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An Innkeeper purchased an odd number of wine kegs, and one keg of beer. These are shown in the picture above. The Innkeeper then sold an amount of the wine to one woman and twice this amount to another, but kept the beer to herself.
Can you figure out which keg contains the beer? (the kegs were sold just as they were purchased, without manipulating the contents)
[answers can be submitted below in the comments section if you can figure it out. To give everyone a fair chance, however, will keep submissions under wraps for about 2 days. Thanks]
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Total gallons = 119.
To divide the wine as described, the gallons of wine must be evenly divisible by 3. The 20 gallons keg is the only one that can be subtracted from 119 to leave a number divisible by 3, so the 20 gallon keg is the beer, leaving 99 gallons of wine (to go with the 99 bottles of beer on the wall). So, 33 gallons must have been sold to one woman, and 66 to another. Hence, the 18 and 15 gallon kegs were sold to the first woman, the 31, 19, and 16 gallon kegs to the second, and the 20 gallon keg of beer was kept.
The 20 gallon barrel contains the beer. The Innkeeper sold the 15 and 18 gallon barrels to the first woman and the remaining wine barrels to the second woman.
31+19+16 = 2(18+15) So the 20gal. keg has all the beer. Mmm, beer…
The first woman bought the 15 & 18 gall. barrels (33 gal); the second woman bought the 31, 19, and 16 gall. barrels (66 gall.). The 20 gallon, which is left has the beer.
The beer keg is the 20 gallon keg.
Pretty much solved it with trial and error except using the following guides:
Sold amount of wine#1 (even or odd #)
Sold amount of wine#2 (even #, because 2 times any number is always even)
Started with the smaller numbers for wine#1, since 2X would need to be a large number.
Guessed 2 kegs for wine#1, 3 kegs for wine #2.
Looked for “even” sums for wine#2
My Second interation found:
Wine#1 = 15 and 18 gals = 33 gals
Wine#2 = 19, 31 and 16 gals = 66 gals
Remainder keg = 20 gals.
this problem is really cool, mostly because of the way of approaching to the answer:
1) Sum all the kegs: 15+31+20+19+16+18=119
now you now that subtracting the content of a Keg( the beer keg) you had to have a relationship 2:1 in the quantities of the remainig kegs:
then you get this mathematically:
W + 2W + B = 119
3W =119 - B so you know that 119 - B have to by multiply of 3 so the only number that satisfy that is: 20 119-20= 99
W= 33 (18+15 ); 2W= 66(31+19+16)
B= 20
the keg that contains the beer is the 20 keg
The beer is in the 20 gallon container. The equation is x + 2x +y =119 (total gallons in all the kegs). Simplify to 3x + y = 119. Then tried replacing each of the quantities of 6 kegs with y. The only one that worked leaving a number divisible by 3, was 20. He sold 33 to the first person, 66 to the second - which leaves 20 for himself.
the 20 Gal keg contains the beer
She sold the 16, 19, and 31 Gal to the first lady. This was 66 gallons.
She sold the 15 and the 18 Gal to the second lady. This was 33 gallons, half as much as the first.
This leaves the remaining keg as the 20.
The 20 gallon keg contains the beer. For each 3-combination of numbers, compare the sum to the sum of possible 2-combinations made out of the remaining 3 kegs which aren’t included in the current 3-combo. The solution is unique {31, 19, 16} { 15, 18 }.
31 + 19 + 16 = 66
15 + 18 = 33
By process of elimination, 20 is the answer.
20
sum of the barrels is 119
amount of wine must be a multiple of 3 (1+2)
so amount of beer must be 2 more than a multiple of 3
This explanation could be made fancier and more concise with modular arithmetic, but this isn’t a math site.
The 20 gal keg is the beer.
The 20 Gallon keg contains the beer. Suineg, Notovitz, Lath, & Tommy show how you can narrow in on the answer.
Suineg-sorry about that; we’re still tweaking the comment system.
>>Can you figure out which keg contains the beer?
No, you cannot with the information given. Besides the answer given above, the innkeeper could have also sold a wine keg containing 18 gallons to one woman and wine kegs containing 20 and 16 gallons to another, leaving three kegs, one of which contains beer. We do not know what the other two may contain. The original question does not specify *five* kegs of wine, only an odd number, all of which are shown in the picture (which may contain other kegs in addition to those specified).
While technically I would agree with you, mad_pear, I think you might be splitting hairs a bit too fine.
Since the puzzle states that the purchased kegs are shown in the picture, I think it is more reasonable to assume that picture exactly represents the purchased kegs than to assume that the picture shows the purchased kegs plus some others thrown in as well. However, to be most precise it is good to make that assumption explicit.
To be even more explicit about assumptions, the puzzle states that the innkeeper kept the beer to herself, so I think it reasonable to assume that the innkeeper kept only the beer.
I agree with mad_pear. If she kept the 15, 31, and 19, one of them could be the beer and the other two wine. It only says she sold “an amount” of the wine, not necessarily all of it.
I think it’s a bad idea to add extra assumptions to the problem just to make it work out. They should be there in the first place.