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I am having a hard time solving the following puzzle. Could you please me to figure it out?

A chemist has a set of five weights. She knows that it includes one 1-gram weight, and also one each 2-, 3-, 4-, and 5-gram weights, but because they are unmarked, she has no way of telling them apart except by placing them on a balance. She may place any combination of weights on each of the two pans and determine if one side is heavier than the other or if they balance. Show how in five weighings she can identify each of the weights.

Thank you.

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  • $\begingroup$ Welcome to Math.SE. Please show your efforts so that someone can guide you. $\endgroup$ – Shailesh Jul 1 '16 at 8:36
  • $\begingroup$ Isn't there an SE for puzzles? $\endgroup$ – Oscar Lanzi Jul 1 '16 at 10:34
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Choose four weights $a$, $b$, $c$, $d$, leaving $e$ out, and use three weighings to compare each pair of them against the other pair. There are three possible outcomes.

  1. One of the three weighings balances, say $a+d = b+c$, and in the remaining two weighings $a$ is always heavier than $d$. In the next weighing, pit $b$ against $c$. Without loss of generality, $b > c$. We know $a>b>c>d$ and $e$ is odd, so there are three possibilities left for $(a,b,c,d,e)$: $(5,4,3,2,1)$, $(5,4,2,1,3)$, $(4,3,2,1,5)$. Weighing $a+c$ against $b+e$ will distinguish between them.
  2. One weight, say $a$, is always on the heavier side. Since none of the first three weighings were equal, we know $e$ is even. From this we can deduce $a = 5$ and $e = 4$. Now weigh $b+c$ against $e$, which is 4. This gives enough info to identify $d$. Then weigh $b$ against $c$ to finish.
  3. One weight, say $a$, is always on the lighter side. As before, we deduce $a = 1$ and $e = 2$. Now weigh $b$ against $c$, and without loss of generality $c > b$. Next, weigh $c$ against $a+d$. If $c$ is heavier, then $(a,b,c,d,e) = (1,4,5,3,2)$. If $c = a+d$, then $(a,b,c,d,e) = (1,3,5,4,2)$. If $c$ is lighter, then $(a,b,c,d,e) = (1,3,4,5,2)$.
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  • $\begingroup$ In 3. it should be "say $a$", but a good answer. $\endgroup$ – Ross Millikan Jul 6 '16 at 4:07
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Okay, this isn't a complete solution as this does seem intricate but maybe this is a good starting point.

Note there are $5! = 120$ possible ways the weights can be determined. Each weighing will have $3$ possible outcomes. The weight balances, the left is heavier, and the right is heavier. If you can figure out a way to weigh things so that each outcome covers the same number of possibilities, the first weigh will reduce your point distributions to $40$ the second weighing to $14$ the third to $5$, the fourth to $2$ and the fifth to $1$.

That would be ideal.

But it can't be done.

So I evaluated how the distributions are accounted for by different combinations of weigh.

Example: If I weigh one weight (call it A) and another (call it B) then A < B accounts for 60 cases. B > A for 60 cases. But B = A will never occur. This is not a very desirable distribution.

If I weigh one weight (A) against two weights (BC) then, A > B account for A = 4; B= 1;C=2 OR A=4; B=2; C=1 OR A=5;B=1;C=2 OR A=5;B=2; C= 1; etc. basically A > BC has 6 possible solutions. A = BC has 8 possible solutions and A < BC has 106 solutions. This is a TERRIBLE distribution.

I won't bore you but if I weigh two against two I get $AB < CD$ has $48$ solutions. $AB = CD$ has $24$ solutions and $AB > CD$ has $48$ solutions. This is the best distribution.

Here's where I gave up and took a leap of faith. I figured that the second weighing between $AC$ and $BE$ would be the best second weighing for even distributions. It intuitively seem to redistribute the light pair among the heavy pair and to get the 5th unknown in early.

But I dunno.

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