Math problem :)

Pre-College Issues High School Life
21

<p>Separate into 3 sets of 4. Weigh 4 vs. 4. If they're equal, the one that's different is in the separated remaining 4. So separate that remaining four into 1 vs. 3. Weigh the 3 vs. 3 balls from the first weighing. If they're equal, then the separated 1 is the lighter/heavier. To find out whether it's heavy or light, weigh it with one of the balls from the first trial.</p>

<p>If not equal, then you already know if the set is lighter or heavier. Take the 3 and separate into 1 vs 2. Weigh the two as 1 vs. 1. If equal, then the bad ball is the separated 1. If not equal, then you can tell which one it is because you already know which is heavier/lighter.</p>

<p>I'm not sure how to do it if the initial weighing is not equal, though.</p>

22

<p>[- indicates weighing]</p>

<ol>
<li>(a)OOOO (b)OOOO-(c)OOOO (equal)</li>
<li>(a1)O (a2)OOO</li>
<li>(a2)OOO-(3 from b)OOO
4a. <a href="a1">if equal</a>O-(1 from b or c)o
4b. <a href="a2,%20either%20heavier%20or%20lighter%20than%20b,%20separate%20into%20a2a%20and%20a2b">if not equal</a> (a2a)O (a2b)OO</li>
<li>Separate a2b into one and one: (a2b)O-O</li>
<li>If equal, then (a2a) is the lighter or heavier depending on what you observed in step 4b. If not equal, then you already know which one in (a2b) is lighter or heavier depending on step 4b.</li>
</ol>

23

<p>
[quote]

  1. (a)OOOO (b)OOOO-(c)OOOO (equal)

[/quote]

Can you explain this part? :)</p>

24

<p>Haha ok. Well, I separated the 12 balls into groups of 4. I named the three groups a, b, and c. The dash indicates that I weighed b and c and they turned out to be equal. Sorry if my wording is difficult to understand.</p>

25

<p>The O's are balls. Visual explanation. :)</p>

26

<p>Oh, I see :) Should I post the good solution already or wait? :)</p>

27

<p>wait, what's wrong with my solution?</p>

28

<p>My way works though, right? But only if the first weighing turns out to be equal. I haven't figured out how to do it if they're not equal.</p>

29

<p>I'll post it. There is a number of solutions (including very fancy ones :) but this is the one I love the most. Also, it's not possible to find right ball with a 100% accuracy for fewer then 3 attempts.</p>

<p>Label balls with next numbers:
002
011
020
022
101
112
120
100
201
210
212
221</p>

<p>Then weight balls with 2xx and 0xx:
a) Equal: write 1 somewhere.
b) 0xx heavier: write 0.
c) 2xx heaveri: write 2.
Now repeat the same thing for x0x and x2x and xx0 and xx2.
You must have a three-digit-number. Search for it, if you found it -- that's the odd, if not -- change 0 to 2 or 2 to 0 in a number and search again (that's for heavy/lighter biases).</p>

30

<p>Ah, I just googled it and I came up with this:
"I had already worked this problem out for someone else using pills.
The setting of the problem, as I remember it, goes something like
this. </p>

<p>A wise man has committed a capital crime and is to be put to death.<br>
The king decides to see how wise the man is. He gives to the man 12
pills which are identical in size, shape, color, smell, etc. However,
they are not all exactly the same. One of them has a different
weight. The wise man is presented with a balance and informed that all
the pills are deadly poison except for the one which has a different
weight. The wise man can make exactly three weighings and then must
take the pill of his choice. </p>

<p>Remember that the only safe pill has a different weight, but the
wise man doesn't know whether it is heavier or lighter.</p>

<p>Most people seem to think that the thing to do is weigh six pills
against six pills, but if you think about it, this would yield you no
information concerning the whereabouts of the only safe pill. </p>

<p>So that the following plan can be followed, let us number the pills
from 1 to 12. For the first weighing let us put on the left pan pills
1,2,3,4 and on the right pan pills 5,6,7,8. </p>

<p>There are two possibilities. Either they balance, or they don't. If
they balance, then the good pill is in the group 9,10,11,12. So for
our second weighing we would put 1,2 in the left pan and 9,10 on the
right. If these balance then the good pill is either 11 or 12. </p>

<p>Weigh pill 1 against 11. If they balance, the good pill is number 12.<br>
If they do not balance, then 11 is the good pill. </p>

<p>If 1,2 vs 9,10 do not balance, then the good pill is either 9 or 10.<br>
Again, weigh 1 against 9. If they balance, the good pill is number
10, otherwise it is number 9. </p>

<p>That was the easy part. </p>

<p>What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then
any one of these pills could be the safe pill. Now, in order to
proceed, we must keep track of which side is heavy for each of the
following weighings. </p>

<p>Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against
2,7,8. If they balance, then the good pill is either 3 or 4.<br>
Weigh 4 against 9, a known bad pill. If they balance then the good
pill is 3, otherwise it is 4. </p>

<p>Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side,
then either 7 or 8 is a good, heavy pill, or 1 is a good, light pill. </p>

<p>For the third weighing, weigh 7 against 8. Whichever side is heavy is
the good pill. If they balance, then 1 is the good pill. Should the
weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then
either 5 or 6 is a good heavy pill or 2 is a light good pill. Weigh 5
against 6. The heavier one is the good pill. If they balance, then 2
is a good light pill.</p>

<p>I think that one could write all of this out in a nice flow chart,
but I'm not sure that the flow chart would show up correctly over
e-mail. </p>

<p>-Doctor Robert, The Math Forum"
(source: <a href="http://mathforum.org/library/drmath/view/55618.html%5B/url%5D"&gt;http://mathforum.org/library/drmath/view/55618.html&lt;/a&gt;)&lt;/p>

31

<p>i honestly dont do logic until next year when i take math 2h but i already know it
like biconditional</p>

<pre><code>contrapositive
converse
</code></pre>

<p>inverse
something else
something else and something else so who here is a math smart person!</p>