The two most populous cities and the non-urban region (NUR) of each of three states, Whimshire, Fogglia, and Humbleset, are assigned Pollution Measures (PMs).

These nine PMs are all distinct multiples of 10, ranging from 10 to 90. The six cities in increasing order of their PMs are: Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo.

The Pollution Index (PI) of a state is a weighted average of the PMs of its NUR and cities, with a weight of 50% for the NUR, and 25% each for its two cities.

There is only one pair of an NUR and a city (considering all cities and all NURs) where the PM of the NUR is greater than that of the city. That NUR and the city both belong to Humbleset.
The PIs of all three states are distinct integers, with Humbleset and Fogglia having the highest and the lowest PI respectively.

For how many of the cities and NURs is it possible to identify their PM and the state they belong to?

The correct answer is 9.

Here is the step-by-step logical deduction to arrive at this answer:

Step 1: Determine the PMs of the three Non-Urban Regions (NURs) and six cities.

We are given nine distinct Pollution Measures (PMs) that are multiples of 10, ranging from 10 to 90:

$10,20,30,40,50,60,70,80,90$

The six cities, arranged in increasing order of their PMs, are Blusterburg (B), Noodleton (N), Splutterville (S), Quackford (Q), Mumpypore (M), and Zingaloo (Z):

$B<N<S<Q<M<Z$

The problem states there is exactly one pair of an NUR and a city across all states where the PM of the NUR is greater than the PM of the city. If there is only one such pair, this specific NUR must have a PM larger than the city with the smallest PM (which is Blusterburg). Furthermore, this NUR must be smaller than the second city (Noodleton), or else it would be larger than two cities, creating more than one pair.

Similarly, the other two NURs must have PMs smaller than all the cities, meaning they must be smaller than Blusterburg. Therefore, the first two PMs in the entire sequence must belong to these two NURs, making Blusterburg the third PM. The fourth PM must be the highest NUR, satisfying the condition that it is greater than exactly one city (Blusterburg).

Thus, we can precisely map the 9 PMs as follows:

${\mathrm{NUR}}_1=10$
${\mathrm{NUR}}_2=20$
$B=30$
${\mathrm{NUR}}_3=40$
$N=50$
$S=60$
$Q=70$
$M=80$
$Z=90$

Step 2: Assign PMs to the state of Humbleset.

The only pair where the NUR's PM is greater than a city's PM is the NUR with 40 and Blusterburg with 30. The problem explicitly tells us that both this NUR and city belong to Humbleset.

We know the Pollution Index (PI) formula based on the weights (50% for NUR, 25% for each city) is:

$\mathrm{PI}=\frac{2\times {\mathrm{PM}}_{\mathrm{NUR}}+{\mathrm{PM}}_{\mathrm{City}1}+{\mathrm{PM}}_{\mathrm{City}2}}{4}$

For Humbleset, we have:

${\mathrm{PI}}_{\mathrm{Humbleset}}=\frac{2\left(40\right)+30+C}{4}=\frac{110+C}{4}$

Because the PI must be an integer, the numerator $110+C$ must be a multiple of 4. The number 110 leaves a remainder of 2 when divided by 4. Therefore, the remaining city's PM ($C$) must also leave a remainder of 2 when divided by 4. Out of the available cities (50, 60, 70, 80, 90), the multiples of 10 that are not divisible by 20 (and thus leave a remainder of 2 mod 4) are 50, 70, and 90.

Step 3: Test the possible cities for Humbleset to find the valid scenario.

We test the options for Humbleset's second city to see which leaves Fogglia and Whimshire with valid integers and ensures Humbleset has the highest PI while Fogglia has the lowest.

Scenario A: Humbleset's second city is 50.

This gives ${\mathrm{PI}}_{\mathrm{Humbleset}}=\frac{110+50}{4}=40$.

The remaining cities would be 60, 70, 80, and 90. To give Fogglia the lowest PI, we assign it the lowest available NUR (10). Its numerator would be $2\left(10\right)+C_1+C_2=20+C_1+C_2$. The smallest sum of two remaining cities is $60+70=130$, but $20+130=150$ is not divisible by 4. The next smallest combination divisible by 4 is $60+80=140$. But this would give Fogglia a PI of $\frac{20+140}{4}=40$, tying it with Humbleset. Since the PIs must be distinct and Fogglia must be lowest, this scenario fails.

Scenario B: Humbleset's second city is 70.

This gives ${\mathrm{PI}}_{\mathrm{Humbleset}}=\frac{110+70}{4}=45$.

The available cities for Fogglia and Whimshire are 50, 60, 80, and 90. We pair them so the sums are divisible by 4. The only pairs that sum to a multiple of 4 (so that $20+\mathrm{Sum}$ and $40+\mathrm{Sum}$ are divisible by 4) are (50, 90) which sum to 140, and (60, 80) which sum to 140. However, both pairs give the same sum! This means whichever pair Whimshire gets (with NUR 20), its PI will be $\frac{2\left(20\right)+140}{4}=45$. This again clashes with Humbleset's PI of 45, making it invalid.

Scenario C: Humbleset's second city is 90.

This gives ${\mathrm{PI}}_{\mathrm{Humbleset}}=\frac{110+90}{4}=50$.

The remaining cities are 50, 60, 70, and 80. To find Fogglia's and Whimshire's cities, we assign Fogglia NUR 10 and Whimshire NUR 20.

For Fogglia, we need two cities that sum to a multiple of 4. We use 50 and 70 (summing to 120).

${\mathrm{PI}}_{\mathrm{Fogglia}}=\frac{2\left(10\right)+50+70}{4}=\frac{140}{4}=35$

For Whimshire, the remaining cities are 60 and 80 (summing to 140).

${\mathrm{PI}}_{\mathrm{Whimshire}}=\frac{2\left(20\right)+60+80}{4}=\frac{180}{4}=45$

Checking the constraints: 35, 45, and 50 are distinct integers. Fogglia (35) is the lowest, and Humbleset (50) is the highest. All conditions are perfectly met!

Step 4: Final count of identified entities.

Through logical deduction, we have found exactly one working configuration. We successfully identified the specific PM and corresponding state for all 3 NURs and all 6 cities:

Humbleset: NUR (40), Cities Blusterburg (30) & Zingaloo (90)
Whimshire: NUR (20), Cities Splutterville (60) & Mumpypore (80)
Fogglia: NUR (10), Cities Noodleton (50) & Quackford (70)

The question asks: "For how many of the cities and NURs is it possible to identify their PM and the state they belong to?"

Since we have matched all 9 entities without any ambiguity, the final count is 9.