Difference between revisions of "2023 AIME II Problems/Problem 2"
MRENTHUSIASM (talk | contribs) |
(→Video Solution 1 by SpreadTheMathLove) |
||
(8 intermediate revisions by 4 users not shown) | |||
Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
+ | Assuming that such palindrome is greater than <math>777_8 = 511,</math> we conclude that the palindrome has four digits when written in base <math>8.</math> Let such palindrome be <cmath>(\underline{ABBA})_8 = 512A + 64B + 8B + A = 513A + 72B.</cmath> | ||
+ | It is clear that <math>A=1,</math> so we repeatedly add <math>72</math> to <math>513</math> until we get palindromes less than <math>1000:</math> | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | ~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 | ||
== 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}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
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
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.