Difference between revisions of "2023 AIME II Problems/Problem 2"
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==Video Solution 1 by SpreadTheMathLove== | ==Video Solution 1 by SpreadTheMathLove== | ||
https://www.youtube.com/watch?v=_JTFiqczLvk | https://www.youtube.com/watch?v=_JTFiqczLvk | ||
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| + | ==Video Solution by the Power of Logic(#1 and #2)== | ||
| + | https://youtu.be/VcEulZ3nvSI | ||
== See also == | == See also == | ||
Latest revision as of 19:32, 8 June 2023
Contents
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 
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 less than 
~MRENTHUSIASM
Video Solution
~MathProblemSolvingSkills.com
Video Solution 1 by SpreadTheMathLove
https://www.youtube.com/watch?v=_JTFiqczLvk
Video Solution by the Power of Logic(#1 and #2)
See also
| 2023 AIME II (Problems • Answer Key • Resources) | ||
| 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. 
