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# 3 Men and a little ladder

18 replies to this topic

### #1 Nep Ton

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Posted 23 January 2012 - 11:12 PM

3 men: Arnold, Bryan and Colin decide to purchase a ladder between them.

They each contribute 5 coins and give these 15 coins to David who is going to purchase the ladder for them.

David buys the ladder for only 10 coins and feels guilty about keeping all 5 coins change.
He decides to give one coin each back and keep only 2 for himself.

On giving one each back this means that Arnold Bryan and Colin have now spent 4 coins each and David kept 2

3x4=12
David’s share of 2 coins means a total of 14

But we started this question with 15 coins where did the other one go?

I will not accept quantitive easing as a solution. LOL
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### #2 bekabeh

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Posted 23 January 2012 - 11:54 PM

Spoiler for Counting
A much better version of this puzzle has been posted using a hotel and busboy scenario that introduces a fourth party. This puzzle is directly clear.
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### #3 davight

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Posted 23 January 2012 - 11:55 PM

Spoiler for My opinion

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### #4 Nep Ton

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Posted 24 January 2012 - 12:10 AM

Spoiler for Counting
A much better version of this puzzle has been posted using a hotel and busboy scenario that introduces a fourth party. This puzzle is directly clear.

If he gave each 2 back he would need to return 6 yet only has 5
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### #5 Snappyblue

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Posted 24 January 2012 - 12:59 AM

It says that Arnold, Bryan and Colin have only spent 4 coins each and Daniel kept 2. e.g 3x4=12,+ 2=14
Arnold, Bryan or Colin have actually spent only 3 coins each because they each got one back: 3x3=9
The three of them kept 1 coin each: 3x1=3
Daniel kept 2 coins: 1x2=2
The two coins Daniel kept were from two of the other three. e.g Arnold & Bryan. The remaining coin is from one of the other three. e.g Colin
3x3=9, so the missing coins must havin(g been spent on the ladder: 9+1=10

I hope this isn't too confusing, I couldn't find a way to put it into context. LOL)
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### #6 commentator

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Posted 24 January 2012 - 01:04 AM

they spend \$12, ladder only costs 10\$, so daniel keeps the extra two. there is no reason to add 2 to 12.

ps i <3 you asdf
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### #7 nso4th

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Posted 24 January 2012 - 01:05 AM

There are no missining coins because David contributed nothing. Arnold, Bryan and Colin each paid 4 coins of that 10 went for the ladder and 2 were kept by David. Perhaps as a commission? 10 for the ladder + 2 for David = 12 coins
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### #8 Nep Ton

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Posted 24 January 2012 - 01:10 AM

But initially 15 coins were on the playing field. Now there are only 14. One must have gone somewhere
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### #9 xander919

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Posted 24 January 2012 - 01:33 AM

So...
The ladder real price, divided for the three, is 3,333(3).
Because there isn't 0,333(3) coins, one of the coins had to be spent to pay the ladder : 0,33(3)*3=1 coin
So, one of the coins from one of the people had to be spent to pay this decimal part of the price.
We can't use a determined number of coins equal from each buyer. There is one of them that will have to spend one more on this buying.
(of course, the rest of the problem covers the whole problem of these)
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### #10 Snappyblue

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Posted 24 January 2012 - 01:38 AM

The other coin went towards the ladder because:
2 for David
3 for Arnold, Bryan and Colin (one for each)
10+2+3=15
Therefore there is no missing coin.

(Earlier when I said Daniel it was meant to be david. LOL)
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