# I can't figure out this riddle



## z105 (22 Jul 2008)

*3 Men Go  Into A Hotel. The Man Behind The Desk Said The Room Is £30.00 So

Each Man  Paid £10.00 And Went To The Room. 

A While Later The Man Behind The Desk  Realized The Room Was Only £25.00

So He Sent The Bellboy To The 3 Guys'  Room With £5.00 

On The Way The Bellboy Couldn't Figure Out How To Split  £5.00 Evenly 

Between 3 Men, So He Gave Each Man A £1.00 And Kept The  Other £2.00 For

Himself. 

This Meant That The 3 Men Each Paid  £9.00 For The Room, Which Is A Total

Of £27.00, Add The £2.00 That The  Bellboy Kept = £29.00 

Where Is The Other Pound? 
*


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## ClubMan (22 Jul 2008)

Havealaugh said:


> *
> This Meant That The 3 Men Each Paid  £9.00 For The Room, Which Is A Total
> 
> Of £27.00, Add The £2.00 That The  Bellboy Kept = £29.00
> ...


In the beginning:

Man behind desk has £0
Bellboy has £0
Guest 1 has £10
Guest 2 has £10
Guest 3 has £10
Total £30

At the end

Man behind desk has £25
Bellboy has £2
Guest 1 has £1
Guest 2 has £1
 Guest 3 has £1
Total £30

In the end each guest paid £9 for the room = £27 and they each have £1 in their pockets. Total €30.

Of the £27 paid for accommodation the man behind the desk has £25 and the bellboy has stolen £2.

The £27 paid includes the £2 stolen by the bellboy and ignores the £3 that the three guests still have.


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## Simeon (23 Jul 2008)

Yes, this is an old chestnut ........ like switching the suitcase. When you get to totting up you use mental sleigh of hand ........ substituting a plus for a minus ....... a minus for a plus. 3 x 9 = 27 ....... here you should subtract the 2 that the bellhop swiped and add that (plus the 1 x 3) = 30.


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## z106 (23 Jul 2008)

ClubMan said:


> In the beginning:
> 
> Man behind desk has £0
> Bellboy has £0
> ...


 
That is only partially correct in my book.
i.e. your accounts of how the 30 is totalled both before and after the transaction is correct

However your line below is incorrect:
"In the end each guest paid £9 for the room = £27 and they each have £1 in their pockets. Total €30."

THE 9 they have paid out takes into account the 1 they have in their pocket
i.e. the 9 is net (i..e 10 paid out - 1 received back)

Also - your line above doesnt take into account the 2 the bellboy has.
If you included that you would have had 32.

Therefore these should not be added.

The thing with this riddle is the phrasing suggests you should be trying to get a total of 30 - it attempts this by adding the 27 and the 2 - but this gets 29 - which obviously is not 30.

However - what you should be doing is trying to reach 25 - not 30.
This is done by subtracting teh 2 from teh 27 to get 25.

i.e. The net outgoings should be 25.

Therefore
Net Outgoings = Gross outgoing - gross incomings
Or 25 = Gross outgoing - gross incomings
=> 25 = 30 - 5
However - the 5 is made up of the 3 they got back and the 2 the bellboy took
i.e. the 2 shoudl also have been theirs but teh bellboy took it instead.
Otherwise it too would have been included as an incoming.

To sum up - the riddle adds 27 and 2 trying to get 30.
It should in fact be looking to reach 25.
This is done by subtracting 2 from the 27.


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## ClubMan (23 Jul 2008)

qwertyuiop said:


> However your line below is incorrect:
> "In the end each guest paid £9 for the room = £27 and they each have £1 in their pockets. Total €30."
> 
> THE 9 they have paid out takes into account the 1 they have in their pocket
> ...


You are correct and I am wrong in this instance.


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## z106 (23 Jul 2008)

While were at it i have a good one.

A statistician is doing a survey in an area.

He knocks on a door and asks the man who answers how many kids he has and what ages they are.
The man says he has 3 kids and  the product of their ages is 36 and that the sum of their ages is equal to the number of his house (At this points he points to teh number on teh door for the statistician to see)

Teh statistician goes away and thinks about it and comes back to say he has not been given enough information.
Throughout all this both men can hear someone playing the piano.

At this point the man poiints upstairs and says "That is my eldest playing the piano" upon which the statistician then knows the ages of teh children.

WHat age are the kids?


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## z103 (23 Jul 2008)

I know what teh answer is.

3, 3, 4

We know one child is older than teh other two. The product must equal 36, so by brute force; 3x3x4 = 36


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## z106 (23 Jul 2008)

leghorn said:


> I don't know what teh answer is.


 
Id PM ypu if i could leghorn but for some reason your profile doent allow it


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## z106 (23 Jul 2008)

leghorn said:


> I know what teh answer is.
> 
> 3, 3, 4
> 
> We know one child is older than teh other two. The product must equal 36, so by brute force; 3x3x4 = 36


 
Incorrect


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## z103 (23 Jul 2008)

Does it have more than one answer?


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## z106 (23 Jul 2008)

leghorn said:


> Does it have more than one answer?


 
no


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## z103 (23 Jul 2008)

How is my answer incorrect?
Three ages, with product of 36, one number greater than teh others. Where does it fall over?


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## ClubMan (23 Jul 2008)

Is the misspelling of "the" as "teh" significant?


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## z103 (23 Jul 2008)

1, 4, 9 
2, 2, 9
1, 2, 18
also seems to work.


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## z106 (23 Jul 2008)

leghorn said:


> How is my answer incorrect?
> Three ages, with product of 36, one number greater than teh others. Where does it fall over?


 
Trying to answer that may give it away.

But anyway - Your answer would mean the number of teh house was 10.

In that scenario, why woud the statistician have to come back saying he didnt have enough info?
He would have had enough info in teh first visit.

The question you have to ask yourself is under what circumstances would the statistician have to come back looking for more info?


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## ClubMan (23 Jul 2008)

leghorn said:


> I know what teh answer is.
> 
> 3, 3, 4
> 
> We know one child is older than teh other two. The product must equal 36, so by brute force; 3x3x4 = 36


Based on your logic 2 x 3 x 6 would work equally well. Or any other three factors of 36!


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## z106 (23 Jul 2008)

leghorn said:


> 1, 4, 9
> 1, 2, 18
> also seems to work.


 Also incorrect


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## z106 (23 Jul 2008)

ClubMan said:


> Is the misspelling of "the" as "teh" significant?


 
No significance 

Thats a bad habit i have for some reason when i type quickly.


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## z103 (23 Jul 2008)

> In that scenario, why woud the statistician have to come back saying he didnt have enough info?


Ah, so we need an answer where the house number can only be added up multiple ways.

2, 2, 9 is the answer.
The sum is 13.

He had to get more info because 1, 6, 6 also works, but there has to be one eldest.


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## ClubMan (23 Jul 2008)

I've seen this problem stated differently and mentioning the house opposite. Are you sure that you have it right above?


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## ClubMan (23 Jul 2008)

leghorn said:


> Ah, so we need an answer where the house number can only be added up multiple ways.
> 
> The sum is 13.


How do we know the house number is 13?


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## z106 (23 Jul 2008)

ClubMan said:


> I've seen this problem stated differently and mentioning the house opposite. Are you sure that you have it right above?


 
ya - i definitely have it right.

i suppose you could also phrase whereby the sum of the ages is equal to the house accross teh road.

all that matters is that the statistician knows the number the ages add up to.


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## ClubMan (23 Jul 2008)

qwertyuiop said:


> all that matters is that the statistician knows the number the ages add up to.


But how do we know that given the information above?


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## z103 (23 Jul 2008)

please tell me it's 2, 2, 9 so I can go to bed...



> But how do we know that given the information above?


We don't need to know. We only need to know there is multiple solutions to x+y+z = housenumber


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## z106 (23 Jul 2008)

leghorn said:


> Ah, so we need an answer where the house number can only be added up multiple ways.
> 
> 2, 2, 9 is the answer.
> The sum is 13.
> ...


 
Well done leghorn - gold star for you !

Yes - for all comboinations of numbers that mutiply to give 36, the sum is unique other than 13.

E.g. 3,3,4 = 36 with a sum of 10.

If the house number was 10,then there would be only 1 possible combination.
Therefore the statistician would have known on teh spot.

However - he says he doesnt have enough infor.

Therefore clearly the house number was 13 as 6,6,1 and 9,2,2, are valid answers.

Whewn he is told the eldest is playing teh piano then it has to be 9,2,2,


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## ClubMan (23 Jul 2008)

Ah - got it. You can go to bed _leghorn_. You have it. There are only three (integral) factors of 36 which add up to the same number 2 x 2 x 9 and 1 x 6 x 6 so to resolve which was correct the statistician needed some more info.

Update: too late - crossed with previous post.


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## z106 (23 Jul 2008)

ClubMan said:


> But how do we know that given the information above?


 
In teh original post i stated 

"(At this points he points to teh number on teh door for the statistician to see)"


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## z103 (23 Jul 2008)

> Well done leghorn - gold star for you


I only deserve silver because you had to give clues.
good night everyone!


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## z106 (23 Jul 2008)

If anyone is hungry for more,I do have another one which is by far the toughest i have ever come accross.
Definitely teh daddy of all these riddles !

Anyone who gets this seriously deserves a gold star in my book !

You have 8 balls all of equal size.
7 are the same weight - 1 rogue ball is a different weight.
Note; We dont know if it is heavier or lighter - all we know is that it is a different weight.

You have a balancing scales.

You have 3 weighs to determine which is the rogue ball.

How do you do it

Anyway - i'm off to bed on that note.

By teh way - I can't remember the answer to thsi right now.
I'll have a think about t tomorrow.

I definitely have the question correct though in case anyone is questioning it.


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## dem_syhp (23 Jul 2008)

I'm assuming that this is a balance where you put say 2 balls on one side and 2 on the other.  I'm going to label the two sides of the balance L and R

*Measure 1 *
Take 3 balls (a, b, c) on one L side of the balance and take 3 balls (d, e, f) and put them R side of the balance.  
Is L side heavier/lighter/balanced? 

IF Balanced, solve for g, h: 
    If it is balanced then you know that none of these balls are the errant ball.  
     - Take ball a (now known to be good) and put it on the balance and ball g on the other side.  
     - If it balances - the errant ball is h, if not then it's g.  
     - Solved in 2 measures. 

*Measure 2 [ where un-balanced, so now know that g, h are good]*
However if after that first balance it was not even, then take balls (a, b) on L side and (c, d) on R side. 
Is L side heavier/lighter/balanced? 

IF balanced, solve for e, f
 - If it balances - the errant ball is either e, or f
 - Put ball e on the balance with g (known good ball).  If balanced errant ball is f.  Similarly if not balanced, errant ball is e.  
 - Solved in 3 measures

*Step 3 - Note no measure taken here [Now know that e, f, g, h are good] this is where Measure 1 and 2 were not balanced. *
If for measure 1 and 2 L side was lighter each time or heavier each time.  Then ball c is good - it moved from one side to the other and the direction did not change.  
If the L side was lighter in one and heavier in the other (that is it changed) then the errant ball moved sides.  Hence c is the errant ball.  
Otherwise, we now know balls c, e, f, g, h are good

*Measure 3  - solving for one of a, b, d *
Put a on L side and b on R side.  If balanced then errant ball is d
If L side is still either Lighter or heavier (same direction as previous two measures) then the errant ball has not moved from that side and hence the errant ball is a. 
Or if now the balance has swung the opposite direction, then the errant ball has moved and hence the errant ball is b. 

Solved in 3!

The insomniac - and no I didn't google or know the answer before.  I've had the weighing scales out for the last two hours


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## z106 (23 Jul 2008)

dem_syhp said:


> I'm assuming that this is a balance where you put say 2 balls on one side and 2 on the other. I'm going to label the two sides of the balance L and R
> 
> *Measure 1 *
> Take 3 balls (a, b, c) on one L side of the balance and take 3 balls (d, e, f) and put them R side of the balance.
> ...


 
Excellent.
That's a tough puzzle.

For anyone who is reading the answer and thinks "Ah...it's not that hard", it only seems that way when the answer is written down as conscisely as it was above.

Definite gold star for you dem_shyp.


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## Teabag (23 Jul 2008)

......OK so the bellboy stole the €2 from the €5 and the lads paid €27, no €25 and the hotel manager said €30 but........oh my head hurts....


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## z106 (23 Jul 2008)

Teabag said:


> ......OK so the bellboy stole the €2 from the €5 and the lads paid €27, no €25 and the hotel manager said €30 but........oh my head hurts....


 
Yep - Blame it on the bellboy !


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## ClubMan (23 Jul 2008)

qwertyuiop said:


> Definite gold star for you dem_shyp.


Well done. I went to bed thinking about this one and woke up still thinking about it!


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## Gordanus (25 Jul 2008)

Havealaugh said:


> *3 Men Go  Into A Hotel. The Man Behind The Desk Said The Room Is £30.00 So
> 
> Each Man  Paid £10.00 And Went To The Room.
> 
> *


*

Why were 3 men sharing a hotel room?*


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## z103 (25 Jul 2008)

Well for a tenner each, I doubt they'd get their own rooms.


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## S.L.F (25 Jul 2008)

qwertyuiop said:


> If anyone is hungry for more,I do have another one which is by far the toughest i have ever come accross.
> Definitely teh daddy of all these riddles !
> 
> Anyone who gets this seriously deserves a gold star in my book !
> ...



Changing the amounts now to nine balls of equal size and one of the balls is slightly heavier and you still can only use the scales three times.

Find the heavy ball!


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## dem_syhp (25 Jul 2008)

Ah now, that ones easier than the 8 balls one


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## z106 (25 Jul 2008)

S.L.F said:


> Changing the amounts now to nine balls of equal size and one of the balls is slightly heavier and you still can only use the scales three times.
> 
> Find the heavy ball!


 
well if you know that one is heavier (unlike the earlier puzzle where all you knew was that it was a different weight i.e. could be either heavier or lighter)), then i think it can be done in 2 weighs.

You have 9 balls each numered 1 to 9.

Weigh balls 1,2,3 against 4,5,6.

Depending on the result of that you then know what group of 3 the heavy ball is in.
i.e. if they weigh the same then it's in the group 7,8,9.
If not then it is in the heavier side from the first weigh.

Next - take the group of 3 the heavy ball is - call them a,b,c - and use the same mewthod from the first weigh to determine the heavy ball


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## miselemeas (27 Jul 2008)

Ages = 36 1 1?


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## dem_syhp (27 Jul 2008)

Bit of a stretch on the womans reproductive system?  A baby at 14, and then twins at 50?  

Though, I suppose possible - see here 

With those sorts of figures, you could have the baby the same age as it's great aunt!  Adds a whole new meaning to the concept of having several generations of the one family under the one roof!


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## miselemeas (28 Jul 2008)

dem_syhp said:


> Bit of a stretch on the womans reproductive system?  A baby at 14, and then twins at 50?
> 
> Though, I suppose possible - see here
> 
> With those sorts of figures, you could have the baby the same age as it's great aunt!  Adds a whole new meaning to the concept of having several generations of the one family under the one roof!



Wouldn't have to be the same mother!


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## dem_syhp (28 Jul 2008)

Okay, you're too smart for me now


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## macnas (28 Jul 2008)

Leghorn is correct

  The ages must be one of the following

1,1,36  house number = 38
1,4,9                        = 14
1,6,6                           13
2,3,6                           11
2,2,9                           13
3,3,4                           10

as he wasnt sure then it was either 1,6,6  or 2,2,9   but the eldest had to be 9
so answer is 2,2,9


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## miselemeas (28 Jul 2008)

Finally I get it,   as two possible answers add up to the same sum (13).
Then he gave the information that he had an eldest child and this automatically eliminated the other combination.


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## Cabaiste (30 Jul 2008)

macnas said:


> Leghorn is correct
> 
> The ages must be one of the following
> 
> ...


 
If the answer is based on 1 kid being the eldest then couldn't it still be the 1 / 6 / 6  combination. The 6 year olds could be Irish Twins! In fact, even if they were regular twins, 1 of them still had to be born first and is therefore eldest?

Suppose I am just being pedantic about it!


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