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

m (→‎Solution 2 (Detailed Version of Principle of Inclusion-Exclusion): Kept the page clean.)
(→‎Solution 3 (Condensed Version of Principle of Inclusion-Exclusion): Sol 2 already does a good job explaining the solution. The conclusion of 144 is somewhat unclear in this solution. So, I will delete this solution.)
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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>
 
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>
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==Solution 3 (Condensed Version of Principle of Inclusion-Exclusion)==
 
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For this problem, we use the Principle of Inclusion-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 <math>4</math>, and every <math>5</math> 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>.
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:51, 2 November 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 (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 2 (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. 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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