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Difference between revisions of "2021 AIME II Problems/Problem 11"

(→‎Solution 1: I have discussed with Arcticturn. The author agrees that this solution is written in a hurry. We will think of using Mathjam's solution. I will type that up when I have time. For now, we will delete this solution.)
(→‎Solution 2 (Illustrations))
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However, upon hearing that all four students replied no, each student was able to determine the elements of <math>S</math>. Find the sum of all possible values of the greatest element of <math>S</math>.
 
However, upon hearing that all four students replied no, each student was able to determine the elements of <math>S</math>. Find the sum of all possible values of the greatest element of <math>S</math>.
  
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==Solution 2 (Illustrations)==
+
==Solution 1==
 
Note that <math>\operatorname{lcm}(6,7)=42.</math> It is clear that <math>42\not\in S</math> and <math>84\not\in S,</math> otherwise the three other elements in <math>S</math> are divisible by neither <math>6</math> nor <math>7.</math>
 
Note that <math>\operatorname{lcm}(6,7)=42.</math> It is clear that <math>42\not\in S</math> and <math>84\not\in S,</math> otherwise the three other elements in <math>S</math> are divisible by neither <math>6</math> nor <math>7.</math>
  

Revision as of 18:20, 7 September 2021

Problem

A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $6$, and a different number in $S$ was divisible by $7$. The teacher then asked if any of the students could deduce what $S$ is, but in unison, all of the students replied no.

However, upon hearing that all four students replied no, each student was able to determine the elements of $S$. Find the sum of all possible values of the greatest element of $S$.

Solution 1

Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$

In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtain cycles: If $n$ is a possible maximum value of $S,$ then $n+42$ must be another possible maximum value of $S,$ and vice versa. By observations, we circle all possible maximum values of $S.$ [asy] /* Made by MRENTHUSIASM */ size(20cm); fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow); fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow); fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow); fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow); fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow); fill((35,1)--(36,1)--(36,3)--(35,3)--cycle,yellow); fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,green); fill((13,0)--(14,0)--(14,3)--(13,3)--cycle,green); fill((20,1)--(21,1)--(21,3)--(20,3)--cycle,green); fill((27,1)--(28,1)--(28,3)--(27,3)--cycle,green); fill((34,1)--(35,1)--(35,3)--(34,3)--cycle,green); fill((42,3)--(41,3)--(41,2)--cycle,yellow); fill((42,2)--(41,2)--(41,1)--cycle,yellow); fill((42,3)--(42,2)--(41,2)--cycle,green); fill((42,2)--(42,1)--(41,1)--cycle,green); for (real i=9.5; i<=41.5; ++i) { label("$"+string(i+0.5)+"$",(i,2.5),fontsize(9pt)); } for (real i=0.5; i<=41.5; ++i) { label("$"+string(i+42.5)+"$",(i,1.5),fontsize(9pt)); } for (real i=0.5; i<=14.5; ++i) { label("$"+string(i+84.5)+"$",(i,0.5),fontsize(9pt)); } draw(circle((6.5,1.5),0.45)); draw(circle((6.5,0.5),0.45)); draw(circle((7.5,1.5),0.45)); draw(circle((7.5,0.5),0.45)); draw(circle((8.5,1.5),0.45)); draw(circle((8.5,0.5),0.45)); draw(circle((13.5,2.5),0.45)); draw(circle((13.5,1.5),0.45)); draw(circle((13.5,0.5),0.45)); draw(circle((14.5,2.5),0.45)); draw(circle((14.5,1.5),0.45)); draw(circle((14.5,0.5),0.45)); draw(circle((20.5,2.5),0.45)); draw(circle((20.5,1.5),0.45)); draw(circle((23.5,2.5),0.45)); draw(circle((23.5,1.5),0.45)); draw(circle((29.5,2.5),0.45)); draw(circle((29.5,1.5),0.45)); draw(circle((30.5,2.5),0.45)); draw(circle((30.5,1.5),0.45)); draw(circle((35.5,2.5),0.45)); draw(circle((35.5,1.5),0.45)); draw(circle((36.5,2.5),0.45)); draw(circle((36.5,1.5),0.45)); draw(circle((37.5,2.5),0.45)); draw(circle((37.5,1.5),0.45)); draw((9,3)--(42,3)); draw((0,2)--(42,2)); draw((0,1)--(42,1)); draw((0,0)--(15,0)); for (real i=0; i<9; ++i) { draw((i,2)--(i,0)); } for (real i=9; i<16; ++i) { draw((i,3)--(i,0)); } for (real i=16; i<=42; ++i) { draw((i,3)--(i,1)); } [/asy] From the second row of the table above, we perform casework on the possible maximum value of $S:$ \[\begin{array}{c||c|c|l} & & & \\ [-2.5ex] \textbf{Max Value} & \boldsymbol{S} & \textbf{Valid?} & \hspace{16.25mm}\textbf{Reasoning/Conclusion} \\ [0.5ex] \hline & & & \\ [-2ex] 49 & \{46,47,48,49\} & & \text{The student who gets } 46 \text{ will reply yes.} \\ 50 & \{47,48,49,50\} & \checkmark & \text{Another possibility is } S=\{89,90,91,92\}. \\ 51 & \{48,49,50,51\} & & \text{The student who gets } 51 \text{ will reply yes.} \\ 56 & \{53,54,55,56\} & & \text{The student who gets } 53 \text{ will reply yes.} \\ 57 & \{54,55,56,57\} & & \text{The student who gets } 57 \text{ will reply yes.} \\ 63 & \{60,61,62,63\} & & \text{The students who get } 60,61,62 \text{ will reply yes.} \\ 66 & \{63,64,65,66\} & & \text{The students who get } 64,65,66 \text{ will reply yes.} \\ 72 & \{69,70,71,72\} & & \text{The student who gets } 69 \text{ will reply yes.} \\ 73 & \{70,71,72,73\} & & \text{The student who gets } 73 \text{ will reply yes.} \\ 78 & \{75,76,77,78\} & & \text{The student who gets } 75 \text{ will reply yes.} \\ 79 & \{76,77,78,79\} & \checkmark & \text{Another possibility is } S=\{34,35,36,37\}. \\ 80 & \{77,78,79,80\} & & \text{The student who gets } 80 \text{ will reply yes.} \end{array}\] Finally, all possibilities for $S$ are $\{34,35,36,37\}, \{47,48,49,50\}, \{76,77,78,79\},$ and $\{89,90,91,92\},$ from which the answer is $37+50+79+92=\boxed{258}.$

Remarks

  1. Alternatively, we can reconstruct the second table in this solution as follows, where Y and N denote the replies of "yes" and "no", respectively. Notice that this table has some kind of symmetry! [asy] /* Made by MRENTHUSIASM */ size(20cm); for (real j=5.5; j<41.5; j+=6) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,yellow); } for (real j=6.5; j<41.5; j+=7) { fill((j+0.5,4)--(j-0.5,4)--(j-0.5,3)--(j+0.5,3)--cycle,green); } fill((4,1)--(8,1)--(8,2)--(4,2)--cycle,mediumgray); fill((33,1)--(37,1)--(37,2)--(33,2)--cycle,mediumgray); fill((42,4)--(41,4)--(41,3)--cycle,yellow); fill((42,4)--(42,3)--(41,3)--cycle,green); for (real i=0.5; i<=41.5; ++i) { label("$"+string(i+42.5)+"$",(i,3.5),fontsize(9pt)); } draw(circle((6.5,3.5),0.45)); draw(circle((7.5,3.5),0.45)); draw(circle((8.5,3.5),0.45)); draw(circle((13.5,3.5),0.45)); draw(circle((14.5,3.5),0.45)); draw(circle((20.5,3.5),0.45)); draw(circle((23.5,3.5),0.45)); draw(circle((29.5,3.5),0.45)); draw(circle((30.5,3.5),0.45)); draw(circle((35.5,3.5),0.45)); draw(circle((36.5,3.5),0.45)); draw(circle((37.5,3.5),0.45)); label("Y",(3.5,2.5),blue); label("N",(4.5,2.5),blue); label("N",(5.5,2.5),blue); label("N",(6.5,2.5),blue); label("N",(4.5,1.5),blue); label("N",(5.5,1.5),blue); label("N",(6.5,1.5),blue); label("N",(7.5,1.5),blue); label("N",(5.5,0.5),blue); label("N",(6.5,0.5),blue); label("N",(7.5,0.5),blue); label("Y",(8.5,0.5),blue); label("Y",(10.5,2.5),blue); label("N",(11.5,2.5),blue); label("N",(12.5,2.5),blue); label("N",(13.5,2.5),blue); label("N",(11.5,1.5),blue); label("N",(12.5,1.5),blue); label("N",(13.5,1.5),blue); label("Y",(14.5,1.5),blue); label("Y",(17.5,2.5),blue); label("Y",(18.5,2.5),blue); label("Y",(19.5,2.5),blue); label("N",(20.5,2.5),blue); label("N",(20.5,1.5),blue); label("Y",(21.5,1.5),blue); label("Y",(22.5,1.5),blue); label("Y",(23.5,1.5),blue); label("Y",(26.5,2.5),blue); label("N",(27.5,2.5),blue); label("N",(28.5,2.5),blue); label("N",(29.5,2.5),blue); label("N",(27.5,1.5),blue); label("N",(28.5,1.5),blue); label("N",(29.5,1.5),blue); label("Y",(30.5,1.5),blue); label("Y",(32.5,2.5),blue); label("N",(33.5,2.5),blue); label("N",(34.5,2.5),blue); label("N",(35.5,2.5),blue); label("N",(33.5,1.5),blue); label("N",(34.5,1.5),blue); label("N",(35.5,1.5),blue); label("N",(36.5,1.5),blue); label("N",(34.5,0.5),blue); label("N",(35.5,0.5),blue); label("N",(36.5,0.5),blue); label("Y",(37.5,0.5),blue); for (real i=0; i<=42; ++i) { draw((i,4)--(i,3)); } draw((0,4)--(42,4)); draw((0,3)--(42,3)); [/asy]
  2. As a confirmation, we can verify that each student will be able to deduce what $S$ is upon hearing the four replies of "no" in unison. For example, if $S=\{47,48,49,50\},$ then all students will know that no one gets $46$ or $51,$ otherwise that student will reply yes (as discussed). Therefore, all students will conclude that $S$ has only one possibility.

~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=7jKjilTRhs4

See Also

2021 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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