2024 AIME II Problems/Problem 14

Problem

Let $b\ge 2$ be an integer. Call a positive integer $n$ $b\text-\textit{eautiful}$ if it has exactly two digits when expressed in base $b$ and these two digits sum to $\sqrt n$. For example, $81$ is $13\text-\textit{eautiful}$ because $81 = \underline{6} \ \underline{3}_{13} $ and $6 + 3 = \sqrt{81}$. Find the least integer $b\ge 2$ for which there are more than ten $b\text-\textit{eautiful}$ integers.

Solution

We write the base- $b$ two-digit integer as $\left( xy \right)_b$ . Thus, this number satisfies $\left( x + y \right)^2 = b x + y$ with $x \in \left\{ 1, 2, \cdots , b-1 \right\}$ and $y \in \left\{ 0, 1, \cdots , b - 1 \right\}$ .

The above conditions imply $\left( x + y \right)^2 < b^2$ . Thus, $x + y \leq b - 1$ .

The above equation can be reorganized as $\left( x + y \right) \left( x + y - 1 \right) = \left( b - 1 \right) x .$

Denote $z = x + y$ and $b' = b - 1$ . Thus, we have $z \left( z - 1 \right) = b' x , \hspace{1cm} (1)$ where $z \in \left\{ 2, 3, \cdots , b' \right\}$ and $x \in \left\{ 1, 2, \cdots , b' \right\}$ .

Next, for each $b'$ , we solve Equation (1).

We write $b'$ in the prime factorization form as $b' = \Pi_{i=1}^n p_i^{k_i}$ . Let $\left(A, \bar A \right)$ be any ordered partition of $\left\{ 1, 2, \cdots , n \right\}$ (we allow one set to be empty). Denote $P_A = \Pi_{i \in A} p_i^{k_i}$ and $P_{\bar A} = \Pi_{i \in \bar A} p_i^{k_i}$ .

Because ${\rm gcd} \left( z, z-1 \right) = 1$ , there must exist such an ordered partition, such that $P_A | z$ and $P_{\bar A} | z-1$ .

Next, we prove that for each ordered partition $\left( A, \bar A \right)$ , if a solution of $z$ exists, then it must be unique.

Suppose there are two solutions of $z$ under partition $\left( A, \bar A \right)$ : $z_1 = c_1 P_A$ , $z_1 - 1 = d_1 P_{\bar A}$ , and $z_2 = c_2 P_A$ , $z_2 - 1 = d_2 P_{\bar A}$ . W.L.O.G., assume $c_1 < c_2$ . Hence, we have $\left( c_2 - c_1 \right) P_A = \left( d_2 - d_1 \right) P_{\bar A} .$

Because ${\rm gcd} \left( P_A, P_{\bar A} \right) = 1$ and $c_1 < c_2$ , there exists a positive integer $m$ , such that $c_2 = c_1 + m P_{\bar A}$ and $d_2 = d_1 + m P_A$ . Thus, \begin{align*} z_2 & = z_1 + m P_A P_{\bar A} \\ & = z_1 + m b' \\ & > b' . \end{align*}

However, recall $z_2 \leq b'$ . We get a contradiction. Therefore, under each ordered partition for $b'$ , the solution of $z$ is unique.

Note that if $b'$ has $n$ distinct prime factors, the number of ordered partitions is $2^n$ . Therefore, to find a $b'$ such that the number of solutions of $z$ is more than 10, the smallest $n$ is 4.

With $n = 4$ , the smallest number is $2 \cdot 3 \cdot 5 \cdot 7 = 210$ . Now, we set $b' = 210$ and check whether the number of solutions of $z$ under this $b'$ is more than 10.

We can easily see that all ordered partitions (except $A = \emptyset$ ) guarantee feasible solutions of $z$ . Therefore, we have found a valid $b'$ . Therefore, $b = b' + 1 = \boxed{\textbf{(211) }}$ .

~Shen Kislay Kai and Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution

https://youtu.be/N7rLL1Xt9go

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

Video Solution

https://youtu.be/0FCGY9xfEq0?si=9Fu4owVaSm-WWxFJ

~MathProblemSolvingSkills.com

2024 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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2024 AIME II Problems/Problem 14

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