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Difference between revisions of "2024 AIME II Problems/Problem 14"

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Therefore, <math>b = b' + 1 = \boxed{\textbf{(211) }}</math>.
 
Therefore, <math>b = b' + 1 = \boxed{\textbf{(211) }}</math>.
  
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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~Shen Kislay Kai and Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 12:20, 3 September 2024

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


See also

2024 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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