Difference between revisions of "2023 AMC 10A Problems/Problem 9"
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A digital display shows the current date as an <math>8</math>-digit integer consisting of a <math>4</math>-digit year, followed by a <math>2</math>-digit month, followed by a <math>2</math>-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in <math>2023</math> will each digit appear an even number of times in the 8-digital display for that date? | A digital display shows the current date as an <math>8</math>-digit integer consisting of a <math>4</math>-digit year, followed by a <math>2</math>-digit month, followed by a <math>2</math>-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in <math>2023</math> will each digit appear an even number of times in the 8-digital display for that date? | ||
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== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/xguAy0PV7EA | https://youtu.be/xguAy0PV7EA | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 20:46, 9 November 2023
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
Do careful casework by each month. In the month and the date, we need a 
, a 
, and two digits repeated. After the case work, we get 
, meaning the answer 
.
For those who are wondering, the numbers are:
20230113, 20230131, 20230223, 20230311, 20230322, 20231013, 20231031, 20231103, 20231130.
Video Solution 1 by OmegaLearn
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
