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You are watching: 5 gallon jug and 3 gallon jug

Posted February 4, 2013 through Presh Talwalkar. Read around me, or email me.

This is a trouble that is prominently featured in the movie *Die hard With a Vengeance*.

You have actually a 3-gallon and also a 5-gallon jug that you can fill native a spring of water.

The difficulty is to fill among the jugs with *exactly* 4 gallons of water. Exactly how do you do it?

**Video: Die hard 3 The Water Jug mathematics Riddle**

.** ."All will certainly be fine if you usage your psychic for your decisions, and also mind only your decisions." since 2007, i have dedicated my life to share the delight of game theory and mathematics. betterworld2016.org now has actually over 1,000 cost-free articles v no ads many thanks to community support! aid out and get early accessibility to short articles with a pledge top top Patreon.**

**. . . . . . M i N D . Y O U R . D E C i S i O N S . W E E K l Y . P U Z Z together E . . . . Answer come the water jug riddle**

**Math In Movies: The Water Jug Riddle native Die tough With A Vengeance (1995)**

The an initial attempt human being have is to try and calculation 4 gallons by including 3 gallons the water come 1/3 of the 3 gallon jug. But the riddle is asking for a an exact measurement and so this solution cannot work.

The trick is to realize that:

5 – 3 = 2** and also 5 – (3 – 2) = 4**

**Here is one way to discover the answer:**

**Incidentally, the reason we can find a solution is since the two numbers 5 and also 3 are fairly prime–that is, they have no usual divisors. We have the right to actually generate any kind of volume the water native 1 to 5 (in fact, we did gain measurements the 1, 2, 3, 4, and also 5 along the means in our solutions).**

**The more general trouble is finding integer remedies for the equation ax + by = c. Solutions (x,y) exist when the greatest usual divisor the a and b is a element of c. This is an old problem. For more, read this article: http://mathforum.org/library/drmath/view/51595.html ——————————————————— (This is the solution in plain text)**

**1. Fill up the 5-gallon jug 2. Fill up the 3-gallon jug using the water indigenous the 5-gallon jug (leaving 2 gallons in the 5-gallon jug) 3. Pour out the 3-gallon jug right into the spring 4. Carry the 2 gallons native the 5-gallon jug into the 3-gallon jug 5. Fill up the 5-gallon jug 6. Transport water indigenous the 5-gallon jug until the 3-gallon jug is full. Because the 3-gallon jug currently had 2 gallons the water, over there is room for just 1 gallon. 7. The lot of water in the 5-gallon jug is exactly 4 gallons**

**If we denote the components of the jugs together the pair (5-gallon jug amount, 3-gallon jug amount), the sequence of events is:**

**(5, 0)–>(2, 3)–>(2, 0)–>(0, 2)–>(5, 2)–>(4, 3)**

**That’s no the only path. We can additionally consider the route:**

**1. To fill up the 3-gallon jug 2. Carry to the 5-gallon jug 3. Fill up the 3-gallon jug again 4. Move water to to fill up the 5-gallon jug, leaving 1 gallon in the 3-gallon jug 5. Empty out the 5-gallon jug 6. Transfer the 1 gallon come the 5-gallon jug 7. To fill up the 3-gallon jug and transfer that to the 5-gallon jug 8. The 5-gallon jug contains 4 gallons the water**

**The sequence here is:**

**(0, 3)–>(3, 0)–>(3, 3)–>(5, 1)–>(0, 1)–>(1, 0)–>(1, 3)–>(4,0)**

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