What is the PI of Fogglia?

City	PM	State
Blusterburg	30
Noodleton	50
Splutterville	60
Quackford	70
Mumpypore	80
Zingaloo	90

State	NUR PM	PI
Whimshire
Fogglia
Humbleset

All six city PMs are filled from the single NUR$>$city rule.

Let the cities in increasing order be $c_1<c_2<\dots<c_6$ and the NURs be $n_1<n_2<n_3$.

The number of (NUR, city) pairs with NUR $>$ city equals, for each NUR, the count of cities below it, summed up. This total must be $1$.

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If two or more NURs were above the smallest city $c_1$, the total would be $\ge 2$.

If no NUR were above $c_1$, the total would be $0$.

So exactly one NUR lies above $c_1$, and it must lie below $c_2$ (otherwise it would exceed $\ge 2$ cities).

The ordering of all nine values is therefore forced:

$n_1<n_2<c_1<n_3<c_2<c_3<c_4<c_5<c_6$.

Mapping to $10,\dots,90$:

$n_1=10,\ n_2=20,\ c_1=30,\ n_3=40,\ c_2=50,\ c_3=60,\ c_4=70,\ c_5=80,\ c_6=90$.

So the cities are

$\text{Blusterburg}=30,\ \text{Noodleton}=50,\ \text{Splutterville}=60,\ \text{Quackford}=70,\ \text{Mumpypore}=80,\ \text{Zingaloo}=90$,

and the three NUR PMs are $\{10,20,40\}$.

City	PM	State
Blusterburg	30	Humbleset
Noodleton	50
Splutterville	60
Quackford	70
Mumpypore	80
Zingaloo	90

State	NUR PM	PI
Whimshire
Fogglia
Humbleset	40

Blusterburg is assigned to Humbleset, and Humbleset's NUR PM is set to $40$.

The only pair with NUR $>$ city is $n_3=40$ over $c_1=30$, i.e. over Blusterburg.

This NUR and this city both belong to Humbleset.

So Humbleset's NUR PM $=40$, and Blusterburg ($30$) is one of Humbleset's two cities.

The remaining NUR PMs $\{10,20\}$ belong to Whimshire and Fogglia, in some order.

City	PM	State
Blusterburg	30	Humbleset
Noodleton	50
Splutterville	60	W/F
Quackford	70
Mumpypore	80	W/F
Zingaloo	90

State	NUR PM	PI
Whimshire
Fogglia
Humbleset	40

Splutterville and Mumpypore are pinned to the same state, which is either Whimshire or Fogglia (marked W/F).

Since $0.5\times\text{NUR}$ is an integer for every $\text{NUR}\in\{10,20,40\}$, a PI is an integer only when

$0.25\times(\text{city}_1+\text{city}_2)$ is an integer,

i.e. the two city PMs of a state sum to a multiple of $4$.

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City PMs modulo $4$:

$60\equiv 0,\ 80\equiv 0$, and $30,50,70,90\equiv 2$.

A pair sums to a multiple of $4$ only if both are $\equiv 0$ or both are $\equiv 2$.

Only $60$ and $80$ are $\equiv 0$, so they must be paired together.

Thus one state's cities are Splutterville ($60$) and Mumpypore ($80$).

Humbleset already contains $30$ (which is $\equiv 2$), so this is not Humbleset.

Hence Splutterville and Mumpypore share Whimshire or Fogglia, and Humbleset's second city is one of $\{50,70,90\}$.

City	PM	State
Blusterburg	30	Humbleset
Noodleton	50
Splutterville	60	W/F
Quackford	70
Mumpypore	80	W/F
Zingaloo	90	Humbleset

State	NUR PM	PI
Whimshire
Fogglia
Humbleset	40	50

Zingaloo is fixed as Humbleset's second city, giving Humbleset PI $=50$.

Let Humbleset $=\{30,x\}$ with $x\in\{50,70,90\}$:

$\text{PI}_H = 20 + 0.25(30+x) = 27.5 + 0.25x$,

giving $40,\ 45,\ 50$ for $x=50,70,90$.

The $\{60,80\}$ state has PI $= 35 + 0.5\times\text{NUR} \in \{40,45\}$.

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Humbleset must have the strictly highest PI.

If $x=50$: $\text{PI}_H=40$, but the $\{60,80\}$ state can reach $45$, so Humbleset is not highest. Reject.

If $x=70$: $\text{PI}_H=45$; the leftover cities form $\{50,90\}$ (sum $140$) and $\{60,80\}$ (sum $140$), both giving $35+0.5\times\text{NUR}$. With NURs $10,20$ one of these states becomes $45$, tying Humbleset. Reject.

So $x=90$: Humbleset $=\{30,90\}=$ Blusterburg, Zingaloo, with

$\text{PI}_H = 27.5 + 22.5 = 50$.

City	PM	State
Blusterburg	30	Humbleset
Noodleton	50	Fogglia
Splutterville	60	Whimshire
Quackford	70	Fogglia
Mumpypore	80	Whimshire
Zingaloo	90	Humbleset

State	NUR PM	PI
Whimshire	20	45
Fogglia	10	35
Humbleset	40	50

Whimshire and Fogglia are completed.

The four remaining cities split as $\{60,80\}$ (Splutterville, Mumpypore) and $\{50,70\}$ (Noodleton, Quackford), with NURs $\{10,20\}$.

$\{60,80\}: 35 + 0.5\times\text{NUR} \Rightarrow 40\ (\text{NUR }10)\ \text{or}\ 45\ (\text{NUR }20)$.

$\{50,70\}: 30 + 0.5\times\text{NUR} \Rightarrow 35\ (\text{NUR }10)\ \text{or}\ 40\ (\text{NUR }20)$.

The combination $\{60,80\}{+}10=40$ and $\{50,70\}{+}20=40$ gives equal PIs, which is rejected (PIs must be distinct).

So $\{60,80\}$ takes NUR $20$ (PI $45$) and $\{50,70\}$ takes NUR $10$ (PI $35$).

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Fogglia has the lowest PI, so Fogglia $=\{50,70\}$: Noodleton ($50$), Quackford ($70$), NUR $10$, PI $35$.

Whimshire $=\{60,80\}$: Splutterville ($60$), Mumpypore ($80$), NUR $20$, PI $45$.

City	PM	State
Blusterburg	30	Humbleset
Noodleton	50	Fogglia
Splutterville	60	Whimshire
Quackford	70	Fogglia
Mumpypore	80	Whimshire
Zingaloo	90	Humbleset

State	NUR PM	PI
Whimshire	20	45
Fogglia	10	35
Humbleset	40	50

Every PM, state and PI is now uniquely determined.

Verification of the constraints:

The only NUR $>$ city pair is $40 > 30$ (Humbleset's NUR over Blusterburg) — exactly one such pair, and both lie in Humbleset.

PIs are $50$ (Humbleset), $45$ (Whimshire), $35$ (Fogglia) — distinct integers, with Humbleset highest and Fogglia lowest.

All nine entities (six cities and three NURs) have a known PM and a known state.

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Fogglia (the lowest PI) holds Noodleton ($50$) and Quackford ($70$) with NUR $= 10$.

$\text{PI} = 0.5\times10 + 0.25\times50 + 0.25\times70 = 5 + 12.5 + 17.5 = 35$.

Answer: $35$.