How Much Does the Envelope Cost in the Riddle?

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Discussion Overview

This thread discusses a mathematical riddle involving the cost of an envelope and a stamp, followed by a series of related riddles and responses. The focus includes problem-solving, reasoning about handshakes, and the implications of mutual friendships and strangers in a social context.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents a riddle about the cost of a stamp and an envelope, stating that the stamp costs a dollar more than the envelope.
  • Another participant suggests that the envelope costs 5 cents, but this is challenged by others who question the correctness of that answer.
  • Some participants argue that the correct answer to the handshake problem is 15, using the formula C(6,2) to justify their reasoning.
  • Another participant calculates handshakes differently, suggesting that if each person shakes hands with every other person, the total could be 30, depending on whether repeated handshakes are allowed.
  • Further clarification is provided that in the context of the riddle, each handshake is counted only once, leading back to the conclusion of 15 handshakes.
  • A new riddle is introduced regarding the number of people needed at a party to ensure at least three mutual friends or strangers, prompting a request for clarification on the wording of the problem.

Areas of Agreement / Disagreement

Participants express disagreement regarding the initial answer to the envelope riddle and the handshake problem, with multiple competing views on the correct interpretation and solution. The discussion remains unresolved on the envelope cost, while there is a stronger consensus around the handshake calculation being 15, though some still propose alternative interpretations.

Contextual Notes

Participants note that the interpretation of the handshake problem may vary based on assumptions about repeated interactions, and the wording of the new riddle may require further clarification to avoid ambiguity.

brainmetrix
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1. This thread is about mathematic riddles that we can all share, I will post one riddle and the next person list the solution and a new mathematic riddles as well (if possible), here is mine:
If a stamp and an envelope cost a dollar and a dime (10 cents), and the stamp costs a dollar more than the envelope, how much does the envelope cost?




2. Find the Solution



3. Post a New Mathematic Riddle
 
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note: moved to Brain Teasers - MIH
 
5 cents
 
C(6,2)=15
 
brainmetrix said:
I don't think that's the right answer, anyone?

Try to count it. May be it will help you to accept that correct answer.
 
How about, each can do 2 at the same time, so 6 simultaneous handshakes can happen at once?
Otherwise that can do quite man, pretty much till they start to die off.
 
brainmetrix said:
I don't think that's the right answer, anyone?

15 is the right answer. n/2 * (n-1), where n is the number of participants. The -1 is to account for the participants not shaking their own hand.
 
Couperin said:
15 is the right answer. n/2 * (n-1), where n is the number of participants. The -1 is to account for the participants not shaking their own hand.

This is one-half of a 6x6 matrix with the diagonal center elements removed.
Now, if they all spit at each other, how many spits are there?
 
it's 30.
there are 6 guys, and they each shake hands with 5 other people.
so for the first guy; he shakes hands with 5 people, meaning there are 5 handshakes for that one guy.
for the 2nd guy, there are another 5,
for the 3rd there are another 5,
and so on.
so you do 6 x 5, which gives you 30.
5 handshakes for each of the 6 men.

butttt, i guess it can depend on whether or not a man can shake the same man's hand twice.
then the answer would be less, but since it didn't include that in the riddle i won't include it while solving.
 
  • #10
Actually, in the original problem, "shaking hands with each other" usually means two people shake hands only once. The idea behind C(6,2) is that a handshake between person A and person B is the same as a handshake between person B and person A, and should be counted as only one handshake (hence the division by 2).

Another way to think about it, going along with your line of thought, is that the first guy can shake hands with 5 different people. The second guy can shake hands with only 4 different people, because his handshake with the first guy was already counted. The third guy can only shake hands with 3 different people, because his handshake with the first and the second guys have already been counted, and so forth.

Thus, we have not 5+5+5+5+5+5 = P(6,2) = 30 but 5+4+3+2+1+0 = C(6,2) = 15.

Of course, if the problem really is a brain teaser, it could have some other senseless answer. Who knows?
 
  • #11
15 is the correct answer. 5 handshakes 4...3...2...1 = 15
 
  • #12
its 15
6 people
guy no 6 shakes hands with 5
guy no 5 shakes hands with 4 [he's already done no.6]
guy no 4 shakes hands with 3 [he's already done no.6, and no.5) etc etc
...3 2
...2 1

thats 5+4+3+2+1 = 15

whats the next puzzle?

ok here's an oldie but a goodie

how many people do you have to invite to a party such that:
AT least 3 will be mutual friends
OR at lest 3 will be mutual strangers?

[mutual friends means A knows B and C, B knows C and A, C know A and B]
same difference for mutual strangers!
 
  • #13
um i think you need more wording here... by what you gave i would say i could invite 3 people because 3 could be mutual friends OR they could just be 3 mutual strangers either way it could satisfy your problem based on your guidlines... did i miss something?
 

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