Difference between revisions of "2023 AIME II Problems/Problem 2"

(Created page with "Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than <math>1000</math> that is a palindrome both when written in...")
 
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Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than <math>1000</math> that is a palindrome both when written in base ten and when written in base eight, such as <math>292 = 444_{eight}</math>.
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== Problem ==
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Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than <math>1000</math> that is a palindrome both when written in base ten and when written in base eight, such as <math>292 = 444_{\text{eight}}.</math>
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== Solution ==
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If the palindrome written in base eight has three digits, then we have <cmath>(\underline{ABA})_8 = 64A+8B+A = 65A+8B \leq 511.</cmath>
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If the palindrome written in base eight has four digits, then we have <cmath>(\underline{ABBA})_8 = 512A+64B+8B+A = 513A+72B \geq 513.</cmath>
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== See also ==
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{{AIME box|year=2023|num-b=1|num-a=3|n=II}}
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{{MAA Notice}}

Revision as of 15:10, 16 February 2023

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

If the palindrome written in base eight has three digits, then we have \[(\underline{ABA})_8 = 64A+8B+A = 65A+8B \leq 511.\] If the palindrome written in base eight has four digits, then we have \[(\underline{ABBA})_8 = 512A+64B+8B+A = 513A+72B \geq 513.\]

See also

2023 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AIME Problems and Solutions

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