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
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.professorchenedub.com)
Solution 2 (Simple)
It is clear, that is not integer if
Denote
\[c_{12m + 4} (mod 10) = 9 \cdot c_{12m} (mod 10) – 16 (mod 10) \cdot c{12m – 4} (mod 10) = (9 \cdot 7 – 6 \cdot 9} (mod 10) = (3 – 4) (mod 10) = 9.\] (Error compiling LaTeX. Unknown error_msg)
$c_{12m + 8} mod 10 = 9 \cdot c_{12m+4} mod 10 – 16 mod 10 \cdot c{12m } mod 10 = (9 \cdot 9 – 6 \cdot 7} mod 10 = (1 – 2) mod 10 = 9.$ (Error compiling LaTeX. Unknown error_msg) $c_{12m + 12} mod 10 = 9 \cdot c_{12m+8} mod 10 – 16 mod 10 \cdot c{12m +4} mod 10 = (9 \cdot 9 – 6 \cdot 9} mod 10 = (1 – 4) mod 10 = 7 \implies$ (Error compiling LaTeX. Unknown error_msg)
The condition is satisfied iff or
If then the number of possible n is
For we get
\boxed{167}.$
vladimir.shelomovskii@gmail.com, vvsss
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 | ||
All AIME Problems and Solutions |
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