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

Line 5: Line 5:
 
== Solution ==
 
== Solution ==
  
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>
+
We have two cases:
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>
+
<ol style="margin-left: 1.5em;">
 +
  <li>If the palindrome written in base eight has three digits, then it is at most <math>777_8 = 511.</math></li><p>
 +
  <li>If the palindrome written in base eight has four digits, then it is at least <math>1001_8 = 513.</math></li><p>
 +
</ol>
 +
To maximize the palindrome,
 +
 
 +
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2023|num-b=1|num-a=3|n=II}}
 
{{AIME box|year=2023|num-b=1|num-a=3|n=II}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:17, 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

We have two cases:

  1. If the palindrome written in base eight has three digits, then it is at most $777_8 = 511.$
  2. If the palindrome written in base eight has four digits, then it is at least $1001_8 = 513.$

To maximize the palindrome,


See also

2023 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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. AMC logo.png