2023 AIME II Problems/Problem 15

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 (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

~MathProblemSolvingSkills.com