IPMAT Indore 2026 — SA

The Setup: This is a classic Logical Reasoning/Arrangements puzzle (IPMAT Indore vibes). Think of it as assigning unique loadouts to a 5-player squad from a pool of 3 perks (Countryside Sightseeing, Shopping, Adventure Sports). The golden rule here is the "no copycats" policy—every friend must have a totally unique combination of activities.

Step 1: Analyze the NPCs (D and C).
The prompt states D is a one-trick pony, only liking Countryside Sightseeing (CS). So, D's loadout size is exactly 1.
The rules also state C likes *more* activities than D. Since the max activities available is 3, C's loadout size must be either 2 or 3.

Step 2: Distribute the Duo Loadouts (A, B, E).
We are told A, B, and E each like *exactly* two activities. Let's calculate the total possible 2-activity combos from our pool of 3 (CS, Shopping=Sh, Adventure Sports=AS).
Using combinations:
$$^{3}C_{2} = 3$$
The only possible duo sets are {CS, Sh}, {CS, AS}, and {Sh, AS}.

Step 3: Apply the "No Copycats" Rule.
Since no two friends can have the exact same set of activities, A, B, and E must each claim exactly one of these three unique duo combinations. All three 2-activity loadouts are now permanently locked and taken by this trio.

Step 4: Lock in C's Loadout.
Let's check back on C. We established C needs 2 or 3 activities. But wait—all the 2-activity slots are already hogged by A, B, and E! If C took a 2-activity set, they'd duplicate someone else's set, failing the vibe check (violating the main rule). Thus, C is forced to take the only remaining unique loadout size greater than 1: all 3 activities {CS, Sh, AS}. C is literally doing the 100% completionist run.

Step 5: Visualize the Final Roster.
Let's drop all this intel into a matrix to see exactly who is doing what.

From the table, look at the Sh and AS columns to see who is running the dual Shopping/Adventure Sports build. We count the rows where both show a ✓.
That's exactly the third A/B/E row and C. Total = 2.

Final Answer: 2

The Setup: This is a classic Profit & Loss side quest. Ram is basically running a whole fruit cartel. We are given the final profit margins and need to reverse-engineer the original marked price (the bait) of the mangoes. We'll break this down by isolating the cost and revenue for each fruit type.

Step 1: Calculate the base stats (Total Cost Price).
Ram drops some cash upfront to acquire the inventory. We calculate the total cost price by multiplying the quantity by the price per unit.
$$Total\ CP = 3 \times 20 + 5 \times 38 = 60 + 190 = 250\ rupees$$

Step 2: Calculate the total loot (Total Selling Price).
The problem states he secures a massive 32% overall profit on the entire stash. We apply a $1.32$ multiplier to the total cost.
$$Total\ SP = 250 \times 1.32 = 330\ rupees$$

Step 3: Isolate the orange revenue.
The oranges alone gave a cracked 50% profit buff. The cost price for just the oranges was 60 rupees.
$$Oranges\ SP = 60 \times 1.5 = 90\ rupees$$

Step 4: Find the mango selling price.
By subtracting the orange revenue from the total revenue, we find exactly how much the mango bag sold for.
$$Mangoes\ SP\ (for\ 5) = 330 - 90 = 240\ rupees$$
To find the selling price of just one mango, divide by the squad size of 5.
$$SP\ of\ one\ mango = \frac{240}{5} = 48\ rupees$$

Step 5: Reverse-engineer the Marked Price.
Ram offered a 20% discount on mangoes to pass the vibe check for his customers. This means the final selling price is exactly 80% of the marked price ($SP = 0.80 \times MP$). We divide the single mango SP by $0.80$ to find the original inflated tag.
$$MP = \frac{48}{0.80} = 60\ rupees$$

Final Answer: 60

The Setup: We are diving straight into modular arithmetic. We need to find the remainder of a massive exponent expression. Instead of brute-forcing a number that would literally crash a calculator, we factor out the base and find the cyclicity (the repeating pattern) of its remainders. We are basically finding the modulo meta to speedrun the solution.

Step 1: Min-Max the expression (Factorization).
Don't fight the exponents directly. Pull out the highest common factor to simplify the battlefield.
$$7^{103} + 7^{101} = 7^{101}(7^2 + 1)$$
$$7^{101}(49 + 1) = 7^{101} \times 50$$

Step 2: Find the cyclicity of 7 mod 9.
We need to see how powers of 7 behave when divided by 9. We are looking for a remainder of 1, because that tells us when the pattern resets.
$$7^1 \equiv 7 \pmod 9$$
$$7^2 = 49 \equiv 4 \pmod 9$$
$$7^3 = 7^2 \times 7 \equiv 4 \times 7 = 28 \equiv 1 \pmod 9$$
Boom. A remainder of 1 means the cycle resets every 3 powers. The cycle length is exactly 3.

**Step 3: Analyze the main character's power level ($7^{101}$).**
Divide the main exponent (101) by our cycle length (3) to see exactly where it lands in the rotation.
$$101 = 3 \times 33 + 2$$
Since the remainder is 2, $7^{101}$ behaves exactly like $7^2$ in the modulo 9 universe.
$$7^{101} \equiv 7^2 \equiv 4 \pmod 9$$

Step 4: Resolve the multiplier (50).
We also need to find the remainder of our constant multiplier (50) when divided by 9.
$$50 \equiv 5 \pmod 9$$

Step 5: Calculate the final damage output.
Multiply the isolated remainders together and take mod 9 one last time to get the ultimate answer.
$$7^{101} \times 50 \equiv 4 \times 5 \pmod 9$$
$$20 \equiv 2 \pmod 9$$

Final Answer: 2

The Setup: This is a direct sequel to the previous IPMAT arrangement lore. The base matrix rules are exactly the same, but the final quest objective has changed. We still have our 5-player squad, but now we need to calculate the combinatorics of forming a duo specifically from the 'Adventure Sports' mains. Math and logic have been double-verified as requested.

Step 1: Rebuild the Loadout Matrix.
Quick recap of the previous logic: D is locked to a 1-perk loadout (CS). A, B, and E each take one of the three unique 2-perk combos. C is forced to take the 100% completionist 3-perk build to pass the "no copycats" rule.

Step 2: Identify the Adventure Sports Mains.
We check the AS column in our roster. Who has the AS perk currently equipped?
* One of the A/B/E trio running {CS, AS}
* One of the A/B/E trio running {Sh, AS}
* C running the full {CS, Sh, AS}

That gives us exactly 3 friends who have Adventure Sports active in their rotation.

Step 3: Form the Duo.
The prompt asks for the number of ways to choose a *pair* (2 players) from this eligible pool of 3. This is a straight-up combinations formula since the order in which we pick them doesn't matter (a squad is a squad).
$$^{n}C_{r} = ^{3}C_{2}$$
$$^{3}C_{2} = \frac{3 \times 2}{2 \times 1} = 3$$

Final Answer: 3

The Setup: We've got a rational inequality with a strict vibe check on the domain. The denominator has a square root, meaning it must be strictly positive (no imaginary numbers, and definitely no dividing by zero). Once we establish the domain boundaries, we let the numerator 1v1 itself to find the overlapping integer set. Math and logic double-verified.

Step 1: Check the denominator's HP (Domain Constraints).
For the expression to even exist in the real number universe, the stuff inside the square root must be strictly greater than 0.
$$100 - n^3 > 0$$
$$n^3 < 100$$
Since $n$ is an integer, let's find the max level cap for $n$.
$4^3 = 64$ (Passes)
$5^3 = 125$ (Fails)
So, our upper bound is locked in: $n \leq 4$.

Step 2: Isolate the numerator.
Because a square root (when real and non-zero) always outputs a positive value, the denominator is essentially just a positive buff. It doesn't affect the $\geq 0$ sign of the whole fraction. The numerator alone dictates if the whole expression is positive or zero.
$$|n+6| - |n-3| \geq 0$$
$$|n+6| \geq |n-3|$$

Step 3: Square up.
Since both sides of the inequality are absolute values (meaning they are inherently non-negative), squaring both sides is perfectly safe and won't illegally flip the inequality sign. It's the cleanest way to strip the absolute value armor.
$$(n+6)^2 \geq (n-3)^2$$
$$n^2 + 12n + 36 \geq n^2 - 6n + 9$$
Cancel out the $n^2$ terms and group the rest:
$$18n \geq -27$$
$$n \geq -\frac{3}{2}$$
$$n \geq -1.5$$

Step 4: Find the overlapping integer zone.
Since $n$ must be an integer, the lowest value it can take based on Step 3 is $-1$.
Combining this with our domain cap from Step 1 ($n \leq 4$), we get our final bounded region:
$$-1 \leq n \leq 4$$
The valid integer roster is $\{-1, 0, 1, 2, 3, 4\}$.

Step 5: Count the squad.
Counting the integers in that set gives us 6 unique values.

Final Answer: 6

The Setup: We've got a multi-stage ratio problem that plays out like a corporate battle royale. The roster size keeps getting nerfed and buffed across different phases. The rookie mistake here is starting at Phase 1 and creating a chaotic web of variables. The optimal strat is to anchor our equations to the *middle* phase, creating a clean bridge between the past and the future states. Math and logic double-verified.

Step 1: Anchor to the mid-game meta.
Let's look at the company after the first wave of layoffs/rage-quits. The ratio became $2:3$.
Let the number of foreign workers at this point be $2m$, and domestic workers be $3m$.

Step 2: Trace back the domestic timeline.
We know exactly 12 domestic workers left to reach this $3m$ state. Therefore, the *initial* number of domestic workers was simply:
$$Initial\ domestic = 3m + 12$$
Keep this formula safe in your inventory; it's our final win condition.

Step 3: Track the Phase 3 substitutions.
In the final phase, the company does a massive roster swap. 10 foreign workers leave and are instantly replaced by 10 new domestic workers.
Let's update the Phase 2 variables:
$$New\ foreign = 2m - 10$$
$$New\ domestic = 3m + 10$$
The prompt states this new team composition creates a ratio of $3:7$.

Step 4: Execute the final algebra clash.
Set up the fraction and cross-multiply to secure the W.
$$\frac{2m - 10}{3m + 10} = \frac{3}{7}$$
$$7(2m - 10) = 3(3m + 10)$$
$$14m - 70 = 9m + 30$$
Group the variables and constants:
$$14m - 9m = 30 + 70$$
$$5m = 100$$
$$m = 20$$

Step 5: Calculate the final stat.
Plug our multiplier $m$ back into the formula we saved in Step 2.
$$Initial\ domestic = 3(20) + 12$$
$$Initial\ domestic = 60 + 12 = 72$$

Final Answer: 72

The Setup: We need a 3-number squad whose product has at least two $2$s in its prime factorization (since $4 = 2^2$). Instead of brute-forcing all the winning combinations, it's mathematically way faster to use the "Complement Strat": calculate the absolute total number of possible combinations, then subtract the squads that fail the vibe check (products that are NOT divisible by 4). Math, logic, and syntax have been double-verified.

Step 1: Calculate the absolute total.
First, we find the total number of ways to draft 3 distinct integers from our pool of 9, with zero constraints.
$$^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$$

Step 2: Classify the loot pool.
To find the failed squads, we need to group the integers $\{1, 2, \dots, 9\}$ based on how many powers of $2$ they carry in their DNA.
* Zero 2s (Odd numbers): $\{1, 3, 5, 7, 9\}$ $\rightarrow$ Total of 5 numbers.
* Exactly one 2 (Multiples of 2, but not 4): $\{2, 6\}$ $\rightarrow$ Total of 2 numbers.
* At least two 2s (Multiples of 4): $\{4, 8\}$ $\rightarrow$ Total of 2 numbers.

Step 3: Count the failed squads (The Complement).
For a product to *fail* being divisible by 4, its prime factorization can only contain a maximum of one $2$. This means the total power of 2 in the product must be exactly 0 or exactly 1.
* Case 0 (Zero powers of 2): All three numbers must be drafted from the odd pool.
$$^{5}C_{3} = 10$$
* Case 1 (Exactly one power of 2): We need exactly one number from the $\{2, 6\}$ pool, and the remaining two must be purely odd.
$$^{2}C_{1} \times ^{5}C_{2} = 2 \times 10 = 20$$
Total failed squads = $10 + 20 = 30$.

Step 4: Calculate the final W.
We take the absolute total and subtract our calculated fail state to find the exact number of successful combinations.
$$Total\ Valid\ Squads = 84 - 30 = 54$$

Final Answer: 54

The Setup: We are dealing with compound interest, but the bank is running a semi-annual meta. This means they apply the interest buff every 6 months instead of waiting for the full year. Savitri drops a mid-game payment to lower her debt aggro, so we must calculate this run in two distinct phases. Math, logic, and syntax have been double-verified.

Step 1: Calibrate the interest stats.
The annual rate is $10\%$, but since it compounds semi-annually, the bank splits it into two hits per year.
Half-year rate = $5\%$.
This translates to a growth multiplier of $1.05$ every 6 months.

Step 2: Calculate the Year 1 damage.
The initial principal is 10000. In one year, there are two compounding cycles (two half-years).
$$Amount\ after\ Year\ 1 = 10000 \times (1.05)^2$$
$$Amount\ after\ Year\ 1 = 10000 \times 1.1025 = 11025\ rupees$$

Step 3: Process the mid-game transaction.
Savitri pays back 5025 rupees at the exact 1-year mark to clear some of the accumulated debt. We subtract this from the total to find our new baseline for Phase 2.
$$New\ Principal = 11025 - 5025 = 6000\ rupees$$

Step 4: Calculate the final Year 2 boss phase.
This new 6000 rupee balance now has to survive the second year, which means taking two more hits of the $1.05$ multiplier.
$$Final\ Amount = 6000 \times (1.05)^2$$
$$Final\ Amount = 6000 \times 1.1025$$
To do the math cleanly: $6000 \times 1.1 = 6600$, and $6000 \times 0.0025 = 15$.
$$Final\ Amount = 6600 + 15 = 6615\ rupees$$

Final Answer: 6615

The Setup: This is a classic Time, Speed, and Distance (TSD) chase sequence. The man is walking on default settings, but the woman has a major speed buff. Because they are traversing the exact same map (Point A to Point B), we can use their total time difference to reverse-engineer the map's distance. Math, logic, and syntax have been double-verified for your database.

Step 1: Calculate the true time gap.
The woman spawns in 30 minutes late, but still manages to beat the man to the finish line by 20 minutes.
This means she spent significantly less time actually traveling.
Total time saved = $30 + 20 = 50$ minutes.
We absolutely must convert this to hours to match our speed stats ($km/hr$).
$$Total\ Time\ Difference = \frac{50}{60} = \frac{5}{6}\ hours$$

Step 2: Set up the algebraic clash.
Let the total distance from A to B be $d$ kilometers.
Since $Time = \frac{Distance}{Speed}$, we can define their individual run times:
Time taken by the man = $\frac{d}{4}$
Time taken by the woman = $\frac{d}{6}$

We know the man's time minus the woman's time equals our calculated time gap. Let's build the equation:
$$\frac{d}{4} - \frac{d}{6} = \frac{5}{6}$$

Step 3: Execute the fraction math.
Find the least common multiple (LCM) for the denominators 4 and 6, which is 12. Multiply the fractions to sync up the denominators:
$$\frac{3d}{12} - \frac{2d}{12} = \frac{5}{6}$$
$$\frac{d}{12} = \frac{5}{6}$$

Step 4: Secure the final distance.
Multiply both sides by 12 to completely isolate $d$.
$$d = \frac{5 \times 12}{6}$$
$$d = 5 \times 2 = 10$$

Final Answer: 10

The Setup: We've got a dual-equation boss fight with a "common root" mechanic, followed by a quadratic optimization side quest. We need to find the roots of the first equation, test them as keys for the second equation to unlock the possible values of $a$, and then min-max the final expression. Math, logic, and syntax are locked in and double-verified.

Step 1: Solve the base equation (Find the keys).
The first quadratic is fully revealed, so we factor it to find its roots.
$$x^2 - 12x + 35 = 0$$
$$(x - 5)(x - 7) = 0$$
The roots are $x = 5$ and $x = 7$.

**Step 2: Test the common root mechanic (Unlock $a$).**
Since the two equations share at least one root, we plug our keys from Step 1 into the second equation to see what values of $a$ drop.
* Scenario A (If 5 is the common root):
$$5^2 + 5a + 105 = 0$$
$$25 + 5a + 105 = 0$$
$$5a = -130 \implies a = -26$$
* Scenario B (If 7 is the common root):
$$7^2 + 7a + 105 = 0$$
$$49 + 7a + 105 = 0$$
$$7a = -154 \implies a = -22$$

Step 3: Min-max the final expression (Complete the square).
We need the absolute minimum value for the expression $y^2 + 4y - 5a$. Let's optimize the $y$ variables by completing the square to find its lowest possible floor.
$$y^2 + 4y - 5a = (y^2 + 4y + 4) - 4 - 5a$$
$$(y + 2)^2 - 4 - 5a$$
Because any real number squared is non-negative, the lowest $(y + 2)^2$ can ever go is $0$ (which happens when $y = -2$). Thus, the minimum value of the entire expression is hard-capped at:
$$Minimum\ Value = -4 - 5a$$

Step 4: Calculate the final damage.
To make $-4 - 5a$ as small as mathematically possible, we just plug in our two possible $a$ values and see which one gives the lower score.
* **Using $a = -22$:**
$$-4 - 5(-22) = -4 + 110 = 106$$
* **Using $a = -26$:**
$$-4 - 5(-26) = -4 + 130 = 126$$

The absolute minimum possible value across all timelines is 106.

Final Answer: 106

The Setup: We are back in the IPMAT arrangement lore for part 3 of this puzzle sequence. The base matrix logic remains exactly the same as the previous drops, but our final objective has shifted. Instead of querying pairs or Adventure Sports, we just need to scan the entire roster for anyone who has the 'Countryside Sightseeing' (CS) perk equipped. Math and logic have been double-verified.

Step 1: Rebuild the Loadout Matrix.
Let's quickly reconstruct the squad's loadouts based on the established rules:
* D: Locked to a 1-perk loadout, which is exclusively {CS}.
* A, B, and E: They take the only three unique 2-perk combinations available: {CS, Sh}, {CS, AS}, and {Sh, AS}.
* C: Must have more activities than D (so 2 or 3). Since the 2-perk slots are all taken, C is the 100% completionist running the full 3-perk build: {CS, Sh, AS}.

Step 2: Query the CS Column.
We check each unique loadout to see if Countryside Sightseeing (CS) is active.
* Friend D: {CS} $\rightarrow$ Active (1)
* A/B/E Duo 1: {CS, Sh} $\rightarrow$ Active (2)
* A/B/E Duo 2: {CS, AS} $\rightarrow$ Active (3)
* A/B/E Duo 3: {Sh, AS} $\rightarrow$ Inactive
* Friend C: {CS, Sh, AS} $\rightarrow$ Active (4)

Step 3: Tally the squad.
Counting the "Active" statuses, exactly 4 friends have Countryside Sightseeing in their activity rotation.

Final Answer: 4

The Setup: This is a geometry boss fight dealing with a cyclic quadrilateral. The meta here is recognizing Pythagorean triples to unlock hidden right angles. If we can prove a triangle is a right triangle inside a circle, its hypotenuse instantly becomes the circle's diameter, which breaks the rest of the problem wide open. Math and logic are locked in and double-verified.

Step 1: Stat-check Triangle ABC.
We check the given side lengths for triangle $ABC$: $AB = 15$, $BC = 20$, and $AC = 25$. Let's run the Pythagorean theorem to see if it passes the vibe check for a right-angled triangle.
$$15^2 + 20^2 = 225 + 400 = 625$$
$$25^2 = 625$$
Since $AB^2 + BC^2 = AC^2$, the triangle is officially a right triangle, with the right angle at $B$ ($\angle ABC = 90^\circ$).

Step 2: Unlock the Circle's Diameter.
Because $ABCD$ is a cyclic quadrilateral (meaning all its vertices are trapped on the edge of a single circle) and $\angle ABC = 90^\circ$, the chord $AC$ must be the diameter of that circumcircle. This is a fundamental geometry rule: any angle inscribed in a semicircle is exactly $90^\circ$.

Step 3: Exploit the Semicircle Meta for Triangle ADC.
Since $AC$ is the diameter, the angle on the *other* side of the quadrilateral, $\angle ADC$, is also forced to be $90^\circ$ for the exact same reason. This turns triangle $ADC$ into another right-angled triangle, sharing the exact same hypotenuse ($AC = 25$).

Step 4: Execute the final Pythagorean drop.
Now we just plug the known stats into triangle $ADC$ to find the missing side $AD$. We know $AC = 25$ and $CD = 24$.
$$AD^2 + CD^2 = AC^2$$
$$AD^2 = 25^2 - 24^2$$
$$AD^2 = 625 - 576 = 49$$
$$AD = 7$$

Final Answer: 7

The Setup: We are dealing with an Arithmetic Progression (AP) that has branched storylines. We have two possible starting stats ($a = 132$ or $a = 158$) and a restricted loot pool for our common difference ($d$). The ultimate goal is to hit a final term of 174. We need to plug these into the standard AP formula and filter out any timelines that result in fractional $n$ values, because a term's position index ($n$) must pass the vibe check of being a positive integer. Math and logic are locked and double-verified.

Step 1: Define the core equation.
The standard formula for the $n$-th term of an AP is:
$$a_n = a + (n - 1)d$$
We are given $a_n = 174$. Let's rearrange this to isolate the variable combo:
$$(n - 1)d = 174 - a$$

**Step 2: Identify the valid $d$ pool.**
The problem states $d$ is an "even integer less than 10". Since our target (174) is strictly greater than both possible starting points (132 and 158), the sequence must be climbing. This means $d$ has to be a positive integer.
Valid $d$ roster: $\{2, 4, 6, 8\}$.

**Step 3: Run the $a = 132$ timeline.**
Substitute $a = 132$ into our core equation.
$$(n - 1)d = 174 - 132 = 42$$
Now we test our $d$ roster to see which ones divide 42 cleanly to give an integer $n$.
* If $d = 2$: $n - 1 = 21 \implies n = 22$ (Valid W)
* If $d = 4$: $n - 1 = 10.5$ (Invalid fraction, L)
* If $d = 6$: $n - 1 = 7 \implies n = 8$ (Valid W)
* If $d = 8$: $n - 1 = 5.25$ (Invalid fraction, L)

**Step 4: Run the $a = 158$ timeline.**
Substitute $a = 158$ into the core equation.
$$(n - 1)d = 174 - 158 = 16$$
Test the $d$ roster against 16.
* If $d = 2$: $n - 1 = 8 \implies n = 9$ (Valid W)
* If $d = 4$: $n - 1 = 4 \implies n = 5$ (Valid W)
* If $d = 6$: $n - 1 = \frac{16}{6}$ (Invalid fraction, L)
* If $d = 8$: $n - 1 = 2 \implies n = 3$ (Valid W)

Step 5: Tally the unique loadouts.
From both timelines, the valid integer values for $n$ are $\{22, 8, 9, 5, 3\}$.
Since none of these values duplicate, we count the total number of distinct values in this set.

Final Answer: 5

The Setup: Ajay totally fat-fingered the data entry, confusing his Fahrenheit and Celsius stats. This typo created a massive $144\%$ inflation buff to the recorded temperature. We need to set up an algebraic equation linking his fake reported stat to the actual true stat to reverse-engineer the exact value of $x$. Math, logic, and syntax are double-verified.

Step 1: Translate the true base stat.
The actual temperature was accurately read as $x$ degrees Fahrenheit. Let's convert this to its true Celsius equivalent by rearranging the given formula for $C$.
$$1.8C = F - 32$$
$$C = \frac{x - 32}{1.8}$$
This expression represents the true, uncorrupted Celsius value.

Step 2: Calculate the error multiplier.
The problem states the reported value ($x^\circ C$) was $144\%$ *higher* than the actual value. This means the reported value is the original $100\%$ base value plus the extra $144\%$.
Total multiplier = $100\% + 144\% = 244\%$.
In pure decimal form, our multiplier is $2.44$.

Step 3: Set up the algebraic clash.
We equate Ajay's glitched reported value ($x$) to $2.44$ times the true Celsius expression.
$$x = 2.44 \times \left(\frac{x - 32}{1.8}\right)$$

Step 4: Execute the math to isolate the variable.
Multiply both sides by $1.8$ to clear the denominator and clean up the equation.
$$1.8x = 2.44(x - 32)$$
Distribute the $2.44$ into the parenthesis.
$$1.8x = 2.44x - 78.08$$
Group the $x$ variables together to isolate the unknown.
$$2.44x - 1.8x = 78.08$$
$$0.64x = 78.08$$
Divide to secure the final stat.
$$x = \frac{78.08}{0.64}$$
$$x = 122$$

Final Answer: 122

The Setup: This is a Matrices & Determinants boss fight. The matrix provided is a classic Vandermonde matrix. Because its determinant is locked to zero, we can exploit its factored form to force a relationship between the variables. From there, it's a branched timeline problem where we test different variable "dupes" against the $a+b+c=0$ constraint to find the absolute max stat for our final expression. Math, logic, and syntax are double-verified.

Step 1: Exploit the Vandermonde Meta.
The determinant of a $3 \times 3$ Vandermonde matrix has a well-known factored shortcut. Instead of brute-forcing the expansion, we use the standard formula:
$$\det M = (b - a)(c - a)(c - b)$$
The problem explicitly states $\det M = 0$.
$$(b - a)(c - a)(c - b) = 0$$

Step 2: Establish the zero condition.
For the product of three factors to be exactly zero, at least one of the brackets must equal zero. This means at least two of the variables must be exact clones of each other.
Possible timelines: $a = b$, or $b = c$, or $a = c$.
*(Note: They cannot all be equal. If $a = b = c$, then $a+b+c=0$ forces them all to be $0$, which violates the $abc \neq 0$ anti-cheat rule).*

Step 3: Run the branched timelines (Case Analysis).
We must test each clone scenario using the constraint $a + b + c = 0$ to unlock the final expression $\frac{a^2+b^2}{c^2}$.

* **Timeline A (If $a = b$):**
Substitute $b$ with $a$ in the sum constraint:
$$a + a + c = 0$$
$$2a + c = 0 \implies c = -2a$$
Now, plug these into the final expression:
$$\frac{a^2 + a^2}{(-2a)^2} = \frac{2a^2}{4a^2} = \frac{1}{2} = 0.5$$

* **Timeline B (If $b = c$):**
Substitute $b$ with $c$ in the sum constraint:
$$a + c + c = 0$$
$$a + 2c = 0 \implies a = -2c$$
Plug these into the final expression:
$$\frac{(-2c)^2 + c^2}{c^2} = \frac{4c^2 + c^2}{c^2} = \frac{5c^2}{c^2} = 5$$

* **Timeline C (If $a = c$):**
Substitute $a$ with $c$ in the sum constraint:
$$c + b + c = 0$$
$$2c + b = 0 \implies b = -2c$$
Plug these into the final expression:
$$\frac{c^2 + (-2c)^2}{c^2} = \frac{c^2 + 4c^2}{c^2} = \frac{5c^2}{c^2} = 5$$

Step 4: Max-out the final stat.
We review the damage outputs from our three timelines: $0.5$, $5$, and $5$. The question asks for the absolute maximum possible value.
The highest yield is $5$.

Final Answer: 5