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## #1 Help Me ! » Truncated tetrahedron » 2017-07-31 00:29:35

Replies: 1

Given a tetrahedron IABC where all its edges have equal length x, we take points A1 on IA, such as IA1=1/2 IA, point B1 on IB such as IB1=2/3 IB and point C1 on IC such as IC1=3/4 IC.
Find the volume of tetrahedron IA1B1C1 in relation to a.

Sorry, I accidentally cut off half the wording

## #2 Help Me ! » Two friends - Probability » 2017-06-21 09:15:49

Replies: 0

Alice and Bob are sitting at the beach and each of them has 5 stones in front of him. They start tossing a fair coin and each time they get a head, they throw one of their stones in the sea, otherwise they do nothing. What is the probability that both friends have thrown all their stones after the same number of tosses?

## #3 Re: Help Me ! » Seating arrangements » 2017-06-15 09:43:44

Thanks also for sharing your work!

## #4 Re: Help Me ! » Seating arrangements » 2017-06-14 03:23:19

@Phrontister & thickhead, you are geniuses

## #5 Re: Help Me ! » Seating arrangements » 2017-06-13 18:30:35

Maybe I need to clarify that all Chinese, Russians and Mexicans are treated equally. Any Chinese can sit in any of their 5 seats etc.
Sorry for not having clarified this earlier; now that I saw your numbers I realised that you have this confusion.

## #6 Re: Help Me ! » Seating arrangements » 2017-06-13 11:52:33

Guys, I am getting 2252 ways...
This is probably wrong I guess...

## #7 Help Me ! » Seating arrangements » 2017-05-16 17:48:28

Replies: 11

5 Chinese, 5 Russians and 5 Mexicans are sitting in a row in such a way that the Chinese cannot be seated in the first 5 seats, the Russians cannot be seated in the 5 middle seats and the Mexicans cannot be seated in the last 5 seats. In how many different ways can they sit?

## #8 Help Me ! » Bag with balls » 2017-02-19 01:25:22

Replies: 3

An bag contains 10 black balls and some white ones. We intend to randomly pull  one ball out of the bag. Before we do this, one more ball is added in the bag (presumably by someone else, so that we don't see its color). Then we pull out one ball and it is a black one. What is the probability that the extra ball was black?

## #9 Help Me ! » Ants in a circle » 2016-12-23 02:42:53

Replies: 0

This is an old and well-known problem:

N ants are placed on a circle with a diameter of one meter. Each ant location is chosen independently, uniformly at random. An ant chooses between clockwise or anti-clockwise direction, uniformly at random, and starts scampering along the circle. All ants move at the same speed: one meter per second. When two ants bump into each other, they reverse their direction of travel. One of the ants is named Alice. What is the probability that Alice returns to the same point as she started, one minute after the ants start their scampering?

## #10 Re: Puzzles and Games » Day-off » 2016-12-21 02:17:23

Clarifications: Only one employee can be charged with the leave day. Each employee will come to work ONLY if he is 100% sure that it will be him to be charged with the leave day.
Any further ideas are most welcome

## #11 Re: Help Me ! » Knights and knaves » 2016-12-09 07:58:45

What I would suggest is ask David's question to one person.You would come to know whether he is knight or knave. then you ask him to list who are the knights.Two questions are enough.

All questions must be answered by yes or no.

## #12 Re: Help Me ! » Knights and knaves » 2016-12-08 00:20:00

What if we start with all possible ways to pick 5 persons out of 10, which are 10!/5!*5!=252, then ask one question to one of the inhabitants (the question that David specified, or any other with an obvious answer) so as to identify one knight or knave. Then we will have 9 remaining inhabitants with 5 knights and 4 knaves or the opposite. Then all possible ways to select 4 (or 5) out of 9 will be 9!/4!*5! = 126. Since 126<128=2^8, can we deduce that this can be covered by 7 questions (since, to cover 128 possibilities we need 8 questions) plus the first one. That gives 8 questions in total. Is this correct?
Can anyone work out an example?

## #13 Re: Help Me ! » Knights and knaves » 2016-12-08 00:08:32

Can you ask the same person more than once?

I believe yes.

## #14 Re: Help Me ! » Knights and knaves » 2016-12-07 20:41:41

I believe it can be done with 8... I cannot prove it though.

David wrote:

On a fictional island there are 10 inhabitants, who all know each other, of which 5 are knights, who always tell the truth and the rest of them are knaves, who always lie.
A visitor to the island wants to determine the 5 knaves. What is the minimum number of yes-no questions he must ask the inhabitants in order to find the 5 knaves?

Knights always tell the truth, knaves always lie. Ask them a fact question. => Do you live on an Island?

Knights will always tell the truth, hence, they will all say yes.
Knaves always lie, hence, they will all say no.

The minimum number of yes-no questions that he must ask the inhabitants is 9. (5-yes 4-no/ 5-no 4-yes) At 9, there's enough information to find out the 5 naves.

## #15 Help Me ! » Knights and knaves » 2016-12-06 23:43:33

Replies: 11

On a fictional island there are 10 inhabitants, who all know each other, of which 5 are knights, who always tell the truth and the rest of them are knaves, who always lie.
A visitor to the island wants to determine the 5 knaves. What is the minimum number of yes-no questions he must ask the inhabitants in order to find the 5 knaves?

## #16 Re: Help Me ! » Change of shape » 2016-10-18 01:32:07

Guys you are geniuses! Respect!!
Thank you so much!!

## #17 Re: Help Me ! » Change of shape » 2016-10-16 02:15:47

Thank you, but how about in the general case, where we don't know whether A lies on the circle?

## #18 Re: Help Me ! » Change of shape » 2016-10-14 06:23:01

Exactly, I corrected the wording. Sorry guys but English is not my mother tongue!!

zetafunc wrote:

Maybe he means equal area.

## #19 Help Me ! » Change of shape » 2016-10-14 05:44:33

Replies: 8

How can we transform a parallelogram with sides n and 1.1*n into a rhombus of equal area, just by one cut?

## #20 Re: Puzzles and Games » Prisoners and hats - one more variation » 2016-06-28 07:06:16

Hi Thickhead - I did some edits and think now it's more clear.

The idea is not clear. Needs more explanation.

## #21 Re: Puzzles and Games » Prisoners and hats - one more variation » 2016-06-28 02:13:50

This is the solution I propose:
We have 5 available colors, so, in order to transmit the information to the others, we will use this method:
We have 5 different values of mod(5): 0, 1, 2, 3, 4.
They coordinate and agree the following:
They assign these 5 values to the five hat colors Cyan, Magenta, Yellow, Red, Green.
0: Cyan
1: Magenta
2: Yellow
3: Red
4: Green

They also assign themselves with one of the numbers 0 to 4 (i.e. the 1st prisoner is assigned with number 0, the 2nd with 1 etc).
The above is pre-agreed before they wear the hats.
After they wear the hats, they do the following:
Then each prisoner calculates the sum of the numbers of the hat colors, based on the above table and DEDUCTS this number from his own and then converts the difference to mod(5). Then he finds the color that this number corresponds to, based on the above table.
Example: Suppose the 2nd prisoner sees the following hats to the other four: 1st-3rd-4th-5th: Cyan-Cyan-Red-Green. Let's assume his own hat is Yellow.
Then he calculates the sum 0+0+3+4=7. His own number is 1 (because he is the 2nd and number 1 is assigned to him). He deducts 7 from 1 and converts the difference (6) to mod(5) = 1. Then he guesses that he wears a Magenta hat. Obviously he is wrong, but let's see what the others will guess (because, even if all are wrong and ONLY one guesses correctly, they are saved).
The 1st one sees Yellow-Cyan-Red-Green and calculates the sum 9, which he deducts from his own that is 0, so the remainder is 9 which is 4 mod(5) so he guesses that he wears Green, which is also wrong because he wears Cyan.
Similarly the 3rd guesses Yellow and is wrong because he wears Cyan,
the 4th guesses Red which is CORRECT
and the 5th guesses Magenta and is wrong because he wears Green.
We therefore see that they are saved because always, at least one will guess correctly, based on the above strategy (the key here is that they are each assigned one different number of the 5 mod(5) so at least one will be correct).

Obviously no one knows the sum of all colors, since each of them does not see his own hat, but we know for sure that it will be one of the 5 numbers 0-4 mod(5).
We know, though, that it is sufficient that at least one finds the correct color, then at least one will have the number that will be the same as the correct sum of the hat colors, so he will announce the correct color!

## #22 Re: Help Me ! » More handshakes » 2016-05-02 22:26:39

I agree with your solution guys (not, of course, that I could have figured it out myself!!).
It is close to what I tried to do via calculations and without the use of excel (adding all possible different numbers starting from 62 and dividing by the number of pairs).

Congrats to both of you!

## #23 Re: Help Me ! » More handshakes » 2016-05-02 00:21:13

As per the terms, we only have the following condition: A must have a different number of TOTAL handshakes from B or C etc. It is possible, however, that he has 10 handshakes with P and 10 with someone else, provided that his handshakes in TOTAL are not the same with anybody else's.

## #24 Re: Help Me ! » More handshakes » 2016-05-01 05:43:40

Correct Anna.
A for example, may have 3 handshakes with B, 3 with C and 5 with D and so on.

anna_gg wrote:

We are not told that each people has different number of handshakes with each of the others; only that each one "participates" in a different number of handshakes in total. Therefore, if A participates in 2*31=62 (because we are told that each pair exchanges at least two handshakes), then all others from B to f participate in a different number of handshakes (but I don't know how to find them). I guess their sum must be 496*X*2.

A had 2 handshakes with B,3 handshakes with C, 4 handshakes with D  .... 32 handshakes with f (32nd person).So total handshakes by A=2+3+3+4+.......+31+32=527 handshakes. Add handshakes of all people and divide by 2.

## #25 Re: Help Me ! » More handshakes » 2016-04-30 19:00:43

All the possible pairs of 32 people are 32!/2!30! = 496. If each of them shook hands (i.e. "participated" in handshaking) at least twice and up to X times, then the total number T of handshakes must be from 2*496 to X*496. On the other hand, the total number of "single" handshakes must be a series of different numbers.
Their sum must be 2*T, correct?

Maybe you can combine the above to give a solution through calculations (and also verify your above outcome)?

Much appreciated!!
BTW I don't know the correct answer!!