2023 AIME II Problems/Problem 13
Problem
Let be an acute angle such that Find the number of positive integers less than or equal to such that is a positive integer whose units digit is
Solution 1
Denote . For any , we have
Next, we compute the first several terms of .
By solving equation , we get . Thus, , , , , .
In the rest of analysis, we set . Thus,
Thus, to get an integer, we have . In the rest of analysis, we only consider such . Denote and . Thus, with initial conditions , .
To get the units digit of to be 9, we have
Modulo 2, for , we have
Because , we always have for all .
Modulo 5, for , we have
We have , , , , , , . Therefore, the congruent values modulo 5 is cyclic with period 3. To get , we have .
From the above analysis with modulus 2 and modulus 5, we require .
For , because , we only need to count feasible with . The number of feasible is
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2 (Simple)
It is clear, that is not integer if
Denote
The condition is satisfied iff or
If then the number of possible n is
For we get
vladimir.shelomovskii@gmail.com, vvsss
Note
A key idea in this solution is realizing that to calculate values of is difficult directly, so we try to think about it recursively.
We then see how we can go from to , and learn that .
So, letting , where we set as , and , as well as .
~Bigbrain_2009
Video Solution
~MathProblemSolvingSkills.com
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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