2023 AIME II Problems/Problem 2
Problem
Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than 
that is a palindrome both when written in base ten and when written in base eight, such as 
Solution
Assuming that such palindrome is greater than 
we conclude that the palindrome has four digits when written in base eight. Let such palindrome be ![\[(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.\]](http://latex.artofproblemsolving.com/8/9/0/89097713b51c8d9e9063d1dca16394cd50efe1ef.png)
It is clear that 
so we repeatedly add 
to 
until we get palindromes:
\begin{align*}
513+72\cdot0 &= 513, \\
513+72\cdot1 &= \boxed{585}, \\
513+72\cdot2 &= 657, \\
513+72\cdot3 &= 729, \\
513+72\cdot4 &= 801, \\
513+72\cdot5 &= 873, \\
513+72\cdot6 &= 945, \\
513+72\cdot7 &= 1017. \\
\end{align*}
See also
| 2023 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
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