2024 AIME II Problems/Problem 14
Contents
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- two-digit integer as . Thus, this number satisfies with and .
The above conditions imply . Thus, .
The above equation can be reorganized as
Denote and . Thus, we have where and .
Next, for each , we solve Equation (1).
We write in the prime factorization form as . Let be any ordered partition of (we allow one set to be empty). Denote and .
Because , there must exist such an ordered partition, such that and .
Next, we prove that for each ordered partition , if a solution of exists, then it must be unique.
Suppose there are two solutions of under partition : , , and , . W.L.O.G., assume . Hence, we have
Because and , there exists a positive integer , such that and . Thus, \begin{align*} z_2 & = z_1 + m P_A P_{\bar A} \\ & = z_1 + m b' \\ & > b' . \end{align*}
However, recall . We get a contradiction. Therefore, under each ordered partition for , the solution of is unique.
Note that if has distinct prime factors, the number of ordered partitions is . Therefore, to find a such that the number of solutions of is more than 10, the smallest is 4.
With , the smallest number is . Now, we set and check whether the number of solutions of under this is more than 10.
We can easily see that all ordered partitions (except ) guarantee feasible solutions of . Therefore, we have found a valid . Therefore, .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.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 | ||
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