Apple Logic
Alexandra, Bob, Chelsea, and Don are all logic experts who always tell the truth.
They each ate some numbers of apples: a , b , c , and d , respectively. They don't know the numbers the others ate, but they do know that each person ate at least 1 apple and that they ate 11 apples in total .
Alexandra says, “Bob, I don't know whether you ate more than I did. Did you?”
Bob replies, “I don’t know. Chelsea, I also don't know whether you ate more than I did. Did you?”
Chelsea replies, “I don’t know.”
Then Don says, "Oh, now I know how many each of us ate."
How many apples did each person eat? Give your answer as the concatenation of their numbers: a b c d .
Assume that the people all have perfect knowledge about each others' logical ability and honesty. That is, they all know that they are all logic experts that always tell the truth, and they all know that they know this, and so on.
The answer is 1235.
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6 solutions
This was the best solution and thanks for the table at last, it helped me a lot.
Curious why the name changes on your answer. Alex- > Andy. Chelsea -> Chris.
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The answer has the original names I chose, it seems in the problem names were edited to expand from my limited imagination with names!
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somehow they all became men!
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@Dan Wheatley – I originally chose simple short names, with one syllabul, starting with A,B,C,D, and easy to pronounce (since I teach lots of international students). But I certainly like how Brillaint noticed my unintentional gender bias, and altered the names slightly to give a more even distribution, and will make these changes for my class as well!
Fixed solution to match name changes created!
What is missing in this solution: you are consequently using - correct - implications, but that is not enough. It needed to be equivalences at each step in time for us to be able to determine whether somebody knows something or not. In his case it doesn't matter, but imagine that for some reason we (and they) could have deduced more from the first statement (Alexandra doesn't know) combined with the constraints, say also A<>1. Thus A >=2. Then we would have known other consequences from the second statement (Bob replying "I don't know"): i.e. that B >= 3. The point is that these would not necessarily have been a subset of the previously found open possibilities. THEY have different information than we have, since they know their apple count. Admittedly, in his problem, they actually ARE a subset, since we never know more than they do. But it could still have ended with an empty set of solutions. This problem didn't punish for using (partial) implications (it were subsets, so we were narrowing down by adding knowledge), but they still are no proof of the validity of this solution. It still can be the case that this solution is invalid (and that there is no solution), until you make a solid proof with ONLY equivalences at each step in time before using the next statement.
I mean: adding the deduction from the first statement that A<5 only later on - after already having made deductions from the other statements - could have been a mistake, since the eaters made all possible deductions BEFORE their statements.
Basically it suffices to make the exact same implications, but in the correct order statement per statement, and show that after each statement, there is nothing more to be known (for them).
great logic..I don't agree with the table narrowing though... what if Don uses some other property to pinpoint the right answer (other than the uniqueness of the combination for 5).
if my reasoning got me to ..say.. 2441...how does the setup eliminate this answer? please help
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besides...if don had 5..none else can have 5 nor more..and he should know it since the begining
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If Don had 5 he knew from the beginning that he ate the most, so he already knew that nobody else could know whether they ate more than the others. They might know that they ate less than or equal to what somebody else ate, but they couldn't be certain that the others ate more. However, Don didn't know the numbers the others had eaten, until the response by Chelsea. So, basically, he obtained information by learning what the others did not know, and the way in which that altered the options.
The only way he could know it at that point, was because he had eaten 5 apples. Finding the solution 1235 actually is quite straightforward: given that 1<=a<=4, 2<=b<=4, 3<=c<=4, it is clear that picking the maximum value for d, all of a, b, and c will need to be their minimum values, so that that solution is guaranteed to work. But the problem would be a bad one if Dan could truthfully make his statement while he had eaten some smaller value of apples.
So the reason the table is given is to prove uniqueness of the 1235 solution. If Don did not eat 5, there would either be no solutions if he ate 6 or more (which would make the problem incorrect), or multiple solutions (as shown by the table) if he ate less than 5, in which case he couldn't make his statement. So the problem only has one solution.
For the case 2441 you mention: Don having eaten only 1 apple would not know which of 2441, 3341, 3431, 4241, and 4331 was the case, so couldn't make the statement "Oh, now I know how many each of us ate." as mentioned in the problem description.
I’m confused. In the directions, it says that Alexandra doesn’t know if Bob ate more apples than she did. If she ate one apple and they each had different numbers of apples and they also had at least one apple, then she’d know that Bob ate more apples than she did. My sequence was then 2351.
Yes it is a tricky problem. Nothing is ever explicitly stated or required that they each person ate a different number of apples - and yes I agree if this was state this would change things. Note: The answer, 1235, does have each eating a different number of apples, but again this was not required in problem statement. I also agree that IF the problem stated that "each had different numbers of apples" - your answer is correct and fits this problem, and I really like as a "tweak" to this problem, will be giving it to my class ASAP!
The first part of my reasoning was exactly like Matthew's.
Andy ate at least one apple. If Bob had only eaten one apple, he would have replied "No" to Andy's question, so Bob ate at least two. If Chris had eaten one or two apples, he would have replied "No" to Bob's question, so Chris ate at least three.
At that point, it seems to me, we can go directly to the answer. Between Andy, Bob and Chris, at least six apples have been eaten. So Don ate at most five apples. But if Don had eaten any less than five, then any one of Andy, Bob or Chris could have eaten more than their minimum number of apples, so Don would not have been certain of how many each person ate. So Don must have eaten exactly five, from which it follows that Andy, Bob and Chris could only have eaten one, two and three apples respectively.
As an explicit example of the last part, suppose Don had only eaten four apples. Then it would have been possible that the number of apples eaten by Andy, Bob and Chris were respectively: two, two and three; one, three and three; or one, two and four. Don would have had no way to decide which of those three options was the correct one.
Zico - Yes, I agree - this is the more concise solution. My students typically give me some version of what you have written above. But, those who do not "get it" seem to need/prefer the more detailed (and long-winded!) explanation :)
For a unique solution to this, I think you still need to consider the upper bound of the apples eaten by A, B and C.
If, for the sake of argument, D knew that A had eaten 1-4, B 2-3 and C exactly 3 then there would be two solutions. Either D has eaten 5 apples and the only way A, B, C could eat (as few as) 6 apples is 1, 2, 3. Or D has eaten 1 apple and the only way A, B, C could eat (as many as) 10 apples is 4, 3, 3.
With the upper bound being 4 apples each, the above hypothetical situation doesn't apply and there is only one solution. But we only know this by considering the upper bound as well as the lower bound.
Relevant wiki: K-level thinking
Lemma: If someone ate 5 or more apples, then they hold the most apples among the group.
Proof: Suppose we express 1 1 = A + B + C + D , where A ≥ B ≥ C ≥ D ≥ 1 are positive integers that represent the number of apples ate by these 4 people, in some order. And if A = 5 , then ( B , C , D ) = ( 4 , 1 , 1 ) is the only solution. This shows that A is largest among these 4 integers.
In other words, Andy, Bob and Chris couldn't have said what they have said if any of them ate 5 or more apples. Thus 1 ≤ a < 5 , 1 ≤ b < 5 , 1 ≤ c < 5 .
Suppose Bob only ate one apple, then he definitely knows that he can't eat more than Andy. But that is not possible, because Bob said he doesn't know whether he ate more apples than Andy did. So Bob ate more than one apple. Thus, refining the inequality of b to 2 ≤ b < 5 .
Suppose Chris ate at most 2 apples, then he definitely knows that he can't eat more than Bob. But that is not possible, because Chris said he doesn't know whether he ate more apples than Bob did. So Charlie ate more than two apples. Thus, refining the inequality of c to 3 ≤ b < 5 .
Since, a ≥ 1 , b ≥ 2 , c ≥ 3 , then d = 1 1 − a − b − c ≤ 1 1 − 1 − 2 − 3 = 5 .
Suppose Don ate less than 5 apples, then
a
+
b
+
c
can take values 7, 8, 9 and 10.
Case 1:
If
a
+
b
+
c
=
7
, then there are at least 2 solutions,
(
a
,
b
,
c
)
=
(
1
,
2
,
4
)
,
(
1
,
3
,
3
)
.
Case 2:
If
a
+
b
+
c
=
8
, then there are at least 2 solutions,
(
a
,
b
,
c
)
=
(
2
,
2
,
4
)
,
(
2
,
3
,
3
)
.
Case 3:
If
a
+
b
+
c
=
9
, then there are at least 2 solutions,
(
a
,
b
,
c
)
=
(
2
,
2
,
4
)
,
(
2
,
3
,
3
)
.
Case 4:
If
a
+
b
+
c
=
1
0
, then there are at least 2 solutions,
(
a
,
b
,
c
)
=
(
2
,
4
,
4
)
,
(
3
,
3
,
4
)
.
In other words, if Don ate less than 5 apples, he is still unable to figure out how many apples everyone else has eaten (even after Andy's, Bob's, and Chris' line). But that is impossible, because he finally said that he knew how many each of them ate. Thus Don must have eaten 5 apples. Hence, d = 5 , which forces a = 1 , b = 2 , c = 3 . The answer follows to be 1 2 3 5 .
a. b=a+1 c=b+1=a+2. d=c+1=b+2=a+3 . 4a+6=11. a=5/4
a=1.25 / b=2.25/ c=3.25/ d=4.25
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No .The question already states that a , b , c , d denotes the number of apples ate by these 4 people. So these 4 variables must be integers.
Given that they all eat one apple, let's concentrate on the distribution of the remaining seven apples.
*Alexandra says, “Bob, I don't know whether you ate more than I did. Did you?” *
This implies that Alexandra ate one of (3,2,1,0) apples, because if she ate four or more she would know that she had eaten more than Bob. (because 4+3=7)
*Bob replies, “I don’t know. Chelsea, I also don't know whether you ate more than I did. Did you?” *
This implies that Bob ate one of (3,2,1) apples. because if he had four or more he would know he had eaten more than Alexandra. If he had eaten no apples he would know that he had not eaten more than Alexandra.
*Chelsea replies, “I don’t know.” *
This implies Chelsea ate one of (3,2) apples. Again if she had eaten 4 or more she would know she had eaten more than Bob. If she had eaten zero or one, she would know she had not eaten more than Bob.
*Then Don says, "Oh, now I know how many each of us ate." *
Since all the above reasoning is transparent to everybody, Don knows that A,B and C may have eaten (3,2,1,0), (3,2,1) and (3,2) apples. How can he possibly deduce the answer? Only If he himself has eaten 4 apples! Then he can deduce that each of the others must have eaten the smallest possible numbers; 0,1 and 2 respectively!
Adding back the four apples which we were told they had all eaten gives the solution 1235.
This might not be the best answer mathematically, but this is the only only one that helped me understand the reasoning. Thank you.
If a certain person ate more than 4 apples, then that person must have eaten most apples.
We'll now analyze sentence after sentence:
-
Alexandra says, “ Bob, I don't know whether you ate more than I did. Did you? ” - Alexandra certainly hasn't ate more than 4 apples, so 1 ≤ a ≤ 4
-
Bob replies, “ I don’t know. Chelsea, I also don't know whether you ate more than I did. Did you? ” - Bob hasn't ate more than 4 apples too, but he also hasn't ate only 1 apple either because he would've known that he hasn't ate more than Alexandra. So, 2 ≤ b ≤ 4 .
-
Chelsea replies, “ I don’t know. ” - Chelsea hasn't ate more than 4 apples too, but she also hasn't ate exactly 1 or 2 apples either because she would've known that she hasn't ate more than Bob. So, 3 ≤ b ≤ 4 .
-
Then Don says, " Oh, now I know how many each of us ate. " - The only way Don could've said this is if he'd eaten a number of apples which implies unique combination of others' apples ie. if there had been unique a , b and c so that a + b + c = 1 1 − d . The only unique combination is a minimal sum a + b + c = 1 + 2 + 3 = 6 . Hence, desired concatenation is 1 2 3 5 .
here each person eats at least one apple. so we are left with 7 apples. we solve the new problem assuming that there are 7 apples.we have get information from the statements how these seven apples are distributed among four people. from the statement of alexsandra alexandra has eaten at most 3 apples. from the statement two of bob it is clear that bob has eaten atleast 1 apple. this is obvious. as if he eats 0 apples. it is clear that he cannot eat more than alexasandra. so from bobs statement it is clear that bob has eaten at least 1 apple. from clesea statement it is clear that clesea has eaten at least 2 apples. if chelsea eats <2 apples. it is clear to chelsea that it has not eaten more than bob. but chelsea says she does not know. this means that chelsea has eaten at least 2 apples. from this we get bob has eaten at least 1 apples, chelsea has eaten at least 2 apples. now don is able to conclude things from this data. this means don has eaten 4 apples. thats it. so finally we get alexsandra-0 bob- 1 chelsea-2 don-4 adding 1 to all these we get alexsandra-0+1=1 bob- 1+1=2 chelsea-2+1=3 don-4+1=5
NB. J code
A=.0&{"1 NB. Alexandra
B=.1&{"1 NB. Bob
C=.2&{"1 NB. Chelsey
D=.3&{"1 NB. Don
T=. +/"1 NB. Total
a=.(#~5>C)(#~2<C)(#~1<B)(#~5>B)(#~5>A)(#~11=T)1+(4$8)#:i.8^4 NB. limit the possibilities
10#.(,d=/(1=+/(d=.D a)=/b)#b=.1+i.8)#a NB. select the unique solution
1235
Alexandra obviously ate at least one apple.
If Bob ate one apple, he knows he did not eat more than Alexandra, since the least amount Alexandra could eat is one apple; therefore, Bob would have to answer NO to Alexandra’s question (Bob, did you eat more than me?). Since Bob answered, “I don’t know”, this means that Bob ate at least two apples.
Following the same logic, Chelsea, since he is a logic expert, also knows Bob ate at least two apples. So if Chelsea ate one or two apples he would know he did NOT eat more than Bob, and would answer NO to Bob’s question (Chelsea, did you eat more than me?). Since Chelsea answered, “I don’t know”, this means that Chelsea ate at least three apples.
Let A = Alexandra, B = Bob, C = Chelsea, and D = Don
To summarize what we know so far:
Before the conversation begins it is obvious to all that Alexandra cannot eat more than eight apples, since the other three must eat at least one apple. If Alexandra ate five or more apples, he would know that he ate more apples than Bob, because if Alexandra ate five apples, Chelsea and Don must eat at least two apples total, leaving only four left for Bob to eat. If Alexandra ate six apples, Chelsea and Don must eat at least two apples total, leaving only three for Bob to eat. We can apply the same logic to Alexandra eating seven or eight apples. Since Alexandra asks Bob, “did you eat more than me”, we know that Alexandra did not eat five apples, six apples, seven apples or eight apples because if he ate five or more apples he would know that Bob did not eat more than him, and so Alexandra would not ask Bob the question, “did you eat more than me”. Thus we can now conclude that Alexandra did not eat more than four apples.
Following this same logic and extending it a bit, if Bob ate five, six, seven, or eight apples, he would know he ate more than Alexandra, and he would have to answer YES to Alexandra’s question (“Bob, did you eat more than me?”). Thus we can conclude that Bob did not eat more than four apples. We can also reach this conclusion by realizing that Bob asks Chelsea “Chelsea, did you eat more than me?” and following the logic we used for Alexandra, that is we know that Bob would not ask this question if he ate five, six, seven, or eight apples because Bob would know he ate more than Chelsea. Thus we can now conclude that Bob did not eat more than four apples.
Finally, again following the same logic, we also know that Chelsea did not eat five, six, seven, or eight apples. If he did he would have to answer YES to Bob’s question, “Chelsea, did you eat more than me”. Thus we can now conclude that Chelsea did not eat more than four apples.
To summarize what we know so far:
Alexandra ate one, two, three, or four apples. Bob ate two, three, or four apples. Chelsea ate three or four apples. In total they ate eleven apples.
The amount of apples Don eats depends on the combinations chosen from the above restrictions. There are 20 possible combinations (see tables). Since Don knows exactly how many each person ate, we find that the only amount Don can eat, such that there is a unique combination, is if Don ate five apples (thus Alexandra eats one apple, Bob eats two apples, Chelsea eats three apples). If Don ate four apples, there are three possibilities, (Alexandra eats one, Bob eats two, Chelsea eats four) or (Alexandra eats one, Bob eats three, Chelsea eats three) or (Alexandra eats two, Bob eats two, and Chelsea eats three). See tables below for all combinations, we see that D = 5 has only one row, this is the only value for D that has just one row.
Thus, Don ate five apples!