath> | </cmath> | ||
The answer is <math>353 + 2(283 + 157) = \boxed{1233}</math>. | The answer is <math>353 + 2(283 + 157) = \boxed{1233}</math>. |
Latest revision as of 19:28, 10 March 2015
Problem
A positive integer is a yo-yo if the absolute value of the difference between any two consecutive digits of is at least . Compute the number of -digit yo-yos.
Solution
Note that all of the digits must be . Let be the number of yo-yos with digits and with a leftmost digit of if is odd ( being a placeholder) or if is even, let those with a leftmost digit of if is odd or if is even, and let those with a leftmost digit of if is odd or if is even. By symmetry, the desired answer is , to exclude the integers with leftmost digit .
Note that a yo-yo of digits with leftmost digit of can be formed from a yo-yo of digits with leftmost digits of ; those with a leftmost digit of can be formed by those ending in ; and those with a leftmost digit of can be formed only by those ending in . The same holds true for the leftmost digits of , respectively. Thus, we have the recursions Setting up a table, The answer is .
See also
2008 Mock ARML 1 (Problems, Source) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 |