2021 AIME II Problem 1

Attempt Problem 1 of the 2021 AIME II below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AIME II solutions, or check the answer key.

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1.

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 777777 or 383.383.)

Answer: 550
Concepts:palindromeplace valuemean
Difficulty rating: 1750
Solution:

A three-digit palindrome has the form aba=101a+10b\overline{aba} = 101a + 10b with a{1,,9}a \in \{1, \ldots, 9\} and b{0,,9},b \in \{0, \ldots, 9\}, and every such pair of digits occurs exactly once, so the two digits vary independently over the 9090 palindromes.

By linearity, the mean is 101101 times the average of aa plus 1010 times the average of b,b, namely 1015+1092=505+45=550. \begin{aligned} &101 \cdot 5 \\ &\quad {}+ 10 \cdot \frac{9}{2} = 505 + 45 = 550. \end{aligned}

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