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.

Which pair of cities definitely belong to the same state?

The correct answer is Noodleton, Quackford.

First, we determine the values of the Pollution Measures (PMs) for all cities and non-urban regions (NURs). We are given nine PMs which are distinct multiples of 10 ranging from 10 to 90. Therefore, the set of PMs is {10, 20, 30, 40, 50, 60, 70, 80, 90}. Out of these, six are assigned to cities and three to the NURs.

We are given that there is exactly one pair of an NUR and a city where the PM of the NUR is greater than the PM of the city. To satisfy this condition, the PMs of the NURs must be very small. Let's arrange all nine PMs in ascending order. For there to be exactly one instance where an NUR has a higher PM than a city, the first two smallest values must be NURs, the third must be a city, the fourth must be an NUR, and the remaining five must be cities. This setup ensures that the third NUR (the 4th smallest value) is greater than the first city (the 3rd smallest value), creating exactly one pair, while all other NURs are smaller than all cities.

Following this logic, we can assign the exact values to the NURs and cities:

• NURs: 10, 20, 40
• Cities (in increasing order): 30, 50, 60, 70, 80, 90

Based on the given order of cities (Blusterburg, Noodleton, Splutterville, Quackford, Mumpypore, Zingaloo), their respective PMs are:

• Blusterburg = 30
• Noodleton = 50
• Splutterville = 60
• Quackford = 70
• Mumpypore = 80
• Zingaloo = 90

The problem states that the unique pair where the NUR's PM is greater than the city's PM belongs to Humbleset. This unique pair is the NUR with 40 and the city with 30. Thus, Humbleset's NUR has a PM of 40, and one of its cities is Blusterburg (30).

Next, we analyze the Pollution Index (PI) formula. The PI is calculated as 50% of the NUR's PM plus 25% of each of its two cities' PMs. Let the NUR's PM be P_NUR and the two cities' PMs be P_C1 and P_C2. The formula is:

$\mathrm{PI}=0.5\times P_{\mathrm{NUR}}+0.25\times P_{C1}+0.25\times P_{C2}$

Which simplifies to:

$\mathrm{PI}=\frac{2\times P_{\mathrm{NUR}}+P_{C1}+P_{C2}}{4}$

Since all PMs are multiples of 10, we can divide the values by 10 for simplicity to check for integer results. For the PI to be an integer, the sum of the divided PMs of the two cities must be an even number. This implies that the divided PMs of the two cities in any state must have the same parity (both even or both odd).

The available cities (divided by 10) are: Noodleton (5), Splutterville (6), Quackford (7), Mumpypore (8), and Zingaloo (9). The even values are 6 and 8. The odd values are 5, 7, and 9. Because cities in the same state must have the same parity, Splutterville (60) and Mumpypore (80) must be in the same state.

Humbleset already has Blusterburg (30, an odd parity when divided by 10), so its second city must be one of the remaining odd parity cities: Noodleton (50), Quackford (70), or Zingaloo (90). Let's test these three possibilities to see which one satisfies all conditions, notably that the PIs of all three states are distinct integers, with Humbleset having the highest and Fogglia having the lowest:

Case 1: Humbleset's second city is Noodleton (50).

Humbleset PI is calculated as follows:

$\mathrm{PI}=(2\times 40+30+50)/4=40$

The other two states would have cities (60, 80) and (70, 90). No matter how we assign the remaining NURs (10 and 20), the PIs will be either 40 or 45. This case is invalid because the PIs must be distinct (two would be 40) and Humbleset must have the highest PI.

Case 2: Humbleset's second city is Quackford (70).

Humbleset PI is calculated as follows:

$\mathrm{PI}=(2\times 40+30+70)/4=45$

The other two states would have cities (60, 80) and (50, 90). The possible PIs using NURs 10 and 20 would be 40 and 45. Again, this case is invalid because the PIs are not distinct.

Case 3: Humbleset's second city is Zingaloo (90).

Humbleset PI is calculated as follows:

$\mathrm{PI}=(2\times 40+30+90)/4=50$

The remaining cities are (60, 80) for one state and (50, 70) for the other. We have NURs 10 and 20 left. Let's calculate their PIs:

If a state has NUR 20 and cities 60, 80:

$\mathrm{PI}=(2\times 20+60+80)/4=45$

If a state has NUR 10 and cities 50, 70:

$\mathrm{PI}=(2\times 10+50+70)/4=35$

This case works perfectly! The PIs are distinct integers: 35, 45, and 50. Humbleset has the highest PI (50). The state with the lowest PI (35) must be Fogglia, and it consists of the NUR with 10 and the cities Noodleton (50) and Quackford (70). The remaining state, Whimshire, has the NUR with 20 and the cities Splutterville (60) and Mumpypore (80) with a PI of 45.

Because Noodleton and Quackford are paired together to achieve the valid integer PI for Fogglia, they definitely belong to the same state.