Difference between revisions of "2023 AIME II Problems/Problem 15"
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This encourages us to let <math>b_n = (a_n - 1)/2^n</math>. Rewriting the above equations, we have | This encourages us to let <math>b_n = (a_n - 1)/2^n</math>. Rewriting the above equations, we have | ||
<cmath> b_n = \begin{cases} b_{n-1}/2 & \text{if } 2 | (b_{n-1}+23)/2 \\ 1 &\text{if } 2\not\vert b_{n-1} \end{cases} </cmath> | <cmath> b_n = \begin{cases} b_{n-1}/2 & \text{if } 2 | (b_{n-1}+23)/2 \\ 1 &\text{if } 2\not\vert b_{n-1} \end{cases} </cmath> | ||
− | Now, we start listing. We know that <math>b_1 = 11</math>, so <math>b_2 = 17</math>, <math>b_3 = 20</math>, <math>b_4 = 10</math>, <math>b_5 = 5</math>, <math>b_6 = 14</math>, <math>b_7 = 7</math>, <math>b_8 = 15</math>, <math>b_9 = 19</math>, <math> | + | Now, we start listing. We know that <math>b_1 = 11</math>, so <math>b_2 = 17</math>, <math>b_3 = 20</math>, <math>b_4 = 10</math>, <math>b_5 = 5</math>, <math>b_6 = 14</math>, <math>b_7 = 7</math>, <math>b_8 = 15</math>, <math>b_9 = 19</math>, <math>b_{10} = 21</math>, <math>b_{11} = 22</math>, <math>b_{12} = 1</math>. Hence the sequence is periodic with period 11. |
Note that <math>a_n = a_{n+1}</math> if and only if <math>b_n = b_{n+1}/2</math>, i.e. <math>b_n</math> is even. This occurrs when <math>n</math> is congruent to 0, 3, 4 or 6 mod 11. | Note that <math>a_n = a_{n+1}</math> if and only if <math>b_n = b_{n+1}/2</math>, i.e. <math>b_n</math> is even. This occurrs when <math>n</math> is congruent to 0, 3, 4 or 6 mod 11. | ||
Revision as of 17:39, 16 February 2023
For each positive integer let be the least positive integer multiple of such that Find the number of positive integers less than or equal to that satisfy
Solution
Denote . Thus, for each , we need to find smallest positive integer , such that
Thus, we need to find smallest , such that
Now, we find the smallest , such that . We must have . That is, . We find .
Therefore, for each , we need to find smallest , such that
We have the following results: \begin{enumerate} \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \item If , then and . \end{enumerate}
Therefore, in each cycle, , we have , , , , such that . That is, . At the boundary of two consecutive cycles, .
We have . Therefore, the number of feasible is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
Observe that if is divisible by , . If not, .
This encourages us to let . Rewriting the above equations, we have Now, we start listing. We know that , so , , , , , , , , , , . Hence the sequence is periodic with period 11. Note that if and only if , i.e. is even. This occurrs when is congruent to 0, 3, 4 or 6 mod 11.
From 1 to , there are values of . Since satisfies the criteria, we subtract 1 to get , and we're done! ~Bxiao31415
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.