6 balls, 2 with New York logo. 2 with California logo, and lastly 2 with Texas logo. In each pair, one ball heavier than the other. The heavier ones weigh the same, so do the lighter ones. You are provided a scale. Find all the heavier ones with 2 weighings.
I would start with 3 against 3 since this way there is no possibility two sides will weigh the same. ...
On
Denote by lower case the lighter ball, and by upper case the heavier ball. So our balls are $N,n, C,c, T,t$.
To start, measure a California and a New York against the other New York and a Texas.
Then the possibilities are
If both sides are equal, weight the New Yorks against each other for the reveal (and be content that we know the weights of the two others, even though we never touched them).
WLOG suppose the left side is heavier than the right for the remainder of this explanation, so that we have one of
Notice that the big $N$ is always on the heavier side. This is the extra piece of information that we'll use. Swap the left New York and the right Texas, yielding (in order)
which, as you can see, are easily distinguishable. And thus with two measurements, we have it. (Notice that we never touched the other two balls!)
On
This doesn't really differ from the logic of the answers posted by Alex Zorn and mixedmath, but the presentation might carry some "weight."
Let's label the balls $N$ and $n$ for New York, $C$ and $c$ for California, and $T$ and $t$ for Texas. Our job is to partition these into a "heavy" set $H$ and a "light" set $L$ with the uppercase and lowercase going into different sets for each letter.
For the first weighing, compare $N$ and $C$ to $n$ and $T$. It's convenient to think of the labels as weights and express the outcome as $N+C\approx n+T$, where "$\approx$" is either $>$, $=$, or $<$. By an elementary relabeling symmetry (swapping the labels $N$ and $n$), it suffices to consider only the first two cases.
If $N+C\gt n+T$, we can conclude that $N\in H$ and $C\ge T$. This implies $H$ is one of three possibilities: $\{N,C,T\}$, $\{N,C,t\}$, or $\{N,c,t\}$. We can determine which by comparing $c$ and $T$ in the second weighing.
If $N+C=n+T$, then $N$ and $C$, and likewise $n$ and $T$, must be in different sets, which we can express as
$$\{H,L\}=\{\{N,T,c\},\{n,t,C\}\}$$
It's clear from this that we can determine which set on the right hand side is $H$ and which is $L$ by simply comparing, say $N$ and $n$.
The key to this puzzle is the following idea (common to logic puzzles like these): There are 8 possible configurations of the balls. That means, in 2 weighings, you need to allow for at least 8 different outcomes.
If each weighing you do is unbalanced, there are only four possible outcomes to the weighings, depending on which side the scale tips. This is not enough. That means you need to use weighings where it is possible for the scale to balance, and where you can use that information to your advantage.
The trick is the first weighing, which is as follows:
Ny and Ca on the left, Ny and Tx on the right.
If they are balanced, then you just need to weigh the new york balls against each other to figure out which one is heavy. Then the ball paired with the heavy ball on the first weighing must be light, and the ball paired with the light ball on the first weighing must be heavy.
If they are not balanced, then the heavy side must have had the heavy Ny ball. Assume the left side was heavier (the right side is the same). If Ca1 and Tx1 denote the balls you have already weighed, and Ca2, Tx2 are the others, the remaining possibilities are:
Ca1 > Ca2 and Tx1 > Tx2
Ca1 > Ca2 and Tx1 < Tx2
Ca1 < Ca2 and Tx1 < Tx2
The fourth possibility is not allowed because the scales would have balanced on the first weighing. Now there are three possibilities left, and you should be able to decide between them with one weighing, which is not too hard to figure out:
Ca1 and Tx1 on the left, Ca2 and Tx2 on the right.