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Difference between revisions of "2013 AMC 10A Problems/Problem 17"
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==Problem==
 
==Problem==
  
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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?
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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 <math>365</math>-day period will exactly two friends visit her?
  
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 </math>
 
<math> \textbf{(A)}\ 48\qquad\textbf{(B)}\ 54\qquad\textbf{(C)}\ 60\qquad\textbf{(D)}\ 66\qquad\textbf{(E)}\ 72 </math>
  
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==Solution==
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==Solution 1==
 +
 
 +
The <math>365</math>-day time period can be split up into <math>6</math>, <math>60</math>-day time periods, because after <math>60</math> days, all three of them visit again (Least common multiple of <math>3</math>, <math>4</math>, and <math>5</math>).
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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.
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Remember to subtract <math>1</math>, because you do not wish to count the <math>60</math>th day, when all three visit.
 +
 
 +
A and B visit <math>\frac{60}{3 \cdot 4}-1 = 4</math> times.
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A and C visit <math>\frac{60}{3 \cdot 5}-1 = 3</math> times.
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 +
B and C visit <math>\frac{60}{4 \cdot 5}-1 = 2</math> times.
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 +
 
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This is a total of <math>9</math> visits per <math>60</math> day period.
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Therefore, the total number of <math>2</math>-person visits is <math>9 \cdot 6 = \boxed{\textbf{(B) }54}</math>.
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 +
*Note: We do not have to worry about the numbers over 360: (<math>361,362,363,364,365</math>) having 2 factors. This is because we can rewrite
 +
<math>(361,362,363,364,365) \Rightarrow (361,362, 3\cdot 121, 4 \cdot 91, 5 \cdot 73)</math>. We note that <math>121</math> is not further divisible by 4 or 5, <math>91</math> is not further divisible by 3 or 5, 73 is not further divisible by 3 or 4. Therefore, none of the numbers from <math>361-365</math> have 2 factors of <math>3,4,</math> or <math>5</math>, so we can conclude that the answer is indeed <math>\boxed{54}</math>
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 +
==Solution 2==
 +
 
 +
From the information above, we get that
 +
<math>A=3x</math>
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<math>B=4x</math>
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<math>C=5x</math>
 +
 
 +
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>.
 +
 
 +
Notice that we are trying to find the LCM of each pair.
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 +
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.
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 +
Hence, <math>LCM(A,B,C)=60x</math>
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Now, we add all of the days up(including overcount).
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We get <math>30+18+24=72</math>.
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Now, because <math>60(6)=360</math>, we have to subtract <math>6</math> days from every pair.
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Hence, our answer is <math>72-18=\boxed{54}\implies\boxed{B}</math>.
  
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2013|ab=A|num-b=16|num-a=18}}
 
{{AMC10 box|year=2013|ab=A|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 15:10, 30 August 2023

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 1

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 $60$th day, when all three visit.

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

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

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


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

  • Note: We do not have to worry about the numbers over 360: ($361,362,363,364,365$) having 2 factors. This is because we can rewrite

$(361,362,363,364,365) \Rightarrow (361,362, 3\cdot 121, 4 \cdot 91, 5 \cdot 73)$. We note that $121$ is not further divisible by 4 or 5, $91$ is not further divisible by 3 or 5, 73 is not further divisible by 3 or 4. Therefore, none of the numbers from $361-365$ have 2 factors of $3,4,$ or $5$, so we can conclude that the answer is indeed $\boxed{54}$

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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