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Difference between revisions of "2014 AMC 8 Problems/Problem 23"
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==Problem==
 
Three members of the Euclid Middle School girls' softball team had the following conversation.
 
Three members of the Euclid Middle School girls' softball team had the following conversation.
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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.
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Bethany: And the sum of your two uniform numbers is the date of my birthday earlier this month.
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Bethany : And the sum of your two uniform numbers is the date of my birthday earlier this month.
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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.
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Ashley: And the sum of you two uniform numbers is today's date.
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Ashley: And the sum of your two uniform numbers is today's date.
  
 
What number does Caitlin wear?
 
What number does Caitlin wear?
  
 
<math>\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23</math>
 
<math>\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23</math>
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==Video Solution for Problems 21-25==
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https://www.youtube.com/watch?v=6S0u_fDjSxc
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==Solution 1==
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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>.
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== Video Solution by OmegaLearn ==
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https://youtu.be/6xNkyDgIhEE?t=735
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~ pi_is_3.14
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==Video Solution==
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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

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 $2$-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?

$\textbf{(A) }11\qquad\textbf{(B) }13\qquad\textbf{(C) }17\qquad\textbf{(D) }19\qquad \textbf{(E) }23$

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 $31$, and the smallest two-digit primes are $11, 13,$ and $17$. There are a few different sums that can be deduced from the following numbers, which are $24, 30,$ and $28$, 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 $24$. Similarly, Caitlin says that the other two people's uniform numbers are later, so the sum must add up to $30$. This leaves $28$ as today's date. From this, Caitlin was referring to the uniform wearers $13$ and $17$, telling us that her number is $11$, giving our solution as $\boxed{(A) 11}$.

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. AMC logo.png

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