Splendid Six
A married couple have six children--two girls (Ann and Brenda) and four boys (Craig, Dean, Eulius, and Frederick). A description of them is given below:
- Ann is among the three oldest children.
- Brenda has at least one younger brother.
- Craig isn't the youngest.
- Dean is older than one of the girls but younger than the other.
- Eulius is older than Ann.
- Frederick and Craig are consecutive siblings.
Based on the above information, assign a number to each of the six children according to their order of birth (i.e. 1:oldest, 2:second oldest, ..., 6:youngest). Then concatenate the six numbers in this order: number for Ann, number for Brenda, number for Craig, etc.
The answer is 245316.
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3 solutions
Using diagrams in the way you proposed seems a little bit forced though. It doesn't seem to make easier to follow the argument by the visual help as you initially thought I think.
In some sense it is so because that visual help doesn't provide too much information by themselves and the way of representing isn't the best either so they may be superfluous though the way you solved the problem , the general method/idea is cute , and intended for the purpose of illustrating such a reasoning though not most suitable here is cute. Note that what you do here , by using the (unnecessary) diagrams can be characterized as obtaining some based and refined (by deduction from what is given) hints about the order of the persons and implementing that information in the diagram such that you can deduce , based on all the refined hints at once , the solution of the problem. This idea of characterizing and calculating everything at once is a good example of a anyway different form of reasoning. Anyway I 'm still not very sure if you intended show a new path , and a different form of reasoning by using those diagrams and trying to illustrate some sort of compressing information technique by deducing everything at once or if you thought this method is better than simply making a chain of deductions anyway.
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Thanks for your comments. That was how I solved it. I admit that I like visual. A picture paints a thousand words. You will see that many of my solutions have graphs. I also like to use spreadsheet. I use a spreadsheet to solve it. First putting all clues into dots and then eliminate the dots to solve the problem. It is easier for me to see pictures than writing and reading a sentence. I am a Chinese. If you know the Chinese language every Chinese word is a picture. And I am so used to seeing pictures. It works for me and I am not suggesting that it works for everyone.
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I certainly don't know Chinese but I know you use "pictures" for words. I even asked someone if you also have an alphabet and he said you do haha.
Anyway , I believe using pictures to understand certain ideas better works for most people and especially for the most part of smart persons. But I'm not very sure if here this works so well. I'm not stating it isn't ok to use them or that the method doesn't work at all and should be abandoned and this especially for persons naturally inclined to reason this way as you , what I am instead stating is that it's nature isn't most suitable (objectively taken) for the problem and I'll try to exemplify what I meant anyway. I take this as a good example of analyzing methods for solving problems and understanding better your way of visualizing it by the use of spreadsheets and I'm not sure you are very interested in it but , if you are not , I'll not mind if you do not read it. I bet though on your curiosity regarding this methods and it's limits.
I think you like this "spreadsheet method" , or at least I like it so , because of the fact it can cover synthetically much information which can be easily computed at once based on what you introduced in it and also because of it's clarity and general simplicity and ease in the setting of such information. Your method can be characterized as completing sequentially the sheets based on given information which is a predication for any given sheet that reads about the "possible" places each person occupies in the diagram at once anyway.
Therefore you have something which translates by your spreadsheets in a a list the following way A is 4 or 6 , "B in 5 or 6 or 8 or9" etc but it's firstly deficit is that it doesn't speak about the possible relations between A and B from which this possibilities are covered anyway , deficit which is very signifying in the reasoning. Of course , because you complete the list sequentially starting from a spreadsheet and then considering the "hints" which speak about such relations you modify that spreadsheet to another anyway accordingly you can say that this deficit of the method is solved easily but this still shows , because of the fact that getting from a list to another you make deductions , dependence on sequential reasoning.
This dependence of sequential reasoning shows that the problem can anyway be solved because some liniar organization of deductions between "hints" anyway. In making a list of elements this hints might have no order between them ,order which naturally arrives in a directly considered and somewhat direct deductive reasoning which speaks of a natural order "heuristics" between deductions anyway.
This natural order is lost when you make deductions starting from spreadsheets not to say that by it's dependence of it's actual structure of deductions is not seen. Or simply said because "spreadsheet method" doesn't show the itnerconnections of deductions as they actually are it has a deficiency covered by the actual way of reasoning to which it is reducible even by using a "spreadsheet method" so to say that being that of sequential reasoning.
You can check Noel Lo's solution.
That's why i find the method somehow forced or the generalr easoning for why at times this method seems not sufficiently proper though ok anyway.
Btw you should correct in this part "Now assume, E is 2, then A is 3, E is 4 and B is 5 and" the fact that "E is 4" with "D is 4" and thanks too anyway.
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@A A – Thanks. Anyway, I find my way easier for me to understand. Just let members vote for the solutions.
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@Chew-Seong Cheong – Yeah , but I don't think you completely understood what follows from what and how the hints are related and correlate with each other nor the readers of the solution. The readers of Noel Lo's solution will understand that immediately if they read it as his solution is more natural and as (or maybe slightly more) beautiful anyway.
Your solution is charming because it's visual but the understanding it provides isn't as completely done hence the resulting lack of understanding how hints correlate and seeing the large reasoning , what follows from what and in what way and instead shows that you can arrive at the result by cleverly organizing things. Anyway I just have to say that it isn't correct to decide who's solution is"best" by looking at the number of upvotes but by testing one's understanding of the solution of the problem which here can't be done and also that this isn't the reason for which I wrote what I wrote and pretty much unimportant though.
You still didn't corrected your paragraph in the solution.
You should replace that "E is 4" with "D is 4" in the following phrase I think haha "then A is 3, E is 4 and B is 5 and" proposition 2 , in paragraph 4 anyway.
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@A A – You are right. I didn't read through Noel's solution. I will comment on the my solution.
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@Chew-Seong Cheong – Yes , your comment is quite right but it is not just because all clues lead to Frederik being the youngest anyway but because you can qualitatively see the real structure of the reasoning the solution of the problem implies and can as such understand the subtle connections between the logical construction of the hints. I wanted to make that remark too at a point of writing one of those 2 upper comments your method works generally especially for an input where a lot of information is given while the way Noel proposed would look tedious.
Anyway , even if it is or might seem tedious I still think the a method as used by Noel Lo still works generally. This is because the actual structure of the reasoning is reducible to that sequential form to which the visual support is nothing but a support and not a qualitative change in the problem and that with a lot of work and that even so the understanding would still be improved. I explained this a little bit more elaborate though it should have been even more so in that long comment. In short Noel Lo's should still work for any kind of problem though tediously anyway.
Going through the six statements, it is not difficult to tell that there is only one possible sibling who is the youngest - that is Frederick. All his five siblings indicate that they are not the youngest. From his statement, it follows that Craig is the fifth oldest child. Thus we have ? > ? > ? > ? > Craig > Frederick.
Now we can apply the same strategy of elimination to determine who the fourth oldest child is. Ann is definitely out as she is among the three oldest. Eulius is all the more impossible since he is older than Ann. A little more brainwork would enable us to rule out Dean too. The fourth oldest child has two younger brothers but no younger sisters so this cannot be Dean who has a younger sister. The only possible child who is the fourth oldest must be Brenda. Therefore we have ? > ? > ? > Brenda > Craig > Frederick.
Since Brenda is the third youngest with Craig and Frederick as her younger brothers, we can tell that she must be the girl who is younger than Dean. Since Dean also has an older sister, that must be referring to Ann. Hence Ann is older than Dean. Since Eulius is older than Ann, we have Eulius > Ann > Dean. Now we can easily fill up the remaining three positions. Hence Eulius > Ann > Dean > Brenda > Craig > Frederick.
| 1 | Eulius |
| 2 | Ann |
| 3 | Dean |
| 4 | Brenda |
| 5 | Craig |
| 6 | Frederick |
Our answer is 2 4 5 3 1 6 .
Cute solution and problem. Your other version was cute too but I like that you thought of this one as solvable by a reasoning which is centered on a more synthetic idea of organizing the information given (deducing them in order of birth from the youngest) which is a good exercise (and beautiful) for logic and reasoning anyway.
Let's start by finding the youngest.
Ann is not the youngest since she is one of the three oldest. Neither is Brenda who has a brother younger than her. We know Craig isn't the youngest because it says so. Dean is older than one of the girls and since both the girls are not the youngest, neither is he. Eulius can't be the youngest because he is older than Ann. So Frederick must be the youngest or 6 th oldest.
Frederick and Craig are consecutive siblings so Craig must be 5 th oldest.
If Dean was older than Ann, then he would be in the top 3, as well as Brenda. But that would mean Eulius, who is older than Ann and therefor is part of the top 3, could not be part of the top three. So Dean is younger than Ann, but older than Brenda. We have to have Eulius and Ann to be in the top three, and having Brenda in there as well would result in Dean not being older than Brenda, so Brenda must be 4 th oldest.
If Ann was third, then Dean would have to be older, which can't happen. If Ann was first, Eulius wouldn't be older than Ann. Therefor Ann is 2 nd oldest and Dean is 3 rd oldest and Eulius is 1 st oldest (I know it doesn't work). So the order is Ann - 2, Brenda - 4, Craig - 5, Dean - 3, Eulius - 1, and Frederick - 6; or 2 4 5 3 1 6
- Oh BTW, like the names ;-)
Relevant wiki: Order Theory
@A A is right. Noel Lo's solution is a better solution for this problem because all the clues lead to Frederick being the youngest. But the method I describe here generally apply to all problems with or without special clues.
Let us visualise this with diagrams as shown below. The right column list the siblings alphabetically. The top row denotes rank in age, 1 for oldest to 6 for youngest. Square square in cell indicates the possible rank in age of the person in that row. For example, hint 1: Ann is among the three oldest children; in diagram it is as follows:
\(\begin{array} {} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & \blacksquare & \blacksquare & \blacksquare & & & \\ B & & & & & & \\ C & & & & & & \\ D & & & & & & \\ E & & & & & & \\ F & & & & & & \end{array} \)
Then, hint 5: Eulius is older than Ann. This means that E can either be the oldest, then A second or third oldest, or E is the second oldest and A the third oldest. In diagram, it is as follows:
\(\begin{array} {} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & & \blacksquare & \blacksquare & & & \\ B & & & & & & \\ C & & & & & & \\ D & & & & & & \\ E & \blacksquare & \blacksquare & & & & \\ F & & & & & & \end{array} \)
Hint 4: Dean is older than one of the girls but younger than the other. Let us first consider Brenda is older than Ann. The only possibility is E is the oldest and B is the second oldest or vice versa and A the third oldest. But then E cannot sandwich between A and B. Therefore, B is younger than A. Also E is younger than A but older than B. Therefore, we have E < (older than) A < D < B and the possibilities in diagram is:
\(\begin{array} {} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & & \blacksquare & \blacksquare & & & \\ B & & & & \blacksquare & \blacksquare & \\ C & & & & & & \\ D & & & \blacksquare & \blacksquare & \\ E & \blacksquare & \blacksquare & & & & \\ F & & & & & & \end{array} \)
Note that B is either 4 and 5 and it fits hint 2: Brenda has at least one younger brother. Now assume, E is 2, then A is 3, D is 4 and B is 5 and the only two slot left are 1 and 6 (blank squares in the diagram below). But this does not fit hint 6: Frederick is consecutive siblings with Craig.
\(\begin{array} {} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & & & \blacksquare & & & \\ B & & & & & \blacksquare & \\ C & \square & & & & & \square \\ D & & & & \blacksquare & \\ E & & \blacksquare & & & & \\ F & \square & & & & & \square \end{array} \)
Therefore, E, A, D and B are 1, 2, 3, and 4 respectively. Since hint 3: Craig isn't the youngest. Then C is 5 and F is 6. And in diagram:
\(\begin{array} {} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline A & & \blacksquare & & & & \\ B & & & & \blacksquare & & \\ C & & & & & \blacksquare & \\ D & & & \blacksquare & & \\ E & \blacksquare & & & & & \\ F & & & & & & \blacksquare \end{array} \)
Therefore, the answer is 2 4 5 3 1 6 .