Difference between revisions of "2023 AIME II Problems/Problem 15"
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For each positive integer <math>n</math> let <math>a_n</math> be the least positive integer multiple of <math>23</math> such that <math>a_n \equiv 1 \pmod{2^n}.</math> Find the number of positive integers <math>n</math> less than or equal to <math>1000</math> that satisfy <math>a_n = a_{n+1}.</math> | For each positive integer <math>n</math> let <math>a_n</math> be the least positive integer multiple of <math>23</math> such that <math>a_n \equiv 1 \pmod{2^n}.</math> Find the number of positive integers <math>n</math> less than or equal to <math>1000</math> that satisfy <math>a_n = a_{n+1}.</math> | ||
− | ==Solution== | + | ==Solution 1== |
Denote <math>a_n = 23 b_n</math>. | Denote <math>a_n = 23 b_n</math>. | ||
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Now, we find the smallest <math>m</math>, such that <math>2^m \equiv 1 \pmod{23}</math>. | Now, we find the smallest <math>m</math>, such that <math>2^m \equiv 1 \pmod{23}</math>. | ||
− | + | By Fermat's Theorem, we must have <math>m | \phi \left( 23 \right)</math>. That is, <math>m | 22</math>. | |
We find <math>m = 11</math>. | We find <math>m = 11</math>. | ||
Line 31: | Line 31: | ||
We have the following results: | We have the following results: | ||
− | + | ||
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 0\), then \(k_n = 22\) and \(b_n = 1\). |
− | \ | + | |
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 1\), then \(k_n = 11\) and \(b_n = 1\). |
− | \ | + | |
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 2\), then \(k_n = 17\) and \(b_n = 3\). |
− | \ | + | |
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 3\), then \(k_n = 20\) and \(b_n = 7\). |
− | \ | + | |
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 4\), then \(k_n = 10\) and \(b_n = 7\). |
− | \ | + | |
− | \ | + | If \({\rm Rem} \left( n , 11 \right) = 5\), then \(k_n = 5\) and \(b_n = 7\). |
− | + | ||
+ | If \({\rm Rem} \left( n , 11 \right) = 6\), then \(k_n = 14\) and \(b_n = 39\). | ||
+ | |||
+ | If \({\rm Rem} \left( n , 11 \right) = 7\), then \(k_n = 7\) and \(b_n = 39\). | ||
+ | |||
+ | If \({\rm Rem} \left( n , 11 \right) = 8\), then \(k_n = 15\) and \(b_n = 167\). | ||
+ | |||
+ | If \({\rm Rem} \left( n , 11 \right) = 9\), then \(k_n = 19\) and \(b_n = 423\). | ||
+ | |||
+ | If \({\rm Rem} \left( n , 11 \right) = 10\), then \(k_n = 21\) and \(b_n = 935\). | ||
Therefore, in each cycle, <math>n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}</math>, we have <math>n = 11l</math>, <math>11l + 3</math>, <math>11l + 4</math>, <math>11l + 6</math>, such that <math>b_n = b_{n+1}</math>. That is, <math>a_n = a_{n+1}</math>. | Therefore, in each cycle, <math>n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}</math>, we have <math>n = 11l</math>, <math>11l + 3</math>, <math>11l + 4</math>, <math>11l + 6</math>, such that <math>b_n = b_{n+1}</math>. That is, <math>a_n = a_{n+1}</math>. | ||
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== Solution 2 == | == Solution 2 == | ||
− | Observe that if <math>a_{n-1}</math> is divisible by <math>2^n</math>, <math>a_n = a_{n-1}</math>. If not, <math>a_n = a_{n-1} + 23 \cdot 2^{n-1}</math>. | + | Observe that if <math>a_{n-1} - 1</math> is divisible by <math>2^n</math>, <math>a_n = a_{n-1}</math>. If not, <math>a_n = a_{n-1} + 23 \cdot 2^{n-1}</math>. |
− | This encourages us to let <math>b_n = | + | This encourages us to let <math>b_n = \frac{a_n - 1}{2^n}</math>. Rewriting the above equations, we have |
− | <cmath> b_n = \begin{cases} b_{n-1} | + | <cmath> b_n = \begin{cases} \frac{b_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } b_{n-1} \ \frac{b_{n-1}+23}{2} &\text{if } 2 \not\vert \text{ } b_{n-1} \end{cases} </cmath> |
− | + | The first few values of <math>b_n</math> are <math>11, 17, 20, 10, 5, 14, 7, 15, 19, 21,</math> and <math>22</math>. We notice that <math>b_{12} = b_1 = 11</math>, and thus the sequence is periodic with period <math>11</math>. | |
− | |||
− | + | Note that <math>a_n = a_{n+1}</math> if and only if <math>b_n</math> is even. This occurs when <math>n</math> is congruent to <math>0, 3, 4</math> or <math>6</math> mod <math>11</math>, giving four solutions for each period. | |
+ | From <math>1</math> to <math>1001</math> (which is <math>91 \times 11</math>), there are <math>91 \times 4 = 364</math> values of <math>n</math>. We subtract <math>1</math> from the total since <math>1001</math> satisfies the criteria but is greater than <math>1000</math> to get a final answer of <math>\fbox{363}</math> . | ||
~[[User:Bxiao31415|Bxiao31415]] | ~[[User:Bxiao31415|Bxiao31415]] | ||
+ | (small changes by bobjoebilly and [[User:Iraevid13|IraeVid13]]) | ||
== Solution 3 (Binary Interpretation, Computer Scientists' Playground) == | == Solution 3 (Binary Interpretation, Computer Scientists' Playground) == | ||
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~ cocoa @ https://www.corgillogical.com | ~ cocoa @ https://www.corgillogical.com | ||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ujP-V170vvI | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
+ | |||
== See also == | == See also == |
Revision as of 07:13, 1 February 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 (Binary Interpretation, Computer Scientists' Playground)
We first check that hence we are always seeking a unique modular inverse of
,
, such that
.
Now that we know that is unique, we proceed to recast this problem in binary. This is convenient because
is simply the last
-bits of
in binary, and if
, it means that of the last
bits of
, only the rightmost bit (henceforth
th bit) is
.
Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:
Now note , and recall that our objective is to progressively zero out the
leftmost bits of
except for the
th bit.
Write , we note that
uniquely defines the
th bit of
, and once we determine
,
uniquely determines the
st bit of
, so on and so forth.
For example, satisfies
Next, we note that the second bit of
is
, so we must also have
in order to zero it out, giving
happens precisely when
. In fact we can see this in action by working out
. Note that
has 1 on the
nd bit, so we must choose
. This gives
Note that since the rd and
th bit are
,
, and this gives
.
It may seem that this process will take forever, but note that has
bits behind the leading digit, and in the worst case, the leading digits of
will have a cycle length of at most
. In fact, we find that the cycle length is
, and in the process found that
,
, and
.
Since we have complete cycles of length
, and the last partial cycle yields
and
, we have a total of
values of
such that
~ 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 |
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