Difference between revisions of "2014 AMC 8 Problems/Problem 23"
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Ashley: I just realized that our uniform numbers are all <math>2</math>-digit primes. | Ashley: I just realized that our uniform numbers are all <math>2</math>-digit primes. | ||
| − | + | Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month. | |
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month. | Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month. | ||
| Line 18: | Line 18: | ||
==Solution 1== | ==Solution 1== | ||
| − | The maximum amount of days any given month can have is <math>31</math>, and the smallest two-digit primes are <math>11, 13,</math> and <math>17</math>. There are a few different sums that can be deduced from the following numbers, which are <math>24, 30,</math> and <math>28</math>, all of which represent the three days. Therefore, since | + | The maximum amount of days any given month can have is <math>31</math>, and the smallest two-digit primes are <math>11, 13,</math> and <math>17</math>. There are a few different sums that can be deduced from the following numbers, which are <math>24, 30,</math> and <math>28</math>, all of which represent the three days. Therefore, since Bethany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to <math>24</math>. Similarly, Caitlin says that the other two people's uniform numbers are later, so the sum must add up to <math>30</math>. This leaves <math>28</math> as today's date. From this, Caitlin was referring to the uniform wearers <math>13</math> and <math>17</math>, telling us that her number is <math>11</math>, giving our solution as <math>\boxed{(A) 11}</math>. |
| + | |||
| + | == Video Solution by OmegaLearn == | ||
| + | https://youtu.be/6xNkyDgIhEE?t=735 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
==Video Solution== | ==Video Solution== | ||
| − | |||
| − | https:// | + | |
| + | https://youtu.be/yESMPOzgdug ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2014|num-b=22|num-a=24}} | {{AMC8 box|year=2014|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 08:52, 23 July 2024
Contents
Problem
Three members of the Euclid Middle School girls' softball team had the following conversation.
Ashley: I just realized that our uniform numbers are all 
-digit primes.
Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month.
Caitlin: That's funny. The sum of your two uniform numbers is the date of my birthday later this month.
Ashley: And the sum of your two uniform numbers is today's date.
What number does Caitlin wear?
Video Solution for Problems 21-25
https://www.youtube.com/watch?v=6S0u_fDjSxc
Solution 1
The maximum amount of days any given month can have is 
, and the smallest two-digit primes are 
and 
. There are a few different sums that can be deduced from the following numbers, which are 
and 
, all of which represent the three days. Therefore, since Bethany says that the other two people's uniform numbers are earlier, so that means Caitlin and Ashley's numbers must add up to 
. Similarly, Caitlin says that the other two people's uniform numbers are later, so the sum must add up to 
. This leaves as today's date. From this, Caitlin was referring to the uniform wearers 
and , telling us that her number is 
, giving our solution as 
.
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=735
~ pi_is_3.14
Video Solution
https://youtu.be/yESMPOzgdug ~savannahsolver
See Also
| 2014 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
