Difference between revisions of "2023 AIME II Problems/Problem 15"
m (→Solution 3 (Similar to solution 2 but more explanation)) |
(→Solution 3 (Similar to solution 2 but more explanation)) |
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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>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 | + | 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 |
Revision as of 16:05, 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
.
Furthermore if quotient
is even then
Therefore
if and only if
is even. And, if this is true, then
. Next, if
is odd, we must have
. Solving for
, we have
Therefore, if
is odd,
. In sum, our recursion is
Finally, let us list out
to find a pattern. Because
,
. Through our recursion, we continue like so:
Therefore
repeats with cycle length
. Since
if and only iff
is even, in each cycle, we have 4 satisfactory values of
. There are
complete cycles. There are 3 extra values in the last incomplete cycle. Therefore we obtain
.
Solution 4 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.