2023 AMC 12A Problems/Problem 7
- The following problem is from both the 2023 AMC 10A #9 and 2023 AMC 12A #7, so both problems redirect to this page.
Contents
Problem
A digital display shows the current date as an -digit integer consisting of a -digit year, followed by a -digit month, followed by a -digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in will each digit appear an even number of times in the 8-digital display for that date?
Solution 1 (Casework)
Do careful casework by each month. In the month and the date, we need a , a , and two digits repeated (which has to be and after consideration). After the casework, we get . For curious readers, the numbers (in chronological order) are:
20230113 20230131 20230223 20230311 20230322 20231013 20231031 20231103 20231130
Solution 2
There is one , so we need one more (three more means that either the month or units digit of the day is ). For the same reason, we need one more .
If is the units digit of the month, then the can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match (). For the second (tens digit of the day), we must have the other two be , as a month can't start with or . There are successes this way.
If is the tens digit of the day, then can be either the tens digit of the month or the units digit of the day. For the first case, must go in the other slots. For the second, the other two slots must be as well. There are successes here.
If is the units digit of the day, then could go in any of the remaining slots again. If it's the tens digit of the day, then the other digits must be . If is the units digit of the day, then the other two slots must both be . If is the tens digit of the month, then the other two slots can be either both or both . In total, there are successes here.
Summing through all cases, there are dates.
-Benedict T (countmath1)
Solution 3
We start with we need an extra and an extra . So we have at least one of those extras in the days, except we can have the month . We now have possible months . For month we have two cases, we now have to add in another 1, and the possible days are . For month we need an extra so we can have the day note that we can't use because it is to large. Now for month we can have any number and multiply it by so we have the solution . For October we need a and a so we have as our choices. For November we have two choices which are .Now for December we have options. Summing we get solutions.
~kyogrexu
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/GP-DYudh5qU?si=iDDNwV-Ut5UZSoQu&t=2010
~Math-X
Video Solution (easy to digest) by Power Solve
https://www.youtube.com/watch?v=4TPsTOHKQTw
Video Solution 1 by OmegaLearn
Video Solution by CosineMethod [🔥Fast and Easy🔥]
https://www.youtube.com/watch?v=a5w_1lN3H4s
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.