Difference between revisions of "2023 AIME II Problems/Problem 15"
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So | So | ||
<cmath> b_{n+1} = b_n, b_n + 2^n </cmath> | <cmath> b_{n+1} = b_n, b_n + 2^n </cmath> | ||
− | Since <math> | + | Since <math>0 \le b_n < 2^n</math> and <math>0 \le b_{n+1} < 2^{n+1}</math> as <math>a_n</math> is the *least* positive integer multiple of 23. |
Now suppose <math>b_{n+1} = b_n</math>. Define <math>q_n</math> to be the quotient of <math>23b_n</math> divided by <math>2^n</math>. Then | Now suppose <math>b_{n+1} = b_n</math>. Define <math>q_n</math> to be the quotient of <math>23b_n</math> divided by <math>2^n</math>. Then | ||
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~ cocoa @ https://www.corgillogical.com | ~ cocoa @ https://www.corgillogical.com | ||
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== Video Solution == | == Video Solution == |
Revision as of 16:02, 29 December 2024
Contents
[hide]Problem
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 1
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
.
By Fermat's Theorem, we must have
. That is,
.
We find
.
Therefore, for each , we need to find smallest
, such that
We have the following results:
If
If
If
If
If
If
If
If
If
If
If
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
The first few values of
are
and
. We notice that
, and thus the sequence is periodic with period
.
Note that if and only if
is even. This occurs when
is congruent to
or
mod
, giving four solutions for each period.
From to
(which is
), there are
values of
. We subtract
from the total since
satisfies the criteria but is greater than
to get a final answer of
.
~Bxiao31415
(small changes by bobjoebilly and IraeVid13)
Solution 3 (Similar to solution 2 but more explanation)
Let . Note that if
Then
Also
Therefore
Then
So
Since
and
as
is the *least* positive integer multiple of 23.
Now suppose . Define
to be the quotient of
divided by
. Then
$$ (Error compiling LaTeX. Unknown error_msg) 23b_n = 2^n q_n + 1 \text{ and } 23b_{n+1} = 23b_n = 2^{n+1} q_{n+1} + 1 = 2^n q_n + 1 \implies q_{n+1} = \frac{q_n}{2}
q_n
b_{n+1} = b_n
q_n
q_{n+1} = \frac{q_n}{2}
q_n
b_{n+1} = b_n + 2^n
q_{n+1}
q_n
q_{n+1} = \frac{q_n + 1}{2} + 11
q_n
a_1 = 23
q_1 = 11
q_n
11
a_{n+1} = a_n
q_n
n
\frac{1000 - 10}{11} = 90
90 \cdot 4 + 3 = \fbox{363}$.
== Solution 4 (Binary Interpretation, Computer Scientists' Playground) ==
We first check that$ (Error compiling LaTeX. Unknown error_msg)\gcd(23, 2^n) = 123
b_n
a_n \equiv 23b_n \equiv 1 \mod{2^n}$.
Now that we know that$ (Error compiling LaTeX. Unknown error_msg)b_nx \mod{2^n}
n
x
x \equiv 1 \mod{2^n}
n
x
0
1$.
Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:
<cmath> \begin{align} 10111_2 \times 1011_2 &= 10111_2 \times (1000_2 + 10_2 + 1_2) \
&= 10111000_2 + 101110_2 + 10111_2 \\ &= 11111101_2
\end{align} </cmath>
Now note$ (Error compiling LaTeX. Unknown error_msg)23 = 10111_2n
a_n = 10111_2 \times b_n
0$th bit.
Write$ (Error compiling LaTeX. Unknown error_msg)b_n = \underline{c_{n-1}\cdots c_2c_1c_0}_2c_0
0
a_n
c_0
c_1
1
a_n$, so on and so forth.
For example,$ (Error compiling LaTeX. Unknown error_msg)c_0 = 1a_1 \equiv10111_2 \times 1_2 \equiv 1 \mod{10_2}
a_1
1
c_1 = 1$in order to zero it out, giving
<cmath>a_2 \equiv 10111_2 \times 11_2 \equiv 101110_2 + a_1 \equiv 1000101_2 \equiv 01_2 \mod{100_2}</cmath>$ (Error compiling LaTeX. Unknown error_msg)a_{n+1} = a_{n}c_n = 0
a_3
a_2
2
c_2 = 1$. This gives
<cmath>a_3 \equiv 10111_2 \times 111_2 \equiv 1011100_2 + a_2 \equiv 10100001_2 \equiv 001_2 \mod{1000_2}</cmath>
Note that since the$ (Error compiling LaTeX. Unknown error_msg)34
0
c_3 = c_4 = 0
a_3 = a_4 = a_5$.
It may seem that this process will take forever, but note that$ (Error compiling LaTeX. Unknown error_msg)23 = 10111_24
a_n
16
11
a_3 = a_4 = a_5
a_6 = a_7
a_{11} = a_{12}$.
Since we have$ (Error compiling LaTeX. Unknown error_msg)9011
a_{993} = a_{994} = a_{995}
a_{996} = a_{997}
90 \times 4 + 3 = \boxed{363}
n \le 1000
a_n = a_{n+1}$
~ cocoa @ https://www.corgillogical.com
Video Solution
~MathProblemSolvingSkills.com
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.