Question 636279
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<table border="1" bordercolor="#336600" style="background-color:#CCFF99" width="60%" cellpadding="3" cellspacing="3" style="text-align:center">
	<tr>
		<td> <b>28 L bucket</td>
		<td><b>15 L bucket</td>
		<td><b>13 L bucket</td>
		<td><b>Move</b></td>
	</tr>
	<tr>
		<td>28</td>
		<td>0</td>
		<td>0</td>
		<td>Start</td>
	</tr>
	<tr>
		<td>13</td>
		<td>15</td>
		<td>0</td>
		<td>fill 15 from 28</td>
	</tr>
	<tr>
		<td>13</td>
		<td>2</td>
		<td>13</td>
		<td>fill 13 from 15</td>
	</tr>
	<tr>
		<td>26</td>
		<td>2</td>
		<td>0</td>
		<td>empty 13 into 28</td>
	</tr>
	<tr>
		<td>26</td>
		<td>0</td>
		<td>2</td>
		<td>empty 15 into 13</td>
	</tr>
	<tr>
		<td>11</td>
		<td>15</td>
		<td>2</td>
		<td>fill 15 from 28</td>
	</tr>
	<tr>
		<td>11</td>
		<td>4</td>
		<td>13</td>
		<td>fill 13 from 15</td>
	</tr>
	<tr>
		<td>24</td>
		<td>4</td>
		<td>0</td>
		<td>empty 13 into 28</td>
	</tr>
	<tr>
		<td>24</td>
		<td>0</td>
		<td>4</td>
		<td>empty 15 into 13</td>
	</tr>
</table>


Continue the pattern:


1. Fill 15 from 28


2. Fill 13 from 15


3. Empty 13 into 28


4. Empty 15 into 13


Reducing the amout of oil in the 28 liter container by 2 liters each cycle.  After several more cycles of the pattern, after step 2 you will have 1 liter in the 28 L container, 14 liters in the 15 liter container, and a full 13 liter container.  Then when you perform the last iteration of step 3, you will have the desired 14, 14, 0 configuration.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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