2023 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2023 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

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Concepts:digitsparitycasework

Difficulty rating: 1380

7.

A digital display shows the current date as an 88-digit integer consisting of a 44-digit year, followed by a 22-digit month, followed by a 22-digit date within the month. For example, Arbor Day this year is displayed as 20230428.20230428. For how many dates in 20232023 will each digit appear an even number of times in the 88-digit display for that date?

55

66

77

88

99

Solution:

The year contributes the digits 2,0,2,3,2,0,2,3, so 22 appears twice while 00 and 33 each appear once. For every digit to end up with an even count, the four digits of the month and day must supply an odd number of 00's, an odd number of 33's, and an even number of every other digit.

With only four digits available, the month-day string must use exactly one 0,0, one 3,3, and a repeated pair of some digit. Checking valid months and days leaves nine dates: 01-13,01\text{-}13, 01-31,01\text{-}31, 02-23,02\text{-}23, 03-11,03\text{-}11, 03-22,03\text{-}22, 10-13,10\text{-}13, 10-31,10\text{-}31, 11-03,11\text{-}03, and 11-30.11\text{-}30.

Thus, the correct answer is E.

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