Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AIME II Problems/Problem 15"

m (→‎Solution 3 (Similar to solution 2 but more explanation))
m (→‎Solution 3 (Similar to solution 2 but more explanation))
Line 92: Line 92:
  
 
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
−
<math></math> 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}<math>.
+
<cmath> 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}</cmath>.
−
Furthermore if quotient </math>q_n<math> is even then
+
Furthermore if quotient <math>q_n</math> is even then
 
<cmath> 23b_n = 2^n q_n +1 = 2^{n+1} \frac{q_n}{2} +1 </cmath>
 
<cmath> 23b_n = 2^n q_n +1 = 2^{n+1} \frac{q_n}{2} +1 </cmath>
−
Therefore </math>b_{n+1} = b_n<math> if and only if </math>q_n<math> is even. And, if this is true, then </math>q_{n+1} = \frac{q_n}{2}<math>. Next, if </math>q_n<math> is odd, we must have </math>b_{n+1} = b_n + 2^n<math>. Solving for </math>q_{n+1}<math>, we have
+
Therefore <math>b_{n+1} = b_n</math> if and only if <math>q_n</math> is even. And, if this is true, then <math>q_{n+1} = \frac{q_n}{2}</math>. Next, if <math>q_n</math> is odd, we must have <math>b_{n+1} = b_n + 2^n</math>. Solving for <math>q_{n+1}</math>, we have
 
<cmath> 23b_{n+1} = 2^{n+1} q_{n+1} + 1 \implies 23b_n + 23 \cdot 2^n = 2^{n+1} q_{n+1} + 1 \implies 2^n q_n + 1 + 23 = 2^{n+1} q_{n+1} + 1 \implies q_{n+1} = \frac{q_n + 1}{2} + 11 </cmath>
 
<cmath> 23b_{n+1} = 2^{n+1} q_{n+1} + 1 \implies 23b_n + 23 \cdot 2^n = 2^{n+1} q_{n+1} + 1 \implies 2^n q_n + 1 + 23 = 2^{n+1} q_{n+1} + 1 \implies q_{n+1} = \frac{q_n + 1}{2} + 11 </cmath>
−
Therefore, if </math>q_n<math> is odd, </math>q_{n+1} = \frac{q_n + 1}{2} + 11<math>. In sum, our recursion is  
+
Therefore, if <math>q_n</math> is odd, <math>q_{n+1} = \frac{q_n + 1}{2} + 11</math>. In sum, our recursion is  
 
<cmath>  q_n = \begin{cases} \frac{q_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } q_{n-1} \\ \frac{q_{n-1}+1}{2} + 11 &\text{if } 2 \not\vert \text{ } q_{n-1} \end{cases}  </cmath>
 
<cmath>  q_n = \begin{cases} \frac{q_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } q_{n-1} \\ \frac{q_{n-1}+1}{2} + 11 &\text{if } 2 \not\vert \text{ } q_{n-1} \end{cases}  </cmath>
−
Finally, let us list out </math>q_n<math> to find a pattern. Because </math>a_1 = 23<math>, </math>q_1 = 11<math>. Through our recursion, we continue like so:
+
Finally, let us list out <math>q_n</math> to find a pattern. Because <math>a_1 = 23</math>, <math>q_1 = 11</math>. Through our recursion, we continue like so:
 
<cmath> q_1 = 11, q_2 = 17, q_2 = 20, q_3 = 10, q_4 = 5, q_6 = 14, q_7 = 7, q_8 = 15, q_9 = 19, q_10 = 21, q_11 = 22, q_12 = 11, \dots </cmath>
 
<cmath> q_1 = 11, q_2 = 17, q_2 = 20, q_3 = 10, q_4 = 5, q_6 = 14, q_7 = 7, q_8 = 15, q_9 = 19, q_10 = 21, q_11 = 22, q_12 = 11, \dots </cmath>
−
Therefore </math>q_n<math> repeats with cycle length </math>11<math>. Since </math>a_{n+1} = a_n<math> if and only iff </math>q_n<math> is even, in each cycle, we have 4 satisfactory values of </math>n<math>. There are </math>\frac{1000 - 10}{11} = 90<math> complete cycles. There are 3 extra values in the last incomplete cycle. Therefore we obtain </math>90 \cdot 4 + 3 = \fbox{363}<math>.
+
Therefore <math>q_n</math> repeats with cycle length <math>11</math>. Since <math>a_{n+1} = a_n</math> if and only iff <math>q_n</math> is even, in each cycle, we have 4 satisfactory values of <math>n</math>. There are <math>\frac{1000 - 10}{11} = 90</math> complete cycles. There are 3 extra values in the last incomplete cycle. Therefore we obtain <math>90 \cdot 4 + 3 = \fbox{363}</math>.
  
 
== Solution 4 (Binary Interpretation, Computer Scientists' Playground) ==
 
== Solution 4 (Binary Interpretation, Computer Scientists' Playground) ==
  
−
We first check that </math>\gcd(23, 2^n) = 1<math> hence we are always seeking a unique modular inverse of </math>23<math>, </math>b_n<math>, such that </math>a_n \equiv 23b_n \equiv 1 \mod{2^n}<math>.
+
We first check that <math>\gcd(23, 2^n) = 1</math> hence we are always seeking a unique modular inverse of <math>23</math>, <math>b_n</math>, such that <math>a_n \equiv 23b_n \equiv 1 \mod{2^n}</math>.
  
  
−
Now that we know that </math>b_n<math> is unique, we proceed to recast this problem in binary. This is convenient because </math>x \mod{2^n}<math> is simply the last </math>n<math>-bits of </math>x<math> in binary, and if </math>x \equiv 1 \mod{2^n}<math>, it means that of the last </math>n<math> bits of </math>x<math>, only the rightmost bit (henceforth </math>0<math>th bit) is </math>1<math>.
+
Now that we know that <math>b_n</math> is unique, we proceed to recast this problem in binary. This is convenient because <math>x \mod{2^n}</math> is simply the last <math>n</math>-bits of <math>x</math> in binary, and if <math>x \equiv 1 \mod{2^n}</math>, it means that of the last <math>n</math> bits of <math>x</math>, only the rightmost bit (henceforth <math>0</math>th bit) is <math>1</math>.
  
 
Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:
 
Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:
Line 120: Line 120:
 
</cmath>
 
</cmath>
  
−
Now note </math>23 = 10111_2<math>, and recall that our objective is to progressively zero out the </math>n<math> leftmost bits of </math>a_n = 10111_2 \times b_n<math> except for the </math>0<math>th bit.  
+
Now note <math>23 = 10111_2</math>, and recall that our objective is to progressively zero out the <math>n</math> leftmost bits of <math>a_n = 10111_2 \times b_n</math> except for the <math>0</math>th bit.  
  
−
Write </math>b_n = \underline{c_{n-1}\cdots c_2c_1c_0}_2<math>, we note that </math>c_0<math> uniquely defines the </math>0<math>th bit of </math>a_n<math>, and once we determine </math>c_0<math>, </math>c_1<math> uniquely determines the </math>1<math>st bit of </math>a_n<math>, so on and so forth.  
+
Write <math>b_n = \underline{c_{n-1}\cdots c_2c_1c_0}_2</math>, we note that <math>c_0</math> uniquely defines the <math>0</math>th bit of <math>a_n</math>, and once we determine <math>c_0</math>, <math>c_1</math> uniquely determines the <math>1</math>st bit of <math>a_n</math>, so on and so forth.  
  
−
For example, </math>c_0 = 1<math> satisfies </math>a_1 \equiv10111_2 \times 1_2 \equiv 1 \mod{10_2}<math>
+
For example, <math>c_0 = 1</math> satisfies <math>a_1 \equiv10111_2 \times 1_2 \equiv 1 \mod{10_2}</math>
−
Next, we note that the second bit of </math>a_1<math> is </math>1<math>, so we must also have </math>c_1 = 1<math> in order to zero it out, giving
+
Next, we note that the second bit of <math>a_1</math> is <math>1</math>, so we must also have <math>c_1 = 1</math> 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>
 
<cmath>a_2 \equiv 10111_2 \times 11_2 \equiv 101110_2 + a_1 \equiv 1000101_2 \equiv 01_2 \mod{100_2}</cmath>
  
−
</math>a_{n+1} = a_{n}<math> happens precisely when </math>c_n = 0<math>. In fact we can see this in action by working out </math>a_3<math>. Note that </math>a_2<math> has 1 on the </math>2<math>nd bit, so we must choose </math>c_2 = 1<math>. This gives  
+
<math>a_{n+1} = a_{n}</math> happens precisely when <math>c_n = 0</math>. In fact we can see this in action by working out <math>a_3</math>. Note that <math>a_2</math> has 1 on the <math>2</math>nd bit, so we must choose <math>c_2 = 1</math>. 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>
 
<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 </math>3<math>rd and </math>4<math>th bit are </math>0<math>, </math>c_3 = c_4 = 0<math>, and this gives </math>a_3 = a_4 = a_5<math>.
+
Note that since the <math>3</math>rd and <math>4</math>th bit are <math>0</math>, <math>c_3 = c_4 = 0</math>, and this gives <math>a_3 = a_4 = a_5</math>.
  
  
−
It may seem that this process will take forever, but note that </math>23 = 10111_2<math> has </math>4<math> bits behind the leading digit, and in the worst case, the leading digits of </math>a_n<math> will have a cycle length of at most </math>16<math>. In fact, we find that the cycle length is </math>11<math>, and in the process found that </math>a_3 = a_4 = a_5<math>, </math>a_6 = a_7<math>, and </math>a_{11} = a_{12}<math>.
+
It may seem that this process will take forever, but note that <math>23 = 10111_2</math> has <math>4</math> bits behind the leading digit, and in the worst case, the leading digits of <math>a_n</math> will have a cycle length of at most <math>16</math>. In fact, we find that the cycle length is <math>11</math>, and in the process found that <math>a_3 = a_4 = a_5</math>, <math>a_6 = a_7</math>, and <math>a_{11} = a_{12}</math>.
  
−
Since we have </math>90<math> complete cycles of length </math>11<math>, and the last partial cycle yields </math>a_{993} = a_{994} = a_{995}<math> and </math>a_{996} = a_{997}<math>, we have a total of </math>90 \times 4 + 3 = \boxed{363}<math> values of </math>n \le 1000<math> such that </math>a_n = a_{n+1}$
+
Since we have <math>90</math> complete cycles of length <math>11</math>, and the last partial cycle yields <math>a_{993} = a_{994} = a_{995}</math> and <math>a_{996} = a_{997}</math>, we have a total of <math>90 \times 4 + 3 = \boxed{363}</math> values of <math>n \le 1000</math> such that <math>a_n = a_{n+1}</math>
  
 
~ cocoa @ https://www.corgillogical.com
 
~ cocoa @ https://www.corgillogical.com

Revision as of 16:03, 29 December 2024

Problem

For each positive integer $n$ let $a_n$ be the least positive integer multiple of $23$ such that $a_n \equiv 1 \pmod{2^n}.$ Find the number of positive integers $n$ less than or equal to $1000$ that satisfy $a_n = a_{n+1}.$

Solution 1

Denote $a_n = 23 b_n$. Thus, for each $n$, we need to find smallest positive integer $k_n$, such that \[ 23 b_n = 2^n k_n + 1 . \]

Thus, we need to find smallest $k_n$, such that \[ 2^n k_n \equiv - 1 \pmod{23} . \]

Now, we find the smallest $m$, such that $2^m \equiv 1 \pmod{23}$. By Fermat's Theorem, we must have $m | \phi \left( 23 \right)$. That is, $m | 22$. We find $m = 11$.

Therefore, for each $n$, we need to find smallest $k_n$, such that \[ 2^{{\rm Rem} \left( n , 11 \right)} k_n \equiv - 1 \pmod{23} . \]

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, $n \in \left\{ 11 l , 11l + 1 , \cdots , 11l + 10 \right\}$, we have $n = 11l$, $11l + 3$, $11l + 4$, $11l + 6$, such that $b_n = b_{n+1}$. That is, $a_n = a_{n+1}$. At the boundary of two consecutive cycles, $b_{11L + 10} \neq b_{11 \left(l + 1 \right)}$.

We have $1000 = 90 \cdot 11 + 10$. Therefore, the number of feasible $n$ is $91 \cdot 4 - 1 = \boxed{\textbf{(363) }}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

Observe that if $a_{n-1} - 1$ is divisible by $2^n$, $a_n = a_{n-1}$. If not, $a_n = a_{n-1} + 23 \cdot 2^{n-1}$.

This encourages us to let $b_n = \frac{a_n - 1}{2^n}$. Rewriting the above equations, we have \[b_n = \begin{cases} \frac{b_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } b_{n-1} \\ \frac{b_{n-1}+23}{2} &\text{if } 2 \not\vert \text{ } b_{n-1} \end{cases}\] The first few values of $b_n$ are $11, 17, 20, 10, 5, 14, 7, 15, 19, 21,$ and $22$. We notice that $b_{12} = b_1 = 11$, and thus the sequence is periodic with period $11$.

Note that $a_n = a_{n+1}$ if and only if $b_n$ is even. This occurs when $n$ is congruent to $0, 3, 4$ or $6$ mod $11$, giving four solutions for each period.

From $1$ to $1001$ (which is $91 \times 11$), there are $91 \times 4 = 364$ values of $n$. We subtract $1$ from the total since $1001$ satisfies the criteria but is greater than $1000$ to get a final answer of $\fbox{363}$ . ~Bxiao31415

(small changes by bobjoebilly and IraeVid13)

Solution 3 (Similar to solution 2 but more explanation)

Let $a_n = 23b_n$. Note that if \[23 b_{n+1} \equiv 1 \pmod{2^{n+1}}\] Then \[23 b_{n+1} \equiv 1 \pmod{2^{n}}\] Also \[23 b_n \equiv 1 \pmod{2^n}\] Therefore \[b_n \equiv b_{n+1} \equiv 23^{-1} \pmod{2^n}\] Then \[b_{n+1} \equiv b_n, b_n + 2^n \pmod{2^{n+1}}\] So \[b_{n+1} = b_n, b_n + 2^n\] Since $0 \le b_n < 2^n$ and $0 \le b_{n+1} < 2^{n+1}$ as $a_n$ is the *least* positive integer multiple of 23.

Now suppose $b_{n+1} = b_n$. Define $q_n$ to be the quotient of $23b_n$ divided by $2^n$. Then \[23b_n = 2^n q_n + 1 \text{ and } 23b_{n+1} = 23b_n = 2^{n+1} q_{n+1} + 1 = 2^n q_n + 1 \implies q_{n+1} = \frac{q_n}{2}\]. Furthermore if quotient $q_n$ is even then \[23b_n = 2^n q_n +1 = 2^{n+1} \frac{q_n}{2} +1\] Therefore $b_{n+1} = b_n$ if and only if $q_n$ is even. And, if this is true, then $q_{n+1} = \frac{q_n}{2}$. Next, if $q_n$ is odd, we must have $b_{n+1} = b_n + 2^n$. Solving for $q_{n+1}$, we have \[23b_{n+1} = 2^{n+1} q_{n+1} + 1 \implies 23b_n + 23 \cdot 2^n = 2^{n+1} q_{n+1} + 1 \implies 2^n q_n + 1 + 23 = 2^{n+1} q_{n+1} + 1 \implies q_{n+1} = \frac{q_n + 1}{2} + 11\] Therefore, if $q_n$ is odd, $q_{n+1} = \frac{q_n + 1}{2} + 11$. In sum, our recursion is \[q_n = \begin{cases} \frac{q_{n-1}}{2} & \text{if } 2 \text{ } \vert \text{ } q_{n-1} \\ \frac{q_{n-1}+1}{2} + 11 &\text{if } 2 \not\vert \text{ } q_{n-1} \end{cases}\] Finally, let us list out $q_n$ to find a pattern. Because $a_1 = 23$, $q_1 = 11$. Through our recursion, we continue like so: \[q_1 = 11, q_2 = 17, q_2 = 20, q_3 = 10, q_4 = 5, q_6 = 14, q_7 = 7, q_8 = 15, q_9 = 19, q_10 = 21, q_11 = 22, q_12 = 11, \dots\] Therefore $q_n$ repeats with cycle length $11$. Since $a_{n+1} = a_n$ if and only iff $q_n$ is even, in each cycle, we have 4 satisfactory values of $n$. There are $\frac{1000 - 10}{11} = 90$ complete cycles. There are 3 extra values in the last incomplete cycle. Therefore we obtain $90 \cdot 4 + 3 = \fbox{363}$.

Solution 4 (Binary Interpretation, Computer Scientists' Playground)

We first check that $\gcd(23, 2^n) = 1$ hence we are always seeking a unique modular inverse of $23$, $b_n$, such that $a_n \equiv 23b_n \equiv 1 \mod{2^n}$.


Now that we know that $b_n$ is unique, we proceed to recast this problem in binary. This is convenient because $x \mod{2^n}$ is simply the last $n$-bits of $x$ in binary, and if $x \equiv 1 \mod{2^n}$, it means that of the last $n$ bits of $x$, only the rightmost bit (henceforth $0$th bit) is $1$.

Also, multiplication in binary can be thought of as adding shifted copies of the multiplicand. For example:

\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}

Now note $23 = 10111_2$, and recall that our objective is to progressively zero out the $n$ leftmost bits of $a_n = 10111_2 \times b_n$ except for the $0$th bit.

Write $b_n = \underline{c_{n-1}\cdots c_2c_1c_0}_2$, we note that $c_0$ uniquely defines the $0$th bit of $a_n$, and once we determine $c_0$, $c_1$ uniquely determines the $1$st bit of $a_n$, so on and so forth.

For example, $c_0 = 1$ satisfies $a_1 \equiv10111_2 \times 1_2 \equiv 1 \mod{10_2}$ Next, we note that the second bit of $a_1$ is $1$, so we must also have $c_1 = 1$ in order to zero it out, giving

\[a_2 \equiv 10111_2 \times 11_2 \equiv 101110_2 + a_1 \equiv 1000101_2 \equiv 01_2 \mod{100_2}\]

$a_{n+1} = a_{n}$ happens precisely when $c_n = 0$. In fact we can see this in action by working out $a_3$. Note that $a_2$ has 1 on the $2$nd bit, so we must choose $c_2 = 1$. This gives

\[a_3 \equiv 10111_2 \times 111_2 \equiv 1011100_2 + a_2 \equiv 10100001_2 \equiv 001_2 \mod{1000_2}\]

Note that since the $3$rd and $4$th bit are $0$, $c_3 = c_4 = 0$, and this gives $a_3 = a_4 = a_5$.


It may seem that this process will take forever, but note that $23 = 10111_2$ has $4$ bits behind the leading digit, and in the worst case, the leading digits of $a_n$ will have a cycle length of at most $16$. In fact, we find that the cycle length is $11$, and in the process found that $a_3 = a_4 = a_5$, $a_6 = a_7$, and $a_{11} = a_{12}$.

Since we have $90$ complete cycles of length $11$, and the last partial cycle yields $a_{993} = a_{994} = a_{995}$ and $a_{996} = a_{997}$, we have a total of $90 \times 4 + 3 = \boxed{363}$ values of $n \le 1000$ such that $a_n = a_{n+1}$

~ cocoa @ https://www.corgillogical.com

Video Solution

https://youtu.be/ujP-V170vvI

~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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_15&oldid=238301"