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Difference between revisions of "2013 AMC 10A Problems/Problem 17"

(→‎Solution 2)
(→‎Solution 2)
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Now, we want the days in which exactly two of these people meet up
 
Now, we want the days in which exactly two of these people meet up
  
−
The three pairs are <math>{A,B}</math>, <math>{B,C}</math>, <math>{A,C}</math>.
+
The three pairs are <math>(A,B)</math>, <math>(B,C)</math>, <math>(A,C)</math>.
  
 
Notice that we are trying to find the LCM of each pair.
 
Notice that we are trying to find the LCM of each pair.
  
−
Hence, <math>LCM{A,B}=12x</math>, <math>LCM{B,C}=20x</math>, <math>LCM{A,C}=15x</math>
+
Hence, <math>LCM(A,B)=12x</math>, <math>LCM(B,C)=20x</math>, <math>LCM(A,C)=15x</math>
  
 
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
 
Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.
  
−
Hence, <math>LCM{A,B,C}=60x</math>
+
Hence, <math>LCM(A,B,C)=60x</math>
  
 
Now, we add all of the days up(including overcount).
 
Now, we add all of the days up(including overcount).

Revision as of 00:10, 6 June 2017

Problem

Daphne is visited periodically by her three best friends: Alice, Beatrix, and Claire. Alice visits every third day, Beatrix visits every fourth day, and Claire visits every fifth day. All three friends visited Daphne yesterday. How many days of the next 365-day period will exactly two friends visit her?

$\textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72$

Solution

The 365-day time period can be split up into $6$ 60-day time periods, because after $60$ days, all three of them visit again (Least common multiple of $3$, $4$, and $5$). You can find how many times each pair of visitors can meet by finding the LCM of their visiting days and dividing that number by 60. Remember to subtract 1, because you do not wish to count the 60th day, when all three visit.

A and B visit $\frac{60}{3 \cdot 4}-1 = 4$ times. B and C visit $\frac{60}{4 \cdot 5}-1 = 2$ times. C and A visit $\frac{60}{3 \cdot 5}-1 = 3$ times.

This is a total of $9$ visits per $60$ day period. Therefore, the total number of 2-person visits is $9 \cdot 6 = \boxed{\textbf{(B) }54}$.

Solution 2

From the information above, we get that $A=3x$ $B=4x$ $C=5x$

Now, we want the days in which exactly two of these people meet up

The three pairs are $(A,B)$, $(B,C)$, $(A,C)$.

Notice that we are trying to find the LCM of each pair.

Hence, $LCM(A,B)=12x$, $LCM(B,C)=20x$, $LCM(A,C)=15x$

Notice that we want to eliminate when all these friends meet up. By doing this, we will find the LCM of the three letters.

Hence, $LCM(A,B,C)=60x$

Now, we add all of the days up(including overcount).

We get $30+18+24=72$. Now, because $60(6)=360$, we have to subtract $6$ days from every pair. Hence, our answer is $72-18=\boxed{54}\implies\boxed{B}$.

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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

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