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Using only two jugs as measuring devices (one 3 gallon jug and one 5 gallon jug) and a supply of only 9 gallons of water, how can you get EXACTLY 4 gallons into one of the jugs?

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Using only two jugs as measuring devices (one 3 gallon jug and one 5 gallon jug) and a supply of only 9 gallons of water, how can you get EXACTLY 4 gallons into one of the jugs?

Fill the 5 gallon jug, dump it out, then fill it with the remaining 4 gallons.

There is a riddle like this that is very old. But your added constraint of there only being 9gallons makes it even easier.

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Let "S" be the nine gallon supply, "T" be three gallons, and "F" be five gallons...

S T F

9 0 0 - Start

6 3 0 - Pour 3 into T

6 0 3 - Empty T into F

4 0 5 - Fill T from S

-- Four gallons remaining in the S (supply)

now to get it into one of the jugs...

Edited by Scottie
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Um, if your supply is exactly 9 gallons, then fill the 5 gallon jug once, dump it out and pour the remaining supply into it--you don't even need the 3 gallon jug...however, I don't think that is the answer you were looking for so...

answer #2

fill both the 5 and 3 gallon jugs, dump out the 5, pour the 3 into the 5 and then add the remaining gallon of supply to the 5...again, this assumes a supply of exactly 9 gallons, so...

answer #3

really, fill the 3, dump it into the 5. Fill the 3 again, dump what you can into the 5, leaving one gallon in the 3. Dump out the 5 and put the one gallon from the 3 into it. Fill the 3 again and add it to the one gallon already in the 5.

Edited by BTH
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Um, if your supply is exactly 9 gallons, then fill the 5 gallon jug once, dump it our and pour the remaining supply into it--you don't even need the 3 gallon jug...however, I don't think that is the answer you were looking for so...

answer #2

fill both the 5 and 3 gallon jugs, dump out the 5, pour the 3 into the 5 and then add the remaining gallon of supply to the 5...again, this assumes a supply of exactly 9 gallons, so...

answer #3

really, fill the 3, dump it into the 5. Fill the 3 again, dump what you can into the 5, leaving one in the 3. Dump out the 5 and put the one gallon from the 3 into it. Fill the 3 again and add it to the one gallon already in the 5.

:duh:

I assumed we couldn't discard extra water by dumping it out... but OP makes no mention of it. Silly me!

Edited by Scottie
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Um, if your supply is exactly 9 gallons, then fill the 5 gallon jug once, dump it our and pour the remaining supply into it--you don't even need the 3 gallon jug...however, I don't think that is the answer you were looking for so...

answer #2

fill both the 5 and 3 gallon jugs, dump out the 5, pour the 3 into the 5 and then add the remaining gallon of supply to the 5...again, this assumes a supply of exactly 9 gallons, so...

answer #3

really, fill the 3, dump it into the 5. Fill the 3 again, dump what you can into the 5, leaving one in the 3. Dump out the 5 and put the one gallon from the 3 into it. Fill the 3 again and add it to the one gallon already in the 5.

Yeeeaah this is the answer I was looking for...

Your answer #3 is what I was looking for: "fill the 3, dump it into the 5. Fill the 3 again, dump what you can into the 5, leaving one in the 3. Dump out the 5 and put the one gallon from the 3 into it. Fill the 3 again and add it to the one gallon already in the 5" MAKING 4! I guess my wording should have been more specific. Whatever, it's what I could remember from Die Hard 3

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Welcome to the Den, talon. There are many useful information about this forum in rules (see link "Important: READ BEFORE POSTING" at the top). For instance, perform a quick search before you post a riddle that you have not created. This way we ensure that you always get new riddles and don't have to see the same riddle each week.

BTW, this is one of my favorite puzzles and I have dedicated 1 whole page to such weighing and measuring puzzles.

Thread locked.

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