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Topic: Cons with a price on their head (Read 688 times) |
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It occurred to me that we could combine the 7-11 puzzle with the hat puzzle with the various prisoners puzzles to make a real challenge. Here goes:
Our old friend, the sadistic warden takes 6 prisoners and puts a hat on everyone's head with a price on it. We'll go around the room once, he said. Each prisoner can either announce the total of all the prices he can see, or the product of all the prices he can see. Just say the number, not which one it is. Or he can announce that he knows the price on his own hat and say nothing else.
Then we'll go around the room a second time and each person who can name their own price goes free.
But there are a couple problems. One, the prisoners all hate each other and won't give any information if they don't have to. And two, you're blind.
And you're first. The warden takes a little pity on you and answers for you, "$9.99."
The second prisoner says, "I know my number."
The third prisoner says, "I know my number." And on down the line.
It comes back to you.
The warden laughs and says, "Tough luck. Okay, I'll give you one little hint. It's more than a penny, and none of you guys is worth more than 10 bucks."
And you answer...
(and just to show off, you tell him the prices on all 7 hats)
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Re: Cons with a price on their head
« Reply #1 on: Mar 7th, 2003, 5:35pm » |
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Duh, I can't count. Make that last line, and just to show off, you tell him the prices on all 6 hats.
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Re: Cons with a price on their head
« Reply #2 on: Mar 7th, 2003, 9:39pm » |
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One more modification.
I guess you couldn't be sure of all 6 prices on the information given, so ...
Change the warden's answer for you to "$9.57"
And for the warden's final hint, "You're not the cheapest and you're not the highest price."
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Re: Cons with a price on their head
« Reply #3 on: Mar 8th, 2003, 9:03am » |
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It occurs to me now that this whole puzzle would need to be reworked, as the first price mentioned would not have to be of the 7-11 nature in order for the other prisoners to all figure out their own hats. You have my permission to completely ignore this whole thread.
Oh, you have ignored it. Carry on.
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aero_guy
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Re: Cons with a price on their head
« Reply #4 on: Mar 9th, 2003, 12:27pm » |
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Just waiting for things to get straight...
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Re: Cons with a price on their head
« Reply #5 on: Mar 9th, 2003, 6:27pm » |
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Okay, I don't see any way to salvage that riddle, so, just to make some use of the data I put together here's a version of the 7-11 riddle.
Here are the prices of items in the store.
$0.32 $1.30 $3.20
$0.42 $1.75 $4.00
$0.50 $2.00 $5.00
$0.84 $2.08 $6.50
$1.00 $2.50 $7.50
$1.12 $2.60
$1.25 $3.00
One customer bought 3 items. One customer bought 4 items. One customer bought 5 items. One customer bought 6 items.
They all paid the same price. They all would have paid the same price even if the clerk had multiplied the prices instead of adding. Oh, and just because that would have been too easy, the store also had a 17 year old package of Slim Jims that nobody bought, so one price won't be used.
So how much were the Slim Jims?
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Re: Cons with a price on their head
« Reply #6 on: Mar 10th, 2003, 6:13pm » |
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um, you can solve it now. Honest, it's ready. No, wait, change the Slim Jims to a, um, nevermind. It's ready.
And if you're afraid it's just going to be a whole lot of guesswork, the solution actually requires very little guessing. (And answering the Sllim Jim question takes no guessing).
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aero_guy
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Re: Cons with a price on their head
« Reply #7 on: Mar 11th, 2003, 9:15am » |
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You can solve the Slim Jim Part even without knowing that you could add the prices together to get the same as when you multiply them.
answer: Multiply all the numbers by 100. You know that the four prices, which come from multiplying the individual items together, will be identical. Therefore, the multiple of all the prices, minus the Slim Jims, will be n4 where n is an integer. So, we factor all the prices, divide the number of times each factor occurs by four and keep the remainder. This gives us a 22 and a 5, which together make 20. Only 260 and 320 have this as a factor, so one of them are the Slim Jims. 20*p4, where p is a prime number will give us the answer, and the only price that fits this is 320. Therefore the Slim Jims cost $3.20
Multiplying all the other prices together and taking the fourth root gives $10.92 as what each person spent. The price ends in 2, so person A (not necessarily the person who bought 3 items) bought .32, person B .42, C 1.12, and D .84 and 2.08.
Everybody but person A now has a factor of 7, so person A also bought the only other factor of 7, 1.75.
A________B_______C________D
.32______.42_____1.12______.84
1.75______________________2.08
(needed the underscores to make columns come out since table editor doesn't seem to work with hide) Hmm, if A is going to end in a 2, it needs a .05 to add on. The only remaining number to end in that is: 1.25.
A________B_______C________D
.32______.42_____1.12______.84
1.75______________________2.08
1.25
Now we need a nine in the first decimal place. Thus for A we need .6 more, B needs .5, C .8, and D needs nothing. We only have .3, .6, and .5 to work with. We have a bunch of choices where to get the .5 from, but the others are limited to:
A________B_______C________D
.32______.42_____1.12______.84
1.75_____________1.30_____2.08
1.25
2.60
The only entry without a factor of 13 is now B, and that can only be gotten from 6.50.
A________B_______C________D
.32______.42_____1.12______.84
1.75_____6.50____1.30_____2.08
1.25
2.60
A and C need a factor of 3, which must come from 3.00 and 7.50. 7.50 would put A over the top, so it must be 3.00. That is five items, so there can only be one more which must be 2.00:
A________B_______C________D
.32______.42_____1.12______.84
1.75_____6.50____1.30_____2.08
1.25_____________7.50
2.60
3.00
2.00
C is now 1.00 away from target and the lowest we have left are .50 and 1.00, so it must be the 4 items and have 1.00. Either B or D will be the 3 item shopper and only B is close enough to the target price to make it with any of the remaining numbers, so B must have 4.00. The rest now fall to D. No guessing needed.
A________B_______C________D
.32______.42_____1.12______.84
1.75_____6.50____1.30_____2.08
1.25_____4.00____7.50_____5.00
2.60____________1.00______2.50
3.00______________________.50
2.00
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| « Last Edit: Mar 11th, 2003, 9:30am by aero_guy » |
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