Difference between revisions of "2023 AMC 12A Problems/Problem 7"
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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 has to be <math>1</math> and <math>2</math> after consideration). After the casework, we get <math>\boxed{\textbf{(E)}~9}</math>. | 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 has to be <math>1</math> and <math>2</math> after consideration). After the casework, we get <math>\boxed{\textbf{(E)}~9}</math>. | ||
For curious readers, the numbers (in chronological order) are: | For curious readers, the numbers (in chronological order) are: |
Revision as of 12:36, 11 November 2023
- 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:
,
,
,
,
,
,
,
,
.
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)
Video Solution 1 by OmegaLearn
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/cMgngeSmFCY?si=2iAoiLoyeAVrVoDf&t=1930 ~Math-X
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.