Difference between revisions of "2023 AMC 12A Problems/Problem 7"
| Line 35: | Line 35: | ||
We start with <math>2023----</math> we need an extra <math>0</math> and an extra <math>3</math>. So we have at least one of those extras in the days, except we can have the month <math>03</math>. We now have <math>6</math> possible months <math>01,02,03,10,11,12</math>. For month <math>1</math> we have two cases, we now have to add in another 1, and the possible days are <math>13,31</math>. For month <math>2</math> we need an extra <math>2</math> so we can have the day <math>23</math> note that we can't use <math>32</math> because it is to large. Now for month <math>3</math> we can have any number and multiply it by <math>11</math> so we have the solution <math>11,22</math>. For October we need a <math>1</math> and a <math>3</math> so we have <math>13,31</math> as our choices. For November we have two choices which are <math>03,30</math>.Now for December we have <math>0</math> options. Summing <math>2+1+2+2+2</math> we get <math>\boxed{\textbf{(E)}~9}</math> solutions. | We start with <math>2023----</math> we need an extra <math>0</math> and an extra <math>3</math>. So we have at least one of those extras in the days, except we can have the month <math>03</math>. We now have <math>6</math> possible months <math>01,02,03,10,11,12</math>. For month <math>1</math> we have two cases, we now have to add in another 1, and the possible days are <math>13,31</math>. For month <math>2</math> we need an extra <math>2</math> so we can have the day <math>23</math> note that we can't use <math>32</math> because it is to large. Now for month <math>3</math> we can have any number and multiply it by <math>11</math> so we have the solution <math>11,22</math>. For October we need a <math>1</math> and a <math>3</math> so we have <math>13,31</math> as our choices. For November we have two choices which are <math>03,30</math>.Now for December we have <math>0</math> options. Summing <math>2+1+2+2+2</math> we get <math>\boxed{\textbf{(E)}~9}</math> solutions. | ||
| + | ~kyogrexu | ||
== Video Solution 1 by OmegaLearn == | == Video Solution 1 by OmegaLearn == | ||
https://youtu.be/xguAy0PV7EA | https://youtu.be/xguAy0PV7EA | ||
Revision as of 15:58, 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
[hide]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)
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 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. 
