# Is there a quick method to solve a problem involving guessing the correct weighs?

This is a problem I often see in exams, it involves calculating the least number of weighing operations needed to get the weight asked.

It states this:

Allison has a bag of rice which weighs more than 17 ounces and she has a two pan balance and 4 different weights being 3 oz, 4 oz, 7 oz and 11 oz respectively. How many times at least does she has to use the balance to get 17oz of rice?

I tried to solve this problem by summing up all the weighs:

$3\,\textrm{oz}+4\,\textrm{oz}+7\,\textrm{oz}+11\,\textrm{oz}=25\,\textrm{oz}$

Since what it is being asked is to get $17$ ounces what I tried to do is to find the possible combinations which put in each pan so that they sum up to 17.

But grouping either $11$ in one side and $3+4+7$ in the other the resulting weight is $3$. If I choose not to use one of the weights let's say $11$ and $4+7$ it cancels both sides, while $3+7$ produces $1$ ounce in the other side therefore it cannot be used, if it is $3+4$ it produces $4$ ounces. So it seems it is impossible to get the $17$ ounces at once.

The other choice would be just using in the first attempt just $3$ and $7$, and in the second turn just the weight of $7$ ounce. Therefore the least number of times to use the balance would be $2$. But this answer is not correct.

The method I tried to use I don't think it is right, it is prone to errors and more importantly is tedious which is something I can't use at an exam where time is limited.

Can somebody help me to find a more orderly and logical method step by step to solve these kind of problems other than just guessing and starting to plug in numbers randomly as if I was trying to hit a target?.

## 2 Answers

In this case trying to plug in numbers is the fastest way to solve this. Also keep in mind that you can put a weight together with the rice, effectively subtracting. The solution is thus:

$$11 + 7 + 3 - 4 = 17$$

• Yes I thought about that I can subtract the weight from one side with what is on the other and also put that with the rice. But my question was on How to avoid to double count the same number. If you show the solution plain as that, the mental process which you used to get the answer is hidden from the casual observer. It looks as if a magician took a rabbit out of the hat if you know what I mean. My question was intended to know if there was a table that I can build and fill up or a graphic or some representation so that it allows me to avoid those errors. – Chris Steinbeck Bell Mar 29 '18 at 2:44

Here's how I did it. First, I guessed that you had to use all the weights. Then I said suppose I put $x$ ounces in one pan and $y$ in the other. I want \begin{align} x+y&= 25\\ x-y&= 17 \end{align} so $x=21$ and $y=4$.

This method isn't guaranteed to work, of course. It just checks whether it's possible to do it in one weighing by using all the weights.

• Out of curiosity is this problem related with balanced ternary?. As you rightfully pointed out, what if the result from solving both equations isn't an integer, or if one of the equations results a fraction i.e 2.5 oz or something like that?. How to proceed in that situation? Or should I assume that I can't weight that mass in just one turn?. – Chris Steinbeck Bell Mar 29 '18 at 2:56
• No, you can't conclude that you can't weight that mass in one turn. All you can conclude is that you can't that mass in one turn by using all the weights. So, first you note that you can't do it with one weight, because no weight is $17.$ The you note that there aren't two weights whose sum or whose difference is $17$, so it can't be done with $2$ weights. Then you have to find a way to preclude doing it with $3$ weights, perhaps by excluding one weight at a time and using the method in my answer. Then if it also fails by using $4$ weights, you know it can't be done in one weighing. – saulspatz Mar 29 '18 at 3:03
• I couldn't fit everything in one comment. As far as I know, this has nothing to do with balanced ternary. I think I know what you're referring to, but in that puzzle the weights all turn out to be powers of $3$, unlike this question. – saulspatz Mar 29 '18 at 3:05
• So can I conclude that it cannot be done by using $4$ weights if the system of equations you used results a non integer or a fraction? But again, should I worry if one of the weights be let's say $0.5$ oz? or would that method also work?. – Chris Steinbeck Bell Mar 29 '18 at 3:21
• If you apply the method you get the total amount of weight in each pan (other than rice). In this case, it worked out neatly to one of the weights. But if we were trying to measure $5$ ounces of rice, we would get $y=10$ and we would have to realize that we could use a $3$ oz. weight and a $7$ oz. weight. – saulspatz Mar 29 '18 at 3:28