Difference between revisions of "1987 AHSME Problems/Problem 29"
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==Solution== | ==Solution== | ||
− | <math>(\ | + | If <math>n</math> is even, then <math>t_{(n/2)}</math> would be negative, which is not possible. Therefore, <math>n</math> is odd. With this function, backwards thinking is the key. If <math>t_x < 1</math>, then <math>x</math> is odd, and <math>t_{(x-1)} = \frac{1}{t_{x}}</math>. Otherwise, you keep on subtracting 1 and halving x until <math>t_\frac{x}{2^{n}} < 1</math>. |
+ | We can use this logic to go backwards until we reach <math>t_1 = 1</math>, like so: | ||
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+ | <math>t_n=\frac{19}{87}\\t_{n-1} = \frac{87}{19}\\t_{\frac{n-1}{2}} = \frac{68}{19}\\t_{\frac{n-1}{4}} = \frac{49}{19}\\t_{\frac{n-1}{8}} = \frac{30}{19}\\t_{\frac{n-1}{16}} = \frac{11}{19}\\t_{\frac{n-1}{16} - 1} = \frac{19}{11}\\t_{\frac{\frac{n-1}{16} - 1}{2}} = \frac{8}{11}\\t_{\frac{\frac{n-1}{16} - 1}{2} - 1} = \frac{11}{8}\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2}} = \frac{3}{8}\\t_{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1} = \frac{8}{3}\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{2}} = \frac{5}{3}\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4}} = \frac{2}{3}\\t_{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1} = \frac{3}{2}\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2}} = \frac{1}{2}\\t_{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1} = 2\\t_{\frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2}} = t_1 = 1 \Rightarrow \frac{\frac{\frac{\frac{\frac{\frac{n-1}{16} - 1}{2} - 1}{2} - 1}{4} - 1}{2} - 1}{2} = 1 \Rightarrow \boxed{n = 1905~\textbf{(A)}\ 15}</math> | ||
== See also == | == See also == |
Revision as of 22:54, 1 March 2018
Problem
Consider the sequence of numbers defined recursively by and for by when is even and by when is odd. Given that , the sum of the digits of is
Solution
If is even, then would be negative, which is not possible. Therefore, is odd. With this function, backwards thinking is the key. If , then is odd, and . Otherwise, you keep on subtracting 1 and halving x until . We can use this logic to go backwards until we reach , like so:
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 28 |
Followed by Problem 30 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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