Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AIME II Problems/Problem 1"

m (→‎Solution 1)
m (We use underline to denote successive digits, as overline is "overloading".)
Line 8: Line 8:
  
 
==Solution 2==
 
==Solution 2==
−
For any palindrome <math>\overline{ABA}</math>, note that <math>\overline{ABA}</math>,  is 100A + 10B + A which is also 101A + 10B.  
+
For any palindrome <math>\underline{ABA}</math>, note that <math>\underline{ABA}</math>,  is 100A + 10B + A which is also 101A + 10B.  
 
The average for A is 5 since A can be any of 1, 2, 3, 4, 5, 6, 7, 8, or 9. The average for B is 4.5 since B is either 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, the answer is 505 + 45 = <math>\boxed{550}</math>.
 
The average for A is 5 since A can be any of 1, 2, 3, 4, 5, 6, 7, 8, or 9. The average for B is 4.5 since B is either 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, the answer is 505 + 45 = <math>\boxed{550}</math>.
  
Line 14: Line 14:
  
 
==Solution 3 (Symmetry and Generalization)==
 
==Solution 3 (Symmetry and Generalization)==
−
For any three-digit palindrome <math>\overline{ABA},</math> where <math>A</math> and <math>B</math> are digits with <math>A\neq0,</math> note that <math>\overline{(10-A)(9-B)(10-A)}</math> must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to  
+
For any three-digit palindrome <math>\underline{ABA},</math> where <math>A</math> and <math>B</math> are digits with <math>A\neq0,</math> note that <math>\underline{(10-A)(9-B)(10-A)}</math> must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\overline{ABA}+\overline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\
+
\underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\
 
&=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\
 
&=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\
 
&=1000+90+10 \\
 
&=1000+90+10 \\
Line 37: Line 37:
 
==Solution 4==
 
==Solution 4==
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
−
\sum_{A = 1}^9 \sum_{B = 0}^9 \overline{ABA} &= \sum_{A = 1}^9 \sum_{B = 0}^9 \left( 101 A + 10 B \right) \\
+
\sum_{A = 1}^9 \sum_{B = 0}^9 \underline{ABA} &= \sum_{A = 1}^9 \sum_{B = 0}^9 \left( 101 A + 10 B \right) \\
 
&= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\
 
&= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\
 
&= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\
 
&= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\

Revision as of 20:59, 25 March 2021

Problem

Find the arithmetic mean of all the three-digit palindromes. (Recall that a palindrome is a number that reads the same forward and backward, such as $777$ or $383$.)

Solution 1

Recall* the the arithmetic mean of all the $n$ digit palindromes is just the average of the largest and smallest $n$ digit palindromes, and in this case the $2$ palindromes are $101$ and $999$ and $\frac{101+999}{2}=550$ and $\boxed{550}$ is the final answer.

~ math31415926535

Solution 2

For any palindrome $\underline{ABA}$, note that $\underline{ABA}$, is 100A + 10B + A which is also 101A + 10B. The average for A is 5 since A can be any of 1, 2, 3, 4, 5, 6, 7, 8, or 9. The average for B is 4.5 since B is either 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Therefore, the answer is 505 + 45 = $\boxed{550}$.

- ARCTICTURN

Solution 3 (Symmetry and Generalization)

For any three-digit palindrome $\underline{ABA},$ where $A$ and $B$ are digits with $A\neq0,$ note that $\underline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. Therefore, we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \begin{align*} \underline{ABA}+\underline{(10-A)(9-B)(10-A)}&=\left[100A+10B+A\right]+\left[100(10-A)+10(9-B)+(10-A)\right] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*} For instances: \begin{align*} 101+999&=1100, \\ 262+838&=1100, \\ 373+727&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*} and so on.

From this symmetry, the arithmetic mean of all the three-digit palindromes is $\frac{1110}{2}=\boxed{550}.$

~MRENTHUSIASM

Solution 4

\begin{align*} \sum_{A = 1}^9 \sum_{B = 0}^9 \underline{ABA} &= \sum_{A = 1}^9 \sum_{B = 0}^9 \left( 101 A + 10 B \right) \\ &= \sum_{A = 1}^9 \sum_{B = 0}^9 101 A + \sum_{A = 1}^9 \sum_{B = 0}^9 10 B \\ &= 101 \cdot 10 \sum_{A = 1}^9 A + 10 \cdot 9 \sum_{B = 0}^9 B \\ &= 1010 \cdot 45 + 90 \cdot 45 \\ &= \end{align*}

- A bit too complicated of a solution - somebody please fix. - ARCTICTURN

Doriding is the original author. I will wait for him to come back. ~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=jDP2PErthkg

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_1&oldid=150326"