IPMAT Indore 2020 — QA

The Setup: We are dealing with the foundational division algorithm: $Dividend=(Divisor\times Quotient)+Remainder$. The absolute golden rule of division is that the remainder must be strictly non-negative and less than the divisor ($0\leq r<d$). We need to min-max the given sum and product to find the exact values before plugging them in.
Step 1: Formulate the system of equations. Let the quotient be $q$ and the remainder be $r$. The problem gives us their product $q\times r=24$ and their sum $q+r=10$.
Step 2: Solve the quadratic substitution. You can substitute $r=10-q$ into the product equation:

$$q(10-q)=24$$

$$10q-q^2=24$$

$$q^2-10q+24=0$$

Factoring this gives $(q-6)(q-4)=0$. The roots are $6$ and $4$. So, our variables $\{q,r\}$ are exactly $\{6,4\}$ - no other pair of numbers has this sum and product.
Step 3: Enforce the remainder rule. We are told the divisor is $5$. By the laws of division, $r<5$. Between $6$ and $4$, only $4$ satisfies this condition. Therefore, it is a hard lock that $r=4$ and $q=6$.
Step 4: Execute the final calculation. Substitute everything back into the core algorithm:

$$Dividend=(5\times 6)+4$$

$$Dividend=30+4=34$$

Final Answer: 34

The Setup: This is a coordinate geometry problem finding the shortest path from an external point to a circle's boundary. Think of the circle as a forcefield; the shortest route is always a straight line aimed dead center, and we just subtract the forcefield's radius from the total distance.
Step 1: Deconstruct the circle's equation. The given equation is $x^2+y^2=1$. This perfectly matches the standard form $(x-h)^2+(y-k)^2=r^2$. Thus, the center is the origin $(0,0)$ and the radius is $r=\sqrt{1}=1$.
Step 2: Calculate absolute distance to center. We use the Euclidean distance formula between the point $(-4,3)$ and the center $(0,0)$:

$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

$$d=\sqrt{(-4-0)^2+(3-0)^2}$$

$$d=\sqrt{(-4)^2+(3)^2}$$

$$d=\sqrt{16+9}=\sqrt{25}=5$$

Step 3: Confirm the point is outside. This step is not optional - the *subtract the radius* move only works for an external point. Since $d=5>r=1$, the point sits outside the circle. (Had it been inside, the shortest distance would have been $r-d$ instead.)
Step 4: Find the boundary distance. The total distance to the core is $5$ units. Because the circle's radius occupies $1$ unit of that path, the remaining gap from the point to the edge is simply the difference: $5-1=4$.

Final Answer: 4

The Setup: This expression looks incredibly toxic, but we can beat it by breaking it down into two phases: resolving the infinite geometric progression (GP) first, and then applying logarithm exponent rules to clean up the base.
Step 1: Evaluate the infinite GP. Look at the sequence inside the log: $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots$
This is an infinite GP where the first term $a=\frac{1}{4}$ and the common ratio $r=\frac{1}{2}$. Since $|r|<1$ the series converges, so the sum to infinity $S=\frac{a}{1-r}$ applies:

$$S=\frac{\frac{1}{4}}{1-\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times \frac{2}{1}=\frac{1}{2}$$

Step 2: Reconstruct the base and the logarithm. The original expression's base is $0.04$, which translates to the fraction $\frac{4}{100}=\frac{1}{25}$. We can write this as an exponent: $5^{-2}$.
The logarithm's base is $\sqrt{5}$, which is $5^{\frac{1}{2}}$.
So the exponent is $\log_{5^{1/2}}\left(\frac{1}{2}\right)$.
Step 3: Simplify the exponent using log properties. We use the rule $\log_{b^n}(x)=\frac{1}{n}\log_b(x)$:

$$\log_{5^{1/2}}\left(\frac{1}{2}\right)=\frac{1}{\frac{1}{2}}\log_5\left(\frac{1}{2}\right)=2\log_5\left(\frac{1}{2}\right)$$

Bring the multiplier inside as a power: $\log_5\left(\left(\frac{1}{2}\right)^2\right)=\log_5\left(\frac{1}{4}\right)$.
Step 4: Combine the final expression. Substitute the cleaned-up base and exponent back together:

$$\left(5^{-2}\right)^{\log_5\left(\frac{1}{4}\right)}=5^{-2\log_5\left(\frac{1}{4}\right)}$$

Move the $-2$ inside the log as a power:

$$5^{\log_5\left(\left(\frac{1}{4}\right)^{-2}\right)}$$

Using the identity $b^{\log_b(x)}=x$, the $5$ and $\log_5$ cancel out. We are left with:

$$\left(\frac{1}{4}\right)^{-2}=4^2=16$$

Final Answer: 16

The Setup: This is a top-tier matrix determinant manipulation. When a determinant's column contains a sum or difference, it can be split into two separate determinants. The given $a=3$ is literal clickbait meant to throw you off; it has absolutely zero impact on the algebraic proof.
Step 1: Split the original determinant. We separate the third column $\left(a^3-1,\ b^3-1,\ c^3-1\right)$ into two determinants:

$$\Delta=\left|\begin{matrix}a&a^2&a^3\\b&b^2&b^3\\c&c^2&c^3\end{matrix}\right|-\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=0$$

Step 2: Factor the first determinant. In the first matrix, notice that row 1 shares an $a$, row 2 shares a $b$, and row 3 shares a $c$. Let's factor them out completely:

$$abc\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|$$

Step 3: Align the second determinant. We want the second matrix to match the first, so we walk its last column to the front by two swaps.
Swap Column 3 with Column 2 (sign flips):

$$\left|\begin{matrix}a&a^2&1\\b&b^2&1\\c&c^2&1\end{matrix}\right|=-\left|\begin{matrix}a&1&a^2\\b&1&b^2\\c&1&c^2\end{matrix}\right|$$

Now swap Column 2 with Column 1 (sign flips back):

$$=+\left|\begin{matrix}1&a&a^2\\1&b&b^2\\1&c&c^2\end{matrix}\right|$$

Two swaps, two sign flips, so the second determinant is exactly the same object as the first one's cofactor - no leftover minus sign.
Step 4: Factor the entire equation. Let the aligned matrix be $V$ (a Vandermonde determinant). Our original equation $\Delta=0$ now looks like this:

$$abc(V)-V=0 \implies V(abc-1)=0$$

The Vandermonde determinant factors cleanly as $V=(b-a)(c-a)(c-b)$. Because the problem states $a$, $b$ and $c$ are distinct real numbers, every one of those three brackets is non-zero, so $V\neq 0$. Therefore the other multiplier must be zero:

$$abc-1=0\implies abc=1$$

Final Answer: 1

The Setup: We are tasked with finding the global minimum of the function $f(x)=|3-x|+|2+x|+|5-x|$. In absolute value geometry, this represents the sum of distances from a coordinate $x$ to three specific points on a number line, and the minimum lands at the median of those points. Rather than just quoting that rule, we can lock the answer down with a two-line proof.
Step 1: Standardize the terms. First, use the property $|a-b|=|b-a|$ to rewrite the function so $x$ is always first. This prevents sign errors.

$$f(x)=|x-3|+|x-(-2)|+|x-5|$$

The roots (the points on the number line) are exactly $3$, $-2$ and $5$. In ascending order: $-2,\ 3,\ 5$ - so the median is $3$.
Step 2: Bound the outer pair. For the two extreme points, the triangle inequality gives a floor that no $x$ can beat:

$$|x+2|+|x-5|\geq |(x+2)-(x-5)|=7$$

with equality for every $x$ in the interval $[-2,5]$.
Step 3: Bound the middle term. Obviously $|x-3|\geq 0$, with equality only at $x=3$. Adding the two bounds gives $f(x)\geq 7+0=7$ for all $x$.
Step 4: Show the floor is actually reached. The two equality conditions must hold at once, and they do: $x=3$ lies inside $[-2,5]$. Substituting:

$$f(3)=|3-3|+|3-(-2)|+|3-5|=0+5+2=7$$

The bound is achieved, so $7$ is the true minimum and not merely a lower estimate.

Final Answer: 7

The Setup: This is a system of linear equations hiding behind ratios, with a strict Diophantine constraint: the prices of the pens and pencils must be positive integers. We have to decode two separate shopping trips, set up the algebra, and pin down the only integer solution.
Step 1: Formalize the second visit. On the second visit, the ratio of pens to pencils is $4:1$, and the total cost is Rs. 112. We are explicitly told Ashok bought 4 pens. By the ratio, he must have bought exactly 1 pencil.
Let $x$ be the price of a pen, and $y$ be the price of a pencil.

$$4x+1y=112$$

Isolate $y$ so we can use it as a substitution later:

$$y=112-4x$$

Step 2: Formalize the first visit. The ratio of pens to pencils bought is $2:3$, costing Rs. 86. We don't know the exact quantities, just the ratio. So, let the quantities be $2k$ pens and $3k$ pencils, where $k$ is a positive integer constant.

$$2kx+3ky=86 \implies k(2x+3y)=86$$

Step 3: Substitute, then kill every wrong branch. Plug the isolated $y$ from Step 1 into the Step 2 equation:

$$k(2x+3(112-4x))=86$$

$$k(2x+336-12x)=86 \implies k(336-10x)=86$$

So $336-10x=\frac{86}{k}$, which forces $k$ to be a positive factor of $86$. The factors are $1, 2, 43, 86$, giving $\frac{86}{k}\in\{86,\ 43,\ 2,\ 1\}$ respectively. Now the kill shot: the left side $336-10x$ is $6$ short of a multiple of ten, so it always ends in the digit 6 - and of those four candidates, only $86$ does. That single observation eliminates $k=2$, $k=43$ and $k=86$ outright, with no trial-and-error, leaving $k=1$ as the one surviving branch.

$$336-10x=86 \implies 10x=250 \implies x=25$$

Then $y=112-4(25)=112-100=12$. Both prices are positive integers, and since only one branch survived, this pair is unique.
Step 4: Reconcile against the first visit. Uniqueness is worth nothing if the numbers don't actually satisfy the original trip. With $k=1$ he bought $2$ pens and $3$ pencils: $2(25)+3(12)=50+36=86$. That is exactly what he paid, so the pair holds.
Step 5: Calculate the final request. The prompt asks for the total amount paid *specifically for the pens* during the second visit.
He bought 4 pens at Rs. 25 each: $4\times 25=100$.

Final Answer: 100

The Setup: This is a cryptographic math puzzle. We need to build a four-digit integer, let's call it $ABCD$ (where $A$ is thousands, $B$ is hundreds, $C$ is tens, $D$ is units). We will decode the rules one by one to lock in each place value.
**Step 1: Solve the outer digits ($A$ and $D$).** We are given two conditions: $A\times D=0$ and the absolute difference is $7$.
For the product to be zero, either $A=0$ or $D=0$. Since it's a four-digit number, the leading digit $A$ cannot be $0$. Therefore, $D=0$.
Because $A$ must be positive, the difference of $7$ means $A-0=7$, locking in $A=7$.
**Step 2: Analyze the middle digits ($B$ and $C$).** We are told their product is 18: $B\times C=18$.
Since $B$ and $C$ are single digits, the only possible ordered pairs that multiply to 18 are $(2,9)$, $(9,2)$, $(3,6)$ and $(6,3)$.
Step 3: Apply the relational constraint. The prompt states: 'The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit.'
Translated to algebra, this means $A-D=B-C$.
We already know $A-D=7-0=7$, so we need $B-C=7$.
Run that filter across the four candidate pairs: $2-9=-7$, $9-2=7$, $3-6=-3$, $6-3=3$. Exactly one survives - $9-2=7$ - which locks in $B=9$ and $C=2$. Note the filter is a signed difference, not an absolute one, so $(2,9)$ is genuinely rejected rather than being a second answer.
Step 4: Assemble the final integer. We have $A=7$, $B=9$, $C=2$ and $D=0$. Concatenate them in order.

Final Answer: 7920

The Setup: We are min-maxing a 3-circle Venn diagram problem. We have a total population of 80 students across Mathematics (M), Physics (P) and Chemistry (C). We need to maximise the triple intersection (students who passed all three) without violating the 'at most 20' limits placed on the pairwise overlaps.
Step 1: Define the variables and totals. Let $x=n(M\cap P\cap C)$ be the number of students who passed all three. We want to push $x$ as high as possible.
The individual subject totals are $n(M)=50$, $n(P)=30$ and $n(C)=40$.
Step 2: Establish the ceiling. By the definition of intersection, anyone who passed all three also passed each pair, so the triple intersection is a subset of every pairwise intersection:

$$x\leq n(M\cap P), \qquad x\leq n(P\cap C), \qquad x\leq n(M\cap C)$$

The problem caps all three pairwise intersections at 20. Therefore $x\leq 20$ - this is a hard upper bound, no construction can beat it.
Step 3: Build a scenario that hits the ceiling. An upper bound is useless until we show it is reachable. Set $x=20$ and push every pairwise overlap to its cap of 20, which means the number of students who passed *exactly two* subjects is zero in all three regions. The *single-subject-only* groups then fall out:
* Passed M only: $50-20=30$
* Passed P only: $30-20=10$
* Passed C only: $40-20=20$

Step 4: Verify the headcount closes. Total students accounted for = (M only) + (P only) + (C only) + (all three):

$$30+10+20+20=80$$

This lands exactly on the 80 students who appeared, with no region going negative and every pairwise intersection sitting at $20$ rather than above it. The scenario is structurally sound, so the ceiling of 20 is genuinely achievable and is therefore the maximum.

Final Answer: 20

The Setup: This is a relative speed problem on a circular track. When runners start together and run the same way, the faster one *passes* the slower one every time he gains exactly one full lap. So the whole question reduces to: how many whole laps of lead does the winner build up before the race ends?
Step 1: Determine the total race parameters. The track is $300$ metres long and the race is $3$ kilometres, which equals $3000$ metres.

$$\text{Laps to finish}=\frac{3000}{300}=10$$

The winner finishes the instant he completes his 10th lap, and that is when the clock stops for both runners.
Step 2: Leverage the speed ratio. The speeds are in the ratio $3:2$. Both runners are on the track for the same duration, and at constant speed distance is directly proportional to speed, so their distances are in that same $3:2$ ratio.
Step 3: Locate the slower runner at the finish. When the winner has covered 10 laps:

$$\text{Slower runner}=\frac{2}{3}\times 10=\frac{20}{3}\text{ laps}$$

Keep this as an exact fraction - $\frac{20}{3}$ laps, not a rounded $6.67$ - because the final count depends on which side of a whole number the lead lands.
Step 4: Count the overtakes. The lead the winner builds over the full race is the difference in laps covered:

$$\text{Lead}=10-\frac{20}{3}=\frac{30-20}{3}=\frac{10}{3}\text{ laps}$$

The winner overtakes once for each whole lap of lead he gains, i.e. as the lead sweeps past $1$, $2$ and $3$. Since $\frac{10}{3}=3\frac{1}{3}$ is strictly greater than $3$ but short of $4$, the third overtake happens before the finish line and a fourth never does.

$$\text{Number of passes}=\left\lfloor \frac{10}{3}\right\rfloor=3$$

Final Answer: 3

The Setup: This combinatorics problem mixes indistinguishable (identical) objects with distinct ones. The standard combination formula $\binom{n}{r}$ only works for choosing distinct items. When dealing with indistinguishable objects, choosing one of them gives the same exact outcome as choosing another, so there is only 1 way to select any given quantity of them. We must break this down into cases based on how many identical items we draft into the final squad of 4.
Step 1: Catalog the inventory. Total pool $=13$ objects.
Indistinguishable objects $=4$. Distinct objects $=13-4=9$.
We need to form a team of exactly $4$ objects.
Step 2: Calculate combinations case-by-case. We iterate through every possible quantity of indistinguishable objects, from 0 up to 4 (we cannot take more than the 4 that exist).
* Case 1 (0 identical, 4 distinct): $\binom{9}{4}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 1}=126$ ways
* Case 2 (1 identical, 3 distinct): $1\times \binom{9}{3}=\frac{9\times 8\times 7}{3\times 2\times 1}=84$ ways
* Case 3 (2 identical, 2 distinct): $1\times \binom{9}{2}=\frac{9\times 8}{2\times 1}=36$ ways
* Case 4 (3 identical, 1 distinct): $1\times \binom{9}{1}=9$ ways
* Case 5 (4 identical, 0 distinct): $1\times \binom{9}{0}=1$ way

Step 3: Aggregate the total possibilities. The cases are mutually exclusive (they differ in how many identical objects were taken), so we add them:

$$126+84+36+9+1=256$$

Step 4: Cross-check with a symmetry argument. That total is $\binom{9}{0}+\binom{9}{1}+\binom{9}{2}+\binom{9}{3}+\binom{9}{4}$ - the lower half of row 9 of Pascal's triangle. Since $\binom{9}{k}=\binom{9}{9-k}$, the ten entries of that row pair off perfectly into two equal halves, so this sum is exactly half the row total:

$$\frac{2^9}{2}=\frac{512}{2}=256$$

Two independent routes, same number, so the count is confirmed.

Final Answer: 256

The Setup: We are dealing with prime factorization and combinations to calculate probability. We need to find the total number of factors of $10^{19}$ (the sample space) and the total number of factors that carry at least $10^{15}$ in their DNA (the winning outcomes).
Step 1: Break down the base stats. Prime factorize the main number:

$$10^{19}=(2\times 5)^{19}=2^{19}\times 5^{19}$$

The formula for the total number of factors of $p^a\times q^b$ is $(a+1)(b+1)$, since each exponent is chosen freely:

$$\text{Total factors}=(19+1)(19+1)=20\times 20=400$$

Step 2: Isolate the target condition. A factor of $10^{19}$ has the shape $2^x\times 5^y$ with $0\leq x,y\leq 19$. For it to be a multiple of $10^{15}=2^{15}\times 5^{15}$, it must carry at least fifteen of each prime, so $x\geq 15$ and $y\geq 15$.
Combined with the ceiling of 19, the valid range for both $x$ and $y$ is $\{15,16,17,18,19\}$.
Step 3: Count the winning outcomes. There are exactly 5 choices for the power of 2 and 5 choices for the power of 5, chosen independently:

$$\text{Valid factors}=5\times 5=25$$

Step 4: Calculate the final RNG. Probability is winning outcomes over total outcomes, and every factor is equally likely because the choice is uniform over all 400:

$$P=\frac{25}{400}=\frac{1}{16}$$

Final Answer: $\frac{1}{16}$

The Setup: This geometry problem is a triple filter. The sides must pass the triangle inequality vibe check, then survive the acute-angle condition, then fit under the perimeter cap. Whichever filter is strictest wins.
Step 1: Set up the side lengths. Let the three consecutive integer sides be $x-1$, $x$ and $x+1$, with $x-1\geq 1$.
For any triangle the two smaller sides must beat the largest: $(x-1)+x>x+1\implies x>2$.
Step 2: Enforce the acute angle rule. A triangle is acute when *every* angle is acute - but we only have to test one. The largest angle always sits opposite the longest side, so if that angle is acute the two smaller ones are smaller still and acute automatically. The longest side here is $x+1$, so the single condition to check is $a^2+b^2>c^2$:

$$(x-1)^2+x^2>(x+1)^2$$

$$x^2-2x+1+x^2>x^2+2x+1$$

$$x^2>4x$$

Since $x$ is positive we may divide by $x$, giving $x>4$. As $x$ is an integer, the lowest valid value is $x=5$ (the smallest qualifying triangle being $4,5,6$). Note this condition is strictly tighter than Step 1's $x>2$, so the triangle inequality is absorbed and adds nothing further.
Step 3: Apply the perimeter cap. The perimeter is at most 100:

$$(x-1)+x+(x+1)=3x\leq 100 \implies x\leq 33.33\ldots$$

Since $x$ is an integer, the maximum is $x=33$, giving sides $32,33,34$ and a perimeter of $99$.
Step 4: Count the surviving squad. The valid range for $x$ runs from 5 to 33 inclusive:

$$33-5+1=29$$

Final Answer: 29

The Setup: This is a coordinate geometry speedrun using intercept form. The line crosses the x-axis at some point $A(a,0)$ and the y-axis at $B(0,b)$. Saying $M$ divides the portion between the axes into two equal halves is just a fancy way of saying $M$ is the midpoint of segment $AB$.
Step 1: Apply the midpoint formula. The midpoint of $A(a,0)$ and $B(0,b)$ is $\left(\frac{a}{2},\frac{b}{2}\right)$, and we are told that point is $M(-5,4)$:

$$\frac{a}{2}=-5\implies a=-10$$

$$\frac{b}{2}=4\implies b=8$$

Step 2: Construct the line equation. Use the two-intercept form $\frac{x}{a}+\frac{y}{b}=1$:

$$\frac{x}{-10}+\frac{y}{8}=1$$

Step 3: Clear the fractions in one shot. Multiply every term by $80$ (a common multiple of 10 and 8 chosen to land directly on the answer's scale):

$$80\cdot\frac{x}{-10}+80\cdot\frac{y}{8}=80 \implies -8x+10y=80$$

Reordering gives $10y-8x=80$.
Step 4: Verify against all three known points. A midpoint slip is easy to make, so test the line on the two intercepts and $M$ itself:
* $A(-10,0)$: $10(0)-8(-10)=80$ ✓
* $B(0,8)$: $10(8)-8(0)=80$ ✓
* $M(-5,4)$: $10(4)-8(-5)=40+40=80$ ✓

All three sit on the line, and $M$ is genuinely halfway between $A$ and $B$. Running the other three options through $M(-5,4)$ fails every time, so the match is unique.

Final Answer: $10y - 8x = 80$

The Setup: This trig expression looks intimidating, but it has zero plot armour once you spot the symmetry in the angles. We use the supplementary angle identity to fold the expression in half, then the complementary identity to finish it off.
Step 1: Fold the back half. Notice that $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$, and $\cos(\pi-\theta)=-\cos\theta$, so $\cos\left(\frac{7\pi}{8}\right)=-\cos\left(\frac{\pi}{8}\right)$. Squaring kills the minus sign:

$$\cos^2\left(\frac{7\pi}{8}\right)=\cos^2\left(\frac{\pi}{8}\right)$$

Identically, $\frac{5\pi}{8}=\pi-\frac{3\pi}{8}$, so $\cos^2\left(\frac{5\pi}{8}\right)=\cos^2\left(\frac{3\pi}{8}\right)$.
Step 2: Simplify the sum. The four terms collapse into two identical pairs:

$$2\left[\cos^2\left(\frac{\pi}{8}\right)+\cos^2\left(\frac{3\pi}{8}\right)\right]$$

Step 3: Exploit the complementary angles. Notice that $\frac{\pi}{8}+\frac{3\pi}{8}=\frac{4\pi}{8}=\frac{\pi}{2}$. The two angles are complementary, and $\cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta$, so:

$$\cos\left(\frac{3\pi}{8}\right)=\sin\left(\frac{\pi}{8}\right)$$

Step 4: Use the legendary identity. Substitute sine into the folded expression:

$$2\left[\cos^2\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)\right]$$

By the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$, the bracket collapses to 1:

$$2\times 1=2$$

Final Answer: $2$

The Setup: We are manipulating the famous Basel problem. We are handed the sum of *all* inverse squares and asked for the sum of only the *odd* ones. The trick is to split the total into its even and odd halves, then notice the even half is a scaled copy of the whole thing.
Step 1: Define the series. Let the total be $S=\frac{\pi^2}{6}$. Every positive integer is either odd or even, so the total splits cleanly with nothing counted twice and nothing missed:

$$S=\text{Sum(Odds)}+\text{Sum(Evens)}$$

Every term here is positive and the series converges, so splitting and rearranging it like this is legitimate rather than a formal trick.
Step 2: Factor the even series. Write out the even terms and pull out the common factor. Each even number is $2n$, so its square is $4n^2$:

$$\text{Sum(Evens)}=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\ldots=\frac{1}{4}\left(\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\ldots\right)$$

The bracket is literally the original total $S$, so $\text{Sum(Evens)}=\frac{S}{4}$. This self-similarity is the whole engine of the problem.
Step 3: Solve for the odds. Substitute back into the master equation:

$$S=\text{Sum(Odds)}+\frac{S}{4} \implies \text{Sum(Odds)}=S-\frac{S}{4}=\frac{3S}{4}$$

Step 4: Final substitution. Plug in the known value of $S$:

$$\text{Sum(Odds)}=\frac{3}{4}\times \frac{\pi^2}{6}=\frac{3\pi^2}{24}=\frac{\pi^2}{8}$$

Final Answer: $\frac{\pi^2}{8}$

The Setup: This is a classic Bayes' Theorem trap. We have a prior probability (the physical die) mixed with a conditional one (the man's honesty stats). We want the probability that a five *actually* came up, given that he claims it did - which is not the same as his truthfulness rate, and the gap between $\frac{4}{5}$ and the answer is the entire point of the question.
Step 1: Map the base probabilities.
* Probability of rolling a five: $P(A)=\frac{1}{6}$
* Probability of rolling anything else: $P(\text{not }A)=\frac{5}{6}$
* Man tells the truth: $P(T)=\frac{4}{5}$
* Man lies: $P(L)=\frac{1}{5}$

Step 2: Map the conditional claims.
If it genuinely is a five, he reports five precisely when he is being truthful: $P(\text{Claims 5}\mid \text{Is 5})=\frac{4}{5}$.
If it is not a five, he can only report five by lying, so this branch is taken as $P(\text{Claims 5}\mid \text{Not 5})=\frac{1}{5}$ - the exam's convention is that a lie about the target face lands on that face.
*Worth naming the assumption, because it is doing real work.* If instead a liar spread his lie uniformly across the five wrong faces, this branch would be $\frac{1}{5}\times\frac{1}{5}=\frac{1}{25}$ and the answer would come out as $\frac{4}{5}$. That value appears nowhere in the options, whereas the convention above lands exactly on option 1 - so the option set itself tells us which reading the paper intended.
Step 3: Apply Bayes' Theorem.

$$P(\text{Is 5}\mid \text{Claims 5})=\frac{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})}{P(\text{Is 5})\times P(\text{Claims 5}\mid \text{Is 5})+P(\text{Not 5})\times P(\text{Claims 5}\mid \text{Not 5})}$$

Step 4: Crunch the numbers.

$$\text{Numerator}=\frac{1}{6}\times \frac{4}{5}=\frac{4}{30}$$

$$\text{Denominator}=\left(\frac{1}{6}\times \frac{4}{5}\right)+\left(\frac{5}{6}\times \frac{1}{5}\right)=\frac{4}{30}+\frac{5}{30}=\frac{9}{30}$$

$$P=\frac{\ \frac{4}{30}\ }{\frac{9}{30}}=\frac{4}{9}$$

Sanity check on the direction: $\frac{4}{9}<\frac{1}{2}$, which is right - a five is rare enough on a fair die that even a mostly-honest report leaves it more likely than not that he is lying.

Final Answer: $\frac{4}{9}$

The Setup: This is a logarithm inception. We unpack it from the outside in using the fundamental definition $\log_b(A)=C\implies A=b^C$, then check the candidates against the option list.
Step 1: Destroy the outer log.

$$\log_5\left(\log_8(x^2-1)\right)=0$$

Convert from log form to exponential form:

$$\log_8(x^2-1)=5^0=1$$

Step 2: Destroy the inner log. Apply the same rule to what remains:

$$x^2-1=8^1=8$$

Step 3: Solve the quadratic.

$$x^2=9 \implies x=\pm 3$$

Step 4: Pick the value that is actually on the menu. Both $x=3$ and $x=-3$ satisfy the equation - the expression only ever sees $x^2$, so the sign is irrelevant, and $x^2-1=8>0$ keeps every logarithm well defined. This is exactly why the stem asks for *a possible* value rather than *the* value. Of the two, only $3$ appears among the options, so it is the answer.
The other three options all fail, and one of them is a genuine trap:
* $x=2\sqrt{2}$: $x^2-1=7$, and $\log_8 7\approx 0.936$, so the outer log is not 0.
* $x=\sqrt{2}$: $x^2-1=1$, so $\log_8 1=0$ - and $\log_5 0$ is undefined, not zero. Mistaking one for the other is the intended snare.
* $x=2$: $x^2-1=3$, and $\log_8 3\approx 0.528$, so the outer log is not 0.