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

Revision as of 21:45, 18 July 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 that 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 $\frac{101+999}{2}=\boxed{550},$ which is the final answer.

~ math31415926535

Solution 2

For any palindrome $\underline{ABA},$ note that $\underline{ABA}$ is $100A + 10B + A = 101A + 10B.$ 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 every three-digit palindrome $\underline{ABA}$ with $A\in\{1,2,3,4,5,6,7,8,9\}$ and $B\in\{0,1,2,3,4,5,6,7,8,9\},$ 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: $\begin{align*} 171+929&=1100, \\ 262+838&=1100, \\ 303+797&=1100, \\ 414+686&=1100, \\ 545+555&=1100, \end{align*}$ and so on.

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

Remark

By the Multiplication Principle, there are $9\cdot10=90$ three-digit palindromes in total. Their sum is $1100\cdot45=49500,$ as we can match them into $45$ pairs such that each pair sums to $1100.$

~MRENTHUSIASM

Solution 4 (Similar to Solution 2: Very, Very Easy and Quick)

We notice that a three-digit palindrome looks like this: $\underline{aba}.$

And we know $a$ can be any digit from $1$ through $9,$ and $b$ can be any digit from $0$ through $9,$ so there are $9\times{10}=90$ three-digit palindromes.

We want to find the sum of these $90$ palindromes and divide it by $90$ to find the arithmetic mean.

How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: $101a+10b.$

Thus, all of these $90$ palindromes can be broken into this form.

Thus, the sum of these $90$ palindromes will be $101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},$ because each $a$ will be in $10$ different palindromes (since for each $a,$ there are $10$ choices for $b$ ). The same logic explains why we multiply by $9$ when computing the total sum of $b.$

We get a sum of $45\times{1100},$ but don't compute this! Divide this by $90$ and you will get $\boxed{550}.$

~ $\alpha b \alpha$

Solution 5 (Extremely Fast Solution)

The average values of the first and last digits are each $5,$ and the average value of the middle digit is $4.5,$ so the average of all three-digit palindromes is $5\cdot 10^2+4.5\cdot 10+5=\boxed{550}.$

~MathIsFun286

Remark

Visit the Discussion Page for questions and further generalizations.

~MRENTHUSIASM

Video Solution

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

@@ Line 39: / Line 39: @@
 ~MRENTHUSIASM
-==Solution 4 (Very, Very Easy and Quick)==
+==Solution 4 (Similar to Solution 2: Very, Very Easy and Quick)==
-We notice that a three-digit palindrome looks like this <math>\overline{aba}</math>
+We notice that a three-digit palindrome looks like this: <math>\underline{aba}.</math>
-And we know a can be any number from 1-9, and b can be any number from 0-9, so there are <math>9\times{10}=90</math> three-digit palindromes
+And we know <math>a</math> can be any digit from <math>1</math> through <math>9,</math> and <math>b</math> can be any digit from <math>0</math> through <math>9,</math> so there are <math>9\times{10}=90</math> three-digit palindromes.
-We want to find the sum of these 90 palindromes and divide it by 90 to find the arithmetic mean
+We want to find the sum of these <math>90</math> palindromes and divide it by <math>90</math> to find the arithmetic mean.
-How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: 101a+10b
+How can we do that? Instead of adding the numbers up, we can break each palindrome into two parts: <math>101a+10b.</math>
-Thus, all of these 90 palindromes can be broken into this form
+Thus, all of these <math>90</math> palindromes can be broken into this form.
-Thus, the sum of these 90 palindromes will be <math>101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9}</math>, because each a will be in 10 different palindromes (since for each a, there are 10 choices for b). The same logic explains why there is a times 9 when computing the total sum of b.
+Thus, the sum of these <math>90</math> palindromes will be <math>101\times{(1+2+...+9)}\times{10}+10\times{(0+1+2+...+9)}\times{9},</math> because each <math>a</math> will be in <math>10</math> different palindromes (since for each <math>a,</math> there are <math>10</math> choices for <math>b</math>). The same logic explains why we multiply by <math>9</math> when computing the total sum of <math>b.</math>
-We get a sum of <math>45\times{1100}</math>
+We get a sum of <math>45\times{1100},</math> but don't compute this! Divide this by <math>90</math> and you will get <math>\boxed{550}.</math>
-But don't compute this! There's no need. Divide this by 90 and you will get <math>\boxed{550}</math>
 ~<math>\alpha b \alpha</math>

2021 AIME II (Problems • Answer Key • Resources)
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