Here is a puzzle that I have been having difficulty with. Let me know what you think.
Your friend and you are in a standard American mall and see a vending machine. Your friend pulls out a crisp $1 bill to buy a 95 cent candy bar, but then realizes that the machine takes exact change only. Your friend turns to you and asks if you could break a dollar. You reach into your pocket and fish out 6 coins, which add up to $1.15 saying, "Well, with this amount I should be able to break a bill ... but it appears that I cannot!" Upon further examination, you see that you cannot give change for a 50 cent piece, a quarter, a dime, or even a nickel with the 6 coins you have.
Frusturated, your friend asks you to just buy him the candy bar and you realize... you cannot buy the 95 cent candy bar.
What are the 6 coins that you have?
My attempt is in the spoiler. Only read the spoiler if you attempted the puzzle by yourself to garuntee a fresh mind!
Spoiler
What I thought was that it is a 50 cent piece, a quarter, and four dimes. The only way this could work (for him not to be able to buy the 95 cent candy bar) is if at least 3 of the dimes are Canadian currency.
2011-12-28, 07:06 AM
Locked
2 $.25
1 $.50
3 $.05
2011-12-28, 07:35 AM
nRxUs
Quote:
Originally Posted by Locked
2 $.25
1 $.50
3 $.05
Except with those coins you CAN give change for a 50 cent coin, for a one dollar bill, and for a dime.
2011-12-28, 01:48 PM
Riyuran
I'm not sure this is even possible.
You can't have more than 1 $.50, because then you'd be able to break $1.
You can't have more than 1 $.25, because then you'd be able to break $.50.
With 1 $.50 and 1 $.25, you need $.40 more in four coins. The only way this is possible is with 4 $.10. This, however, allows you to add up to $.95.
Without one of the $.50 or $.25 (or both of them), it's impossible to add up to $1.15 in only six coins.
Am I interpreting the question incorrectly?..
2011-12-28, 01:49 PM
Stereo
Gonna put my thinking in spoiler too.
Spoiler
You reach into your pocket and fish out 6 coins, which add up to $1.15 saying, "Well, with this amount I should be able to break a bill ... but it appears that I cannot!" Upon further examination, you see that you cannot give change for a 50 cent piece, a quarter, a dime, or even a nickel with the 6 coins you have.
conditions:
total = 1.15
6 coins
no subset of more than 1 totals $1
no subset of more than 1 totals $0.50
no subset of more than 1 totals $0.25
no subset of more than 1 totals $0.10
no subset of more than 1 totals $0.05
ok.
I am going to assume that, given the question refers to a 50 cent piece, the possible coin values are
0.01 0.05 0.10 0.25 0.50
Basically I start with the largest single coin satisfying the conditions
0.50
Then add the next largest while considering the conditions
0.25
Then the next...
0.10 + 0.10 + 0.10 + 0.10
test it:
total = 1.15
6 coins = true
no subset totals $1
no subset totals $0.50 (except a single 50c coin, which doesn't count)
no subset totals $0.25 (again except the single coin)
no subset of $0.10
of course no subset of $0.05
So:
1 50 cent piece
1 quarter
4 dimes
Unfortunately in doing this I missed the last condition:
No subset can total $0.95
As far as I can tell, this makes the question unsolvable.
With (0.50, 0.25, 0.10) as the first 3 coins:
- another 50 cent coin fails the $1 rule
- another 25 cent coin fails the $0.50 rule
- another 10 cent coin fails the $0.95 rule
- as the current total is 0.85, you cannot make up the rest with 3 nickels
2011-12-28, 01:50 PM
happylight
There's no solution unless there's a coin denomination other than $1, $0.25, $0.10, $0.05 and $0.01.
Here's proof.
Observation 1: $1 coin is not one of the 6 coins. Because adding up to exactly $0.15 from 5 coins is impossible.
Observation 2: You can have at most 1 $0.50 coin. Because if you had more than 1 $0.50 coin then you could change for $1 bill.
Observation 3: You can have at most 1 $0.25 coin. Because if you had more than 1 $0.25 coin then you could change for $0.5.
Theory A: One of the 6 coins is a $0.50 coin.
Implication: The other 5 coins add up to exactly $0.65.
Theory a: One of the 5 coins is a $0.25 coin.
Implication: The other 4 coins add up to exactly $0.40.
Then the 4 remaining coins must be all $0.10 coins. Because if they were a lesser demonination they would add up to less than $0.40.
However this combination can produce exact change for $0.95 with 1 $0.50, 1 $0.25 and 2 $0.10 and therefore is wrong.
Conclusion a: None of the 5 coins is a $0.25 coin.
Implication: The highest possible denomination for the 5 coins is $0.10. However 5 * $0.10 < $0.65.
Conclusion A: None of the 6 coins is a $0.50 coin.
Theory B: One of the 6 coins is a $0.25 coin.
Implication: The other 5 coins add up to exactly $0.90. Since there cannot be more than 1 $0.25, the highest possible denomination of these 5 coins is $0.10.
However 5 * $0.10 < $0.90.
Conclusion B: None of the 6 coins is a $0.25 coin.
Therefore the highest possible denomination for the 6 coins is $0.10.
However 6 * $0.10 < $1.15.
Edit: damnit ninja
2011-12-28, 01:55 PM
Stereo
The unbalanced solution
Canadian dimes are thinner than American ones, and won't work in American vending machines, so if you have 3 of those, plus american $0.50, 0.25, 0.10 it works.
Quarters might be the same.
2011-12-28, 02:03 PM
happylight
-_- then it wouldn't add up to $1.15. It'd add up to $0.85USD + $0.30CAD.
2011-12-28, 02:10 PM
Locked
Quote:
Originally Posted by nRxUs
Except with those coins you CAN give change for a 50 cent coin, for a one dollar bill, and for a dime.
I completely missed that part of the statement, heh.
2011-12-28, 02:24 PM
SaptaZapta
Spoiler
The vending machine is broken, or out of that candy bar.
Nowhere does it say you don't have the 95 cents, it just says you can't buy the candy bar.
(yes, this is somewhat like OP's idea of the Canadian dimes)
2011-12-28, 06:57 PM
Lugin
Or they realize they're Canadians and all the coins they have are Canadian. In an American mall.
The currency type possessed is never mentioned. And looking at user flags...
Anyway, the six coins are a 50 cent piece, a quarter, and 4 dimes.
95 cents is just the goal. The challenge is to exchange the $1 bill for coins (Can't do it).
2011-12-28, 09:36 PM
modular
you realize the confederacy lives on in the south when you pull out a $1 coin and 5 3 cent pieces
...this problem don't wurk in amurrrka
2011-12-29, 07:34 PM
Lugin
Figured out the actual reason you can't buy it.
American vending machines don't accept coins larger than quarters.
Considering how uncommon 50 cent pieces are anymore, it makes sense.
2012-01-03, 05:27 AM
Derosis
Quote:
Originally Posted by modular
you realize the confederacy lives on in the south when you pull out a $1 coin and 5 3 cent pieces
...this problem don't wurk in amurrrka
Hey... F'uck you.
They're all using invalid coins, or coins of a different version i.e: 2 cent coins, etc.
2012-01-03, 12:49 PM
modular
Quote:
Originally Posted by Derosis
Hey... F'uck you.
They're all using invalid coins, or coins of a different version i.e: 2 cent coins, etc.
i'm from tx/la
pineapple you too
2012-01-03, 10:24 PM
Holypie
Quote:
Originally Posted by modular
i'm from tx/la
pineapple you too
Standard american mall. That probably means that there are no 3 cent coins.
2012-01-06, 04:37 PM
modular
what does the mall have to do with what's in your pocket?
2012-01-06, 08:12 PM
Derosis
Quote:
Originally Posted by modular
what does the mall have to do with what's in your pocket?
