2021 AIME II Problems/Problem 1

Revision as of 17:18, 22 March 2021 by Doriding (talk | contribs) (Solution 4)

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 $\overline{ABA}$, note that $\overline{ABA}$, 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)

For any three-digit palindrome $\overline{ABA},$ where $A$ and $B$ are digits and $A\neq0,$ note that $\overline{(10-A)(9-B)(10-A)}$ must be another palindrome by symmetry. This means we can pair each three-digit palindrome uniquely with another three-digit palindrome so that they sum to \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] \\ &=\left[100A+10B+A\right]+\left[1000-100A+90-10B+10-A\right] \\ &=1000+90+10 \\ &=1100. \end{align*} For instances: \begin{align*} 272+828&=1100, \\ 414+686&=1100, \\ 595+505&=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

The number takes the form $\overline{ABA}$, where $A \in \left\{ 1 , 2 , \cdots , 9 \right\}$ and $B \in \left\{ 0 , 1 , \cdots , 9 \right\}$.

Therefore, the sum of all such numbers is equal to

\begin{eqnarray*} \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}^9 101 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{eqnarray*}

~ Steven Chen

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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