Another Handshake Problem

[Clueless Morgan]

Is this one of those trick questions where everyone comes inside and shakes hands and the answer ends up being 0 because no one shook hands when they were outside?

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If all the guests reported a different number of handshakes, and Mrs. D doesn't remember how many handshakes she had, but it might be the same number as a guest (which means it is) then:

Mr. D is a non-entity, being the in the kitchen (what a guy ) so that leaves us with:

G1 shakes no hands
G2 shakes 1 hand
G3 shakes 2 hands
G4 shakes 3 hands
G5 shakes 4 hands
G6 shakes 5 hands

Mrs. D shakes 5 shakes (doesn't shake Mr. D's hand, nor guest 1's hand, nor her own but shakes enough hands to forget how many she's shaken.)

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[BrianSoddingBoru4]

Not answerable as asked.

Boru

[Excited Penguin]

I'm sorry, but I couldn't stop making a face at that:

When Mr Delepine calls them in to eat he asks how many handshakes took place outside.

Why would he even ask such a thing?

Irrelevant question. Solved. Where are my cookies?

[Whateverist]

Not answerable as asked.

Boru

Check out Poca's solution in the third post.  I stand behind it too.  What did you think needed to be specified?

[BrianSoddingBoru4]

We are not told whether Mrs. D shook hands with each of her guests. She may have shaken hands only with the males, only with the females, all of them, or any other number. From the way the question is worded, she could have shaken hands with Mr. B and Mrs. C and no one else.

Also, we cannot tell if all of the guests shook hands with Mrs. D and not with one another. For example, if Mrs. D shook everyone's hand, but Mr. and Mrs. A shook hands only with Mr. C, then the results are off.

Boru

[Joods]

You know, a group hug would have solved this problem nicely.

ijs.

[Aractus]

Brian, simple process of elimination. The guests all gave different numbers, the highest possible number is 5 (because they can't shake their own hand or the hand of their spouse), therefore they gave the numbers: 0, 1, 2, 3, 4, 5. Since 5 can't shake the hand of their spouse that must mean that the person who didn't shake any hands is their spouse, and since there are only 4 other guests they shook hands with all other guests AND Mrs. D. So +1 for Mrs D, eliminate those two numbers. Next is number 4. They can't have shaken their own hand, or their spouse's hand, or the person who shook no hands. That eliminates 3 people out of seven they could have shaken hands with, leaving only 4 remaining so they shook hands with all those four. Their spouse has to be the person who shook 1 hand since they didn't shake it and everyone else, including Mrs D (+1) but excluding person 0 has shaken at least 2 hands by now. That just leaves people 2 and 3. They can't shake hands with people 0 and 1 because they've already shaken their quota. Both of them shook hands with 4 and 5, therefore 2 can't shake hands with any more guests, and 3 has to shake another hand and the only available hand is Mrs D, bringing her total up to 3.

[Whateverist]

We are not told whether Mrs. D shook hands with each of her guests.  She may have shaken hands only with the males, only with the females, all of them, or any other number.  From the way the question is worded, she could have shaken hands with Mr. B and Mrs. C and no one else.

Also, we cannot tell if all of the guests shook hands with Mrs. D and not with one another.  For example, if Mrs. D shook everyone's hand, but Mr. and Mrs. A shook hands only with Mr. C, then the results are off.

Boru

True.  But the solution is all mathematical and logical based only on the fact that each guest shook hands a different number of times.  Figure out how to make that happen and you will know how many times that moral leper Mrs Delepine shakes hands.

[Clueless Morgan]

We are not told whether Mrs. D shook hands with each of her guests.

Um... that's the question that needs to be solved:

With how many people did Mrs Delepine shake hands with?

If you state in the initial problem that Mrs. D shook everyone's hand, then you've answered the riddle immediately after asking everyone to solve it, thus fulfilling my query of whether it's some kind of trick question.

She may have shaken hands only with the males, only with the females, all of them, or any other number.  From the way the question is worded, she could have shaken hands with Mr. B and Mrs. C and no one else.

With whom Mrs. D shakes hands is immaterial to the solution of the problem.  You could say that Couple A were the 5/0 handshakes couple, or that Couple C were, it doesn't make a difference to how many hands Mrs. D shook.  The OP doesn't ask "How many hands did Mrs. D shake, and whose were they?" it just asks for the number of hands shaken by Mrs. D.

For example:


Whether C1-m and C1-f (Couple 1, male and female) are Mr. and Mrs. Anderson or Mr. and Mrs. Carvalho doesn't matter to the answer (how many hands did Mrs. D shake?).  Likewise, you could swap the genders of each couple in this chart so that Mrs. D is shaking the hands of C1-m, C2-m and C3-f, it doesn't matter.  What matters is that one couple shakes 0/5 hands, one shakes 1/4 hands, one shakes 2/3 hands, and Mrs. D shakes the hand of one person from each couple in order that each of the guests shakes a different number of hands than anyone else (which is where I went wrong in my previous post - always make the chart when doing logic problems )

[Tiberius]

I basically worked it out with a table like Clueless Morgan posted above, and got the answer as 3 as well.