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

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<math>\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152</math>
 
<math>\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152</math>
  
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==Solution 1==
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==Solution 1 (Principle of Inclusion-Exclusion)==
 +
We use Principle of Inclusion-Exclusion. There are <math>365</math> days in the year, and we subtract the days that she gets at least <math>1</math> phone call, which is <cmath> \left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor.</cmath>
  
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For this problem, we use the Principle of Inclusion and Exclusion(PIE). Notice every <math>3</math> day a day would be a multiple of <math>3</math>, every <math>4</math> day a day is a multiple of 4, and every 5 days it's a multiple of <math> 144</math> days without calls. Note that in the last five days of the year, days <math>361</math> and <math>362</math> also do not have any calls, as they are not multiples of <math>3</math>, <math>4</math>, or <math>5</math>. Thus our answer is <math>144+2 = \boxed{\textbf{(D)}\ 146}</math>.
+
To this result we add the number of days where she gets at least <math>2</math> phone calls in a day because we double subtracted these days, which is <cmath>\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor.</cmath>
  
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==Solution 2==
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We now subtract the number of days where she gets three phone calls, which is <math>\left \lfloor \frac{365}{60} \right \rfloor.</math> Therefore, our answer is <cmath>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}.</cmath>
 +
 
 +
==Solution 2 (Least Common Multiple)==
 +
Note that <math>\operatorname{lcm}(3,4,5)=60,</math> so there is a cycle every <math>60</math> days.
 +
 
 +
As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green.
 
<asy>
 
<asy>
 
/* Made by MRENTHUSIASM */
 
/* Made by MRENTHUSIASM */
Line 72: Line 78:
 
add(grid(10,6,linewidth(1.25)));
 
add(grid(10,6,linewidth(1.25)));
 
</asy>
 
</asy>
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The year 2017 has <math>365</math> days, or <math>6</math> cycles and <math>5</math> days.
 +
 +
* For each cycle, there are <math>24</math> days that Mrs. Sanders does not receive a phone call, as indicated by the white squares.
 +
 +
* For the last <math>5</math> days, there are <math>2</math> days that Mrs. Sanders does not receive a phone call, as indicated by the first <math>5</math> days in a cycle.
 +
 +
Together, the answer is <math>24\cdot6+2=\boxed{\textbf{(D) }146}.</math>
  
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EDITING IN PROGRESS ...
+
~MRENTHUSIASM
  
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==Solution 3==
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==Solution 3 (Linearity of Expectation)==
−
We use Principle of Inclusion and Exclusion. There are <math>365</math> days in the year, and we subtract the days that she gets at least <math>1</math> phone call, which is <cmath> \left \lfloor \frac{365}{3} \right \rfloor +  \left \lfloor \frac{365}{4} \right \rfloor +  \left \lfloor \frac{365}{5} \right \rfloor</cmath>
+
For any randomly chosen day, there is a <math>\frac{2}{3}</math> chance the first child does not call her, a <math>\frac{3}{4}</math> chance the second child does not call her and a <math>\frac{4}{5}</math> chance the third child does not call her. So, in a randomly chosen day, there is a <math>\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}</math> chance no child calls her.
  
−
To this result we add the number of days where she gets at least <math>2</math> phone calls in a day because we double subtracted these days. This number is <cmath>\left \lfloor \frac{365}{12} \right \rfloor +  \left \lfloor \frac{365}{15} \right \rfloor +  \left \lfloor \frac{365}{20} \right \rfloor</cmath>
+
In particular, this "unrigorous" reasoning becomes rigorous, by linearity of expectation, when we take a sample of <math>360</math> days; over the first <math>360</math> days, <math>360 \times \frac{2}{5} = 144</math> days will go without a phone call. Now, we simply check the remaining days; on day <math>361</math> and <math>362</math>, nobody calls her; on day <math>363</math>, the child that calls every three days will call her; on day <math>364</math>, the child that calls every four days will call her; on day <math>365</math>, the child that calls every five days will call her. Thus, we add two more days to <math>144</math>, to get our answer of <math>\boxed{\textbf{(D) }146}</math>.
  
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We now subtract the number of days where she gets three phone calls, which is <math>\left \lfloor \frac{365}{60} \right \rfloor</math>. Therefore, our answer is
+
~ihatemath123
  
−
<math>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</math>
+
==Video Solutions==
  
−
<math> = 365 - 285+72 - 6 = \boxed{\textbf{(D) } 146}</math>.
 
  
 +
https://youtu.be/9JDGys53Sn8
  
−
==Video Solution==
+
~savannahsolver
−
https://youtu.be/wXI-y_Ns0n0
 
  
−
https://youtu.be/a3rGDEmrxC0 - Happytwin
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https://www.youtube.com/watch?v=6eJ422kMgs0
  
−
https://youtu.be/Zhsb5lv6jCI?t=2797
+
~David
  
 
==See Also==
 
==See Also==

Latest revision as of 09:57, 24 July 2024

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 (Principle of Inclusion-Exclusion)

We use Principle of Inclusion-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, which 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 \[365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor = 365 - 285+72 - 6 = \boxed{\textbf{(D) }146}.\]

Solution 2 (Least Common Multiple)

Note that $\operatorname{lcm}(3,4,5)=60,$ so there is a cycle every $60$ days.

As shown below, all days in a cycle that Mrs. Sanders receives a phone call from any of her grandchildren are colored in red, yellow, or green. [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] The year 2017 has $365$ days, or $6$ cycles and $5$ days.

  • For each cycle, there are $24$ days that Mrs. Sanders does not receive a phone call, as indicated by the white squares.
  • For the last $5$ days, there are $2$ days that Mrs. Sanders does not receive a phone call, as indicated by the first $5$ days in a cycle.

Together, the answer is $24\cdot6+2=\boxed{\textbf{(D) }146}.$

~MRENTHUSIASM

Solution 3 (Linearity of Expectation)

For any randomly chosen day, there is a $\frac{2}{3}$ chance the first child does not call her, a $\frac{3}{4}$ chance the second child does not call her and a $\frac{4}{5}$ chance the third child does not call her. So, in a randomly chosen day, there is a $\frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} = \frac{2}{5}$ chance no child calls her.

In particular, this "unrigorous" reasoning becomes rigorous, by linearity of expectation, when we take a sample of $360$ days; over the first $360$ days, $360 \times \frac{2}{5} = 144$ days will go without a phone call. Now, we simply check the remaining days; on day $361$ and $362$, nobody calls her; on day $363$, the child that calls every three days will call her; on day $364$, the child that calls every four days will call her; on day $365$, the child that calls every five days will call her. Thus, we add two more days to $144$, to get our answer of $\boxed{\textbf{(D) }146}$.

~ihatemath123

Video Solutions

https://youtu.be/9JDGys53Sn8

~savannahsolver

https://www.youtube.com/watch?v=6eJ422kMgs0

~David

See Also

2017 AMC 8 (Problems • Answer Key • Resources)
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Problem 25
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