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>. | ||
− | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | + | ~Shen Kislay Kai and Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) |
==Video Solution== | ==Video Solution== | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/0FCGY9xfEq0?si=9Fu4owVaSm-WWxFJ | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
==See also== | ==See also== | ||
{{AIME box|year=2024|num-b=13|num-a=15|n=II}} | {{AIME box|year=2024|num-b=13|num-a=15|n=II}} | ||
− | [[Category:Intermediate | + | [[Category:Intermediate Number Theory Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:20, 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- 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,
.
~Shen Kislay Kai and Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.