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AIME 2023 II · 第 2 题

AIME 2023 II — Problem 2

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than 10001000 that is a palindrome both when written in base ten and when written in base eight, such as 292=444eight.292 = 444_{\text{eight}}.

解析

Solution

Assuming that such palindrome is greater than 7778=511,777_8 = 511, we conclude that the palindrome has four digits when written in base 8.8. Let such palindrome be

(ABBA‾)8=512A+64B+8B+A=513A+72B.(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B. It is clear that A=1,A=1, so we repeatedly add 7272 to 513513 until we get palindromes less than 1000:1000:

513+72⋅0=513,513+72⋅1=585,513+72⋅2=657,513+72⋅3=729,513+72⋅4=801,513+72⋅5=873,513+72⋅6=945,513+72⋅7=1017.\begin{aligned} 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{aligned} ~MRENTHUSIASM

Video Solution

https://youtu.be/BxXP6s1za0g

~MathProblemSolvingSkills.com

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=_JTFiqczLvk

Video Solution by the Power of Logic(#1 and #2)

https://youtu.be/VcEulZ3nvSI