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2017 AMC 8 Problems/Problem 24

Revision as of 12:11, 26 October 2019 by Vbkris77 (talk | contribs) (→‎Solution 2)

Problem 24

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

Solution 1

In $360$ days, there are \[360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144\] days without calls. Note that in the last five days of the year, days $361$ and $362$ also do not have any calls, as they are not multiples of $3$, $4$, or $5$. Thus our answer is $144+2 = \boxed{\textbf{(D)}\ 146}$.

\[\]

Solution 2

Alternatively, there are $365\cdot \frac45 \cdot \frac 34 \cdot \frac23=\boxed{146}$ days without calls. Multiplying the fractions in this order prevents partial days, as $365$ is a multiple of $5$, $365\cdot \frac45$ is a multiple of $4$ and $365\cdot \frac45 \cdot \frac 34$ is a multiple of $3$.

Solution 3

We use Principle of Inclusion and Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is \[\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor\]

To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days. This number is \[\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor\]

We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor$. Therefore, our answer is

$365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor$

$= 365 - 285+72 - 6 = \boxed{\textbf{(D) } 146}$.

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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

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