Difference between revisions of "2023 AMC 12A Problems/Problem 7"
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==Problem== | ==Problem== | ||
+ | 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? | ||
− | + | <math>\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9</math> | |
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− | ==Solution | + | ==Solution== |
+ | Do careful casework by each month. In the month and the date, we need a <math>0</math>, a <math>3</math>, and two digits repeated (which have to be <math>1</math> and <math>2</math> after consideration). After the case work, we get <math>9</math>, meaning the answer <math>\boxed{\textbf{(E)}~9}</math>. | ||
+ | For those who are wondering, the numbers are: | ||
+ | <math>20230113</math>, <math>20230131</math>, <math>20230223</math>, <math>20230311</math>, <math>20230322</math>, <math>20231013</math>, <math>20231031</math>, <math>20231103</math>, <math>20231130</math>. | ||
− | + | == Video Solution 1 by OmegaLearn == | |
− | + | https://youtu.be/xguAy0PV7EA | |
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==See Also== | ==See Also== | ||
− | {{ | + | {{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}} |
− | {{ | + | {{AMC12 box|year=2023|ab=A|num-b=6|num-a=8}} |
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:24, 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 (which have to be
and
after consideration). After the case work, we get
, meaning the answer
.
For those who are wondering, the numbers are:
,
,
,
,
,
,
,
,
.
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 |
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.