2013 AMC 10A Problems/Problem 17

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 $72-18=\boxed{54}\implies\boxed{B}$ .

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2013 AMC 10A Problems/Problem 17

Contents

Problem

Solution 1

Solution 2

See Also