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 $1000$ that is a palindrome both when written in base ten and when written in base eight, such as $292 = 444_{\text{eight}}.$

Solution

Assuming that such palindrome is greater than $777_8 = 511,$ we conclude that the palindrome has four digits when written in base $8.$ Let such palindrome be $(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.$

It is clear that $A=1,$ so we repeatedly add $72$ to $513$ until we get palindromes less than $1000:$ $\begin{align*} 513+72\cdot0 &= 513, \\ 513+72\cdot1 &= \boxed{585}, \\ 513+72\cdot2 &= 657, \\ 513+72\cdot3 &= 729, \\ 513+72\cdot4 &= 801, \\ 513+72\cdot5 &= 873, \\ 513+72\cdot6 &= 945, \\ 513+72\cdot7 &= 1017. \\ \end{align*}$

~MRENTHUSIASM

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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2023 AIME II Problems/Problem 2

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