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Difference between revisions of "2017 AMC 8 Problems/Problem 24"

(Although Solution 2 gives the correct answer, I don't think 365*2/3*3/4*4/5 is a legit approach. What if we change the 365 to 361 or something? Fractional days and off-by-one always haunt us. I will think of restoring an old solution. PM me if you disagre)
(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
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<asy>
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/* Made by MRENTHUSIASM */
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size(7cm);
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fill((2,6)--(3,6)--(3,5)--(2,5)--cycle,red);
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fill((5,6)--(6,6)--(6,5)--(5,5)--cycle,red);
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fill((8,6)--(9,6)--(9,5)--(8,5)--cycle,red);
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fill((1,5)--(2,5)--(2,4.5)--(1,4.5)--cycle,red);
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fill((4,5)--(5,5)--(5,4.5)--(4,4.5)--cycle,red);
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fill((7,5)--(8,5)--(8,4)--(7,4)--cycle,red);
 +
fill((0,4)--(1,4)--(1,3)--(0,3)--cycle,red);
 +
fill((3,4)--(4,4)--(4,3.5)--(3,3.5)--cycle,red);
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fill((6,4)--(7,4)--(7,3)--(6,3)--cycle,red);
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fill((9,4)--(10,4)--(10,3.5)--(9,3.5)--cycle,red);
 +
fill((2,3)--(3,3)--(3,2)--(2,2)--cycle,red);
 +
fill((5,3)--(6,3)--(6,2.5)--(5,2.5)--cycle,red);
 +
fill((8,3)--(9,3)--(9,2)--(8,2)--cycle,red);
 +
fill((1,2)--(2,2)--(2,1)--(1,1)--cycle,red);
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fill((4,2)--(5,2)--(5,1.5)--(4,1.5)--cycle,red);
 +
fill((7,2)--(8,2)--(8,1.5)--(7,1.5)--cycle,red);
 +
fill((0,1)--(1,1)--(1,0)--(0,0)--cycle,red);
 +
fill((3,1)--(4,1)--(4,0)--(3,0)--cycle,red);
 +
fill((6,1)--(7,1)--(7,0)--(6,0)--cycle,red);
 +
fill((9,1)--(10,1)--(10,2/3)--(9,2/3)--cycle,red);
 +
fill((3,6)--(4,6)--(4,5)--(3,5)--cycle,yellow);
 +
fill((7,6)--(8,6)--(8,5)--(7,5)--cycle,yellow);
 +
fill((1,4.5)--(2,4.5)--(2,4)--(1,4)--cycle,yellow);
 +
fill((5,5)--(6,5)--(6,4)--(5,4)--cycle,yellow);
 +
fill((9,5)--(10,5)--(10,4.5)--(9,4.5)--cycle,yellow);
 +
fill((3,3.5)--(4,3.5)--(4,3)--(3,3)--cycle,yellow);
 +
fill((7,4)--(8,4)--(8,3)--(7,3)--cycle,yellow);
 +
fill((1,3)--(2,3)--(2,2)--(1,2)--cycle,yellow);
 +
fill((5,2.5)--(6,2.5)--(6,2)--(5,2)--cycle,yellow);
 +
fill((9,3)--(10,3)--(10,2.5)--(9,2.5)--cycle,yellow);
 +
fill((3,2)--(4,2)--(4,1)--(3,1)--cycle,yellow);
 +
fill((7,1.5)--(8,1.5)--(8,1)--(7,1)--cycle,yellow);
 +
fill((1,1)--(2,1)--(2,0)--(1,0)--cycle,yellow);
 +
fill((5,1)--(6,1)--(6,0)--(5,0)--cycle,yellow);
 +
fill((9,2/3)--(10,2/3)--(10,1/3)--(9,1/3)--cycle,yellow);
 +
fill((4,6)--(5,6)--(5,5)--(4,5)--cycle,green);
 +
fill((9,6)--(10,6)--(10,5)--(9,5)--cycle,green);
 +
fill((4,4.5)--(5,4.5)--(5,4)--(4,4)--cycle,green);
 +
fill((9,4.5)--(10,4.5)--(10,4)--(9,4)--cycle,green);
 +
fill((4,4)--(5,4)--(5,3)--(4,3)--cycle,green);
 +
fill((9,3.5)--(10,3.5)--(10,3)--(9,3)--cycle,green);
 +
fill((4,3)--(5,3)--(5,2)--(4,2)--cycle,green);
 +
fill((9,2.5)--(10,2.5)--(10,2)--(9,2)--cycle,green);
 +
fill((4,1.5)--(5,1.5)--(5,1)--(4,1)--cycle,green);
 +
fill((9,2)--(10,2)--(10,1)--(9,1)--cycle,green);
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fill((4,1)--(5,1)--(5,0)--(4,0)--cycle,green);
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fill((9,1/3)--(10,1/3)--(10,0)--(9,0)--cycle,green);
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 +
real cur = 1;
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for (real i=6; i>0; --i) {
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  for (real j=0; j<10; ++j) {
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      label("$"+string(cur)+"$",(j+0.5,i-0.5));
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      ++cur;
 +
  }
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}
 +
 +
add(grid(10,6,linewidth(1.25)));
 +
</asy>
 +
 
EDITING IN PROGRESS ...
 
EDITING IN PROGRESS ...
  

Revision as of 13:17, 15 July 2021

Problem

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

For this problem, we use the Principle of Inclusion and Exclusion(PIE). Notice every $3$ day a day would be a multiple of $3$, every $4$ day a day is a multiple of 4, and every 5 days it's a multiple of $144$ 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

[asy] /* Made by MRENTHUSIASM */ size(7cm); fill((2,6)--(3,6)--(3,5)--(2,5)--cycle,red); fill((5,6)--(6,6)--(6,5)--(5,5)--cycle,red); fill((8,6)--(9,6)--(9,5)--(8,5)--cycle,red); fill((1,5)--(2,5)--(2,4.5)--(1,4.5)--cycle,red); fill((4,5)--(5,5)--(5,4.5)--(4,4.5)--cycle,red); fill((7,5)--(8,5)--(8,4)--(7,4)--cycle,red); fill((0,4)--(1,4)--(1,3)--(0,3)--cycle,red); fill((3,4)--(4,4)--(4,3.5)--(3,3.5)--cycle,red); fill((6,4)--(7,4)--(7,3)--(6,3)--cycle,red); fill((9,4)--(10,4)--(10,3.5)--(9,3.5)--cycle,red); fill((2,3)--(3,3)--(3,2)--(2,2)--cycle,red); fill((5,3)--(6,3)--(6,2.5)--(5,2.5)--cycle,red); fill((8,3)--(9,3)--(9,2)--(8,2)--cycle,red); fill((1,2)--(2,2)--(2,1)--(1,1)--cycle,red); fill((4,2)--(5,2)--(5,1.5)--(4,1.5)--cycle,red); fill((7,2)--(8,2)--(8,1.5)--(7,1.5)--cycle,red); fill((0,1)--(1,1)--(1,0)--(0,0)--cycle,red); fill((3,1)--(4,1)--(4,0)--(3,0)--cycle,red); fill((6,1)--(7,1)--(7,0)--(6,0)--cycle,red); fill((9,1)--(10,1)--(10,2/3)--(9,2/3)--cycle,red); fill((3,6)--(4,6)--(4,5)--(3,5)--cycle,yellow); fill((7,6)--(8,6)--(8,5)--(7,5)--cycle,yellow); fill((1,4.5)--(2,4.5)--(2,4)--(1,4)--cycle,yellow); fill((5,5)--(6,5)--(6,4)--(5,4)--cycle,yellow); fill((9,5)--(10,5)--(10,4.5)--(9,4.5)--cycle,yellow); fill((3,3.5)--(4,3.5)--(4,3)--(3,3)--cycle,yellow); fill((7,4)--(8,4)--(8,3)--(7,3)--cycle,yellow); fill((1,3)--(2,3)--(2,2)--(1,2)--cycle,yellow); fill((5,2.5)--(6,2.5)--(6,2)--(5,2)--cycle,yellow); fill((9,3)--(10,3)--(10,2.5)--(9,2.5)--cycle,yellow); fill((3,2)--(4,2)--(4,1)--(3,1)--cycle,yellow); fill((7,1.5)--(8,1.5)--(8,1)--(7,1)--cycle,yellow); fill((1,1)--(2,1)--(2,0)--(1,0)--cycle,yellow); fill((5,1)--(6,1)--(6,0)--(5,0)--cycle,yellow); fill((9,2/3)--(10,2/3)--(10,1/3)--(9,1/3)--cycle,yellow); fill((4,6)--(5,6)--(5,5)--(4,5)--cycle,green); fill((9,6)--(10,6)--(10,5)--(9,5)--cycle,green); fill((4,4.5)--(5,4.5)--(5,4)--(4,4)--cycle,green); fill((9,4.5)--(10,4.5)--(10,4)--(9,4)--cycle,green); fill((4,4)--(5,4)--(5,3)--(4,3)--cycle,green); fill((9,3.5)--(10,3.5)--(10,3)--(9,3)--cycle,green); fill((4,3)--(5,3)--(5,2)--(4,2)--cycle,green); fill((9,2.5)--(10,2.5)--(10,2)--(9,2)--cycle,green); fill((4,1.5)--(5,1.5)--(5,1)--(4,1)--cycle,green); fill((9,2)--(10,2)--(10,1)--(9,1)--cycle,green); fill((4,1)--(5,1)--(5,0)--(4,0)--cycle,green); fill((9,1/3)--(10,1/3)--(10,0)--(9,0)--cycle,green); real cur = 1; for (real i=6; i>0; --i) { for (real j=0; j<10; ++j) { label("$"+string(cur)+"$",(j+0.5,i-0.5)); ++cur; } } add(grid(10,6,linewidth(1.25))); [/asy]

EDITING IN PROGRESS ...

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

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

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


Video Solution

https://youtu.be/wXI-y_Ns0n0

https://youtu.be/a3rGDEmrxC0 - Happytwin

https://youtu.be/Zhsb5lv6jCI?t=2797

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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