IPMAT Indore 2023 — QA

The Setup: We are given relative speed and time constraints across four car gears to complete a uniform distance, which we must use to evaluate the time taken for a mixed-distance journey.

Step 1: Define the speed variables and their relations.
Let the speeds in the four gears be $v_1, v_2, v_3,$ and $v_4$.
We are given $v_4 = 5v_1$.
We are also told the time in the 2nd gear is twice the time in the 3rd gear for a constant distance: $\frac{d}{v_2} = 2\left(\frac{d}{v_3}\right) \implies v_3 = 2v_2$.

Step 2: Express the times for the $100\text{ km}$ uniform journey.
The $100\text{ km}$ journey consists of equal distances in each gear, meaning $25\text{ km}$ per gear.
Let $x = \frac{1}{v_1}$ and $y = \frac{1}{v_2}$.
This means $\frac{1}{v_4} = \frac{x}{5}$ and $\frac{1}{v_3} = \frac{y}{2}$.
The total time equation is:

$$25\left(\frac{1}{v_1}\right) + 25\left(\frac{1}{v_2}\right) + 25\left(\frac{1}{v_3}\right) + 25\left(\frac{1}{v_4}\right) = 585$$

$$25x + 25y + 25\left(\frac{y}{2}\right) + 25\left(\frac{x}{5}\right) = 585$$

$$25x + 5x + 25y + 12.5y = 585$$

$$30x + 37.5y = 585$$

Step 3: Simplify the baseline equation.
Divide the entire equation by $7.5$:

$$4x + 5y = 78$$

Step 4: Calculate the time for the newly requested journey.
The new distances are $4\text{ km}$ (1st), $4\text{ km}$ (2nd), $32\text{ km}$ (3rd), and $60\text{ km}$ (4th).

$$\text{Total Time} = 4\left(\frac{1}{v_1}\right) + 4\left(\frac{1}{v_2}\right) + 32\left(\frac{1}{v_3}\right) + 60\left(\frac{1}{v_4}\right)$$

Substitute $x$ and $y$:

$$\text{Total Time} = 4x + 4y + 32\left(\frac{y}{2}\right) + 60\left(\frac{x}{5}\right)$$

$$\text{Total Time} = 4x + 4y + 16y + 12x$$

$$\text{Total Time} = 16x + 20y$$

Factor out a $4$:

$$\text{Total Time} = 4(4x + 5y)$$

Step 5: Substitute the baseline value to find the final time.

$$\text{Total Time} = 4(78) = 312\text{ minutes}$$

Final Answer: 312

The Setup: We need to find the remainder when a factorial summation sequence ($1!+2!+3!+...+95!$) is divided by $15$.

Step 1: Analyze the divisibility of factorials by $15$.
Since $15 = 3 \times 5$, any factorial $n!$ where $n \ge 5$ will strictly contain both $3$ and $5$ as prime factors.
Therefore, for all $n \ge 5$, $n! \equiv 0 \pmod{15}$.

Step 2: Evaluate the terms that are not divisible by $15$.
The only terms in the sequence that do not contain $5!$ as a factor are the first four terms:

$$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$$

Step 3: Calculate the final remainder.
The entire sequence modulo $15$ reduces simply to $33 \pmod{15}$.

$$33 = 2 \times 15 + 3$$

The remainder is $3$.

Final Answer: 3

The Setup: Given a sum equation $a+b+c+d=2023$ for positive integers and specified ratio pairs, we must maximize the sum $a+c$.

Step 1: Express the variables using their ratios.
Given $a:b=2:5$, let $a = 2x$ and $b = 5x$ for some positive integer $x$.
Given $c:d=5:2$, let $c = 5y$ and $d = 2y$ for some positive integer $y$.

Step 2: Formulate the sum constraint.
Substitute the variables into the total sum equation:

$$2x + 5x + 5y + 2y = 2023$$

$$7x + 7y = 2023$$

$$x + y = 289$$

Step 3: Optimize the target function.
We want to maximize $a + c$, which translates to maximizing $2x + 5y$.
To maximize the value of $2x + 5y$ given the constraint $x + y = 289$, we must make $y$ (which carries the larger coefficient $5$) as large as possible.
Since $a,b,c,d$ are strictly positive integers, the scaling factors $x$ and $y$ must be at least $1$.
The maximum valid integer value for $y$ occurs when $x$ is minimized at $1$.

$$1 + y = 289 \implies y = 288$$

Step 4: Calculate the maximum value.

$$\text{Max}(a+c) = 2(1) + 5(288) = 2 + 1440 = 1442$$

Final Answer: 1442

The Setup: We are asked to determine the total number of voters on a list given percentage-based turnout and fractional vote shares in a two-candidate election.

Step 1: Define the election parameters.
Let $V$ be the total number of eligible voters on the list.
Voters who didn't turn up = $0.15V$.
Total votes cast = $0.85V$.
Valid votes cast = $0.85V - 50$.

Step 2: Represent the candidates' vote shares.
The winner got exactly $44\%$ of *all voters on the voting list* (not just valid votes).

$$W = 0.44V$$

The other candidate (the loser) received the remaining valid votes:

$$L = \text{Valid Votes} - W = (0.85V - 50) - 0.44V = 0.41V - 50$$

Step 3: Apply the winning margin constraint to solve for $V$.
The winner won by exactly $200$ votes.

$$W - L = 200$$

$$0.44V - (0.41V - 50) = 200$$

$$0.03V + 50 = 200$$

$$0.03V = 150$$

$$V = \frac{150}{0.03} = 5000$$

Final Answer: 5000

The Setup: A recursive sequence is defined as $f(n)=3n-f(n-1)$ with base case $f(1)=1$. We must find the value of $f(2023)$.

Step 1: Generate the first few terms to identify a pattern.
$f(1) = 1$
$f(2) = 3(2) - 1 = 5$
$f(3) = 3(3) - 5 = 4$
$f(4) = 3(4) - 4 = 8$
$f(5) = 3(5) - 8 = 7$
$f(6) = 3(6) - 7 = 11$

Step 2: Formulate a generalized relationship.
Let's find the difference between every second term ($f(n)$ and $f(n-2)$):

$$f(n) = 3n - f(n-1)$$

Substitute the recursive definition for $f(n-1)$:

$$f(n) = 3n - (3(n-1) - f(n-2))$$

$$f(n) = 3n - 3n + 3 + f(n-2)$$

$$f(n) = f(n-2) + 3$$

This confirms that the odd-indexed terms and even-indexed terms form independent Arithmetic Progressions, each with a common difference of $3$.

Step 3: Evaluate $f(2023)$.
Since $2023$ is odd, we use the sequence of odd-indexed terms: $1, 4, 7, 10, \dots$
The generic formula for the odd index sequence $f(2k-1)$ is an AP where the first term is $1$ and the difference is $3$.
Let $n = 2023$. The number of steps (or the term position in the separated odd sequence) is $\frac{n-1}{2}$.

$$\text{Term Index} = \frac{2023 - 1}{2} = 1011$$

$$f(2023) = \text{Base} + (\text{Term Index}) \times \text{Difference}$$

$$f(2023) = 1 + 1011 \times 3$$

$$f(2023) = 1 + 3033 = 3034$$

Final Answer: 3034

The Setup: Points $A=(-2,0)$ and $B=(2,0)$ lie on the $x$-axis. We must find all points $C$ on the circle $x^2+y^2=4$ such that the area of $\triangle ABC$ is an integer.

Step 1: Define the area of the triangle.
The base of the triangle is the segment $AB$ connecting $(-2,0)$ and $(2,0)$. Its length is exactly $4$.
Let point $C$ have coordinates $(x, y)$. The height of the triangle relative to base $AB$ is the absolute value of the $y$-coordinate, $|y|$.

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times |y| = 2|y|$$

Step 2: Establish integer constraints for the area.
We require the Area ($2|y|$) to be a positive integer $k$ (a valid geometric triangle must have an area strictly greater than 0).

$$2|y| = k \implies |y| = \frac{k}{2}$$

Since $C$ lies on the circle $x^2+y^2=4$, the maximum possible value for $|y|$ is the radius, $2$.

$$0 < |y| \le 2 \implies 0 < \frac{k}{2} \le 2 \implies k \in \{1, 2, 3, 4\}$$

Step 3: Map the valid area states back to coordinate points.
For each valid $k$, we substitute $|y| = k/2$ back into the circle equation $x^2 = 4 - y^2$ to count the distinct points:
* If $k = 1$: $|y| = 0.5 \implies y = \pm 0.5$. $x^2 = 4 - 0.25 = 3.75$. This yields $2$ values for $x$ per $y$, giving $4$ points.
* If $k = 2$: $|y| = 1 \implies y = \pm 1$. $x^2 = 4 - 1 = 3$. This gives $4$ points.
* If $k = 3$: $|y| = 1.5 \implies y = \pm 1.5$. $x^2 = 4 - 2.25 = 1.75$. This gives $4$ points.
* If $k = 4$: $|y| = 2 \implies y = \pm 2$. $x^2 = 4 - 4 = 0$. Here $x = 0$ uniquely, giving exactly $1$ point for $y=2$ and $1$ point for $y=-2$, totaling $2$ points.

Step 4: Sum the valid points.

$$\text{Total Points} = 4 + 4 + 4 + 2 = 14$$

Final Answer: 14

The Setup: We are given three consecutive coefficients in the binomial expansion of $(x+y)^n$ strictly in the ratio $1:9:63$. We must determine the power $n$.

Step 1: Express the coefficients algebraically.
Let the three consecutive coefficients be $\binom{n}{r-1}$, $\binom{n}{r}$, and $\binom{n}{r+1}$.
We are given two ratio equations:

$$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{9}{1} = 9$$

$$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{63}{9} = 7$$

Step 2: Apply the standard binomial coefficient ratio formula.
The ratio $\frac{\binom{n}{k}}{\binom{n}{k-1}}$ simplifies universally to $\frac{n - k + 1}{k}$.
Apply this to our first equation ($k = r$):

$$\frac{n - r + 1}{r} = 9 \implies n - r + 1 = 9r \implies n = 10r - 1$$

Apply this to our second equation ($k = r + 1$):

$$\frac{n - (r+1) + 1}{r+1} = 7 \implies \frac{n - r}{r + 1} = 7 \implies n - r = 7r + 7 \implies n = 8r + 7$$

Step 3: Solve the linear system for $r$ and $n$.
Equate the two expressions for $n$:

$$10r - 1 = 8r + 7$$

$$2r = 8 \implies r = 4$$

Substitute $r$ back into either equation to find $n$:

$$n = 10(4) - 1 = 39$$

Final Answer: 39

The Setup: A chess tournament consists of four equal groups. Players play matches inside their group and against the other three groups at varying frequencies ($1\times, 2\times, 3\times, 4\times$). We must find the minimum group size to surpass $1000$ total matches.

Step 1: Analyze the match distribution logic.
Let each of the $4$ groups have $k$ players. Total players = $4k$.
Every group must be assigned an inter-group match frequency of $2, 3,$ and $4$ against the other three groups.
This is perfectly modeled by weighting the edges of a $K_4$ graph with values $\{2, 3, 4\}$ such that every vertex touches exactly one of each weight. Such a configuration requires three distinct perfect matchings on $K_4$, meaning the edge weights between the $6$ pairs of groups are exactly: two $2$s, two $3$s, and two $4$s.

Step 2: Calculate total inter-group matches.
Between any two groups of size $k$ playing each other $w$ times, the number of matches is $k \times k \times w = w k^2$.
Summing over the 6 pairs of groups:

$$\text{Inter-Group Matches} = k^2(2 + 2 + 3 + 3 + 4 + 4) = 18k^2$$

Step 3: Calculate total intra-group matches.
Each player plays everyone in their own group exactly once.
For one group, the matches are $\binom{k}{2} = \frac{k(k-1)}{2}$.
Since there are $4$ groups:

$$\text{Intra-Group Matches} = 4 \times \frac{k(k-1)}{2} = 2k(k-1) = 2k^2 - 2k$$

Step 4: Formulate the total matches inequality.

$$\text{Total Matches} = 18k^2 + 2k^2 - 2k = 20k^2 - 2k$$

We are given that there are more than $1000$ matches:

$$20k^2 - 2k > 1000 \implies 10k^2 - k > 500$$

Step 5: Test integer values for $k$ to find the minimum.
* If $k = 7$: $10(49) - 7 = 490 - 7 = 483$ (Fails, $483 \ngtr 500$)
* If $k = 8$: $10(64) - 8 = 640 - 8 = 632$ (Succeeds, $632 > 500$)
The minimum possible number of players in each group is $8$.

Final Answer: 8

The Setup: We need to find the product of the roots for the logarithmic equation $log_{2}2^{(log_{2}x)^{2}}-5~log_{2}x+6=0$.

Step 1: Simplify the first term using logarithm properties.
The term $\log_2(2^y)$ universally simplifies to $y$.
Thus, $\log_2(2^{(\log_2 x)^2})$ simplifies directly to $(\log_2 x)^2$.
The equation becomes:

$$(\log_2 x)^2 - 5\log_2 x + 6 = 0$$

Step 2: Apply a variable substitution.
Let $t = \log_2 x$. The equation is a standard quadratic:

$$t^2 - 5t + 6 = 0$$

Step 3: Solve for $t$ and reverse the substitution.
Factor the quadratic:

$$(t - 2)(t - 3) = 0 \implies t = 2 \text{ or } t = 3$$

Reverse the substitution to solve for the roots in terms of $x$:
If $t = 2 \implies \log_2 x = 2 \implies x = 2^2 = 4$
If $t = 3 \implies \log_2 x = 3 \implies x = 2^3 = 8$

Step 4: Calculate the product of the roots.

$$\text{Product} = 4 \times 8 = 32$$

Final Answer: 32

The Setup: We must find the total number of positive integer solutions to the bounded inequality $21 \le a+b+c \le 25$.

Step 1: Normalize the variables to utilize Stars and Bars.
Since $a, b, c$ must be strictly positive integers ($\ge 1$), let:
$a = x + 1$
$b = y + 1$
$c = z + 1$
where $x, y, z$ are non-negative integers ($\ge 0$).
Substitute these into the inequality:

$$21 \le (x+1) + (y+1) + (z+1) \le 25$$

$$21 \le x + y + z + 3 \le 25$$

$$18 \le x + y + z \le 22$$

Step 2: Apply the combinatorics formula for exact sums.
The number of non-negative integer solutions to $x+y+z = n$ is given by $\binom{n+3-1}{3-1} = \binom{n+2}{2}$.
We need to sum this over $n \in \{18, 19, 20, 21, 22\}$:

$$\text{Solutions} = \binom{20}{2} + \binom{21}{2} + \binom{22}{2} + \binom{23}{2} + \binom{24}{2}$$

Step 3: Evaluate the sum.
Calculate each combination $\binom{k}{2} = \frac{k(k-1)}{2}$:
$\binom{20}{2} = \frac{20 \times 19}{2} = 190$
$\binom{21}{2} = \frac{21 \times 20}{2} = 210$
$\binom{22}{2} = \frac{22 \times 21}{2} = 231$
$\binom{23}{2} = \frac{23 \times 22}{2} = 253$
$\binom{24}{2} = \frac{24 \times 23}{2} = 276$

$$\text{Total} = 190 + 210 + 231 + 253 + 276 = 1160$$

Final Answer: 1160

The Setup: We must find the geometric length of the line segment connecting the two intersection points of the absolute value function $y = 4970 - |x|$ and the parabola $y = x^2$.

Step 1: Set the equations equal to find intersection points.

$$x^2 = 4970 - |x|$$

Rearrange to form a quadratic in terms of $|x|$, recognizing that $x^2 = |x|^2$:

$$|x|^2 + |x| - 4970 = 0$$

Step 2: Factor the quadratic equation.
We need two integers that multiply to $-4970$ and add to $1$.
Note that $4970 = 10 \times 497 = 10 \times 7 \times 71 = 70 \times 71$.

$$(|x| + 71)(|x| - 70) = 0$$

Step 3: Solve for $x$ and locate the coordinates.
Since absolute value $|x|$ must be non-negative, $|x| = -71$ is rejected.

$$|x| = 70 \implies x = 70 \text{ or } x = -70$$

Because the equations are perfectly symmetric across the y-axis, the segment connecting the two points is perfectly horizontal.

Step 4: Calculate the distance.
The length of a horizontal line segment is simply the absolute difference between the $x$-coordinates.

$$\text{Length} = |70 - (-70)| = 140$$

Final Answer: 140

The Setup: Given an arithmetic sum base function $f(n)$, we have a reciprocal series $g(n)$. We must find the least integer $n$ that pushes $g(n)$ above the threshold of $99/100$.

Step 1: Simplify the base function $f(n)$.
The function is the sum of consecutive integers:

$$f(n) = 1 + 2 + 3 + \dots + (n+1) = \frac{(n+1)(n+2)}{2}$$

Step 2: Construct the series $g(n)$ and apply partial fractions.

$$g(n) = \sum_{k=1}^n \frac{1}{f(k)} = \sum_{k=1}^n \frac{2}{(k+1)(k+2)}$$

Split the fraction using partial fraction decomposition:

$$\frac{2}{(k+1)(k+2)} = 2 \left( \frac{1}{k+1} - \frac{1}{k+2} \right)$$

Step 3: Evaluate the telescoping sum.
When summing from $k=1$ to $n$, all intermediate terms cancel out:

$$g(n) = 2 \left[ \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \dots + \left(\frac{1}{n+1} - \frac{1}{n+2}\right) \right]$$

$$g(n) = 2 \left( \frac{1}{2} - \frac{1}{n+2} \right) = 1 - \frac{2}{n+2}$$

Step 4: Apply the threshold inequality.
We need $g(n) > \frac{99}{100}$:

$$1 - \frac{2}{n+2} > 0.99$$

$$0.01 > \frac{2}{n+2}$$

$$\frac{1}{100} > \frac{2}{n+2} \implies n+2 > 200 \implies n > 198$$

Since $n$ must be an integer, the least value fulfilling this is $199$.

Final Answer: 199

The Setup: A polynomial divided by $(x-1)$ leaves a remainder of $2$. We apply the Polynomial Remainder Theorem to find a missing coefficient $c$ and then compute $c+6$.

Step 1: Apply the Remainder Theorem.
If a polynomial $P(x)$ is divided by $(x - a)$, the remainder is exactly $P(a)$.
Here, dividing by $(x - 1)$ gives a remainder of $2$, meaning $P(1) = 2$.

Step 2: Evaluate $P(1)$ algebraically.
Substitute $x = 1$ into the polynomial. For $x=1$, all powers of $x$ resolve to $1$, leaving just the sum of the coefficients.

$$P(1) = 4(1) - 1(1) + 3(1) - 5(1) + c(1) + 2(1) - 1(1) + 1(1) - 4(1) + 6(1) - 2$$

Group the numerical terms:

$$P(1) = 4 - 1 + 3 - 5 + 2 - 1 + 1 - 4 + 6 - 2 + c$$

$$P(1) = 3 + c$$

Step 3: Equate to the remainder and solve.

$$c + 3 = 2 \implies c = -1$$

Step 4: Calculate the final expression.
We need the value of $c + 6$.

$$-1 + 6 = 5$$

Final Answer: 5

The Setup: We evaluate sequential deposits placed at the end of each year with growing principals and scaling simple interest rates over a bounded 5-year timeline from 'today' (Start of Year 1).

Step 1: Chart the timeline and investment parameters.
Five years from today implies the timeline ends exactly at the conclusion of Year 5 (T=5).
Simple Interest Formula: $\text{Interest} = P \times R \times T$.
* Deposit 1 (End of Yr 1, T=1): Earns $10,000$. Rate = $0.5\% = 0.005$. Time invested = $4$ years (T=1 to T=5).
* Deposit 2 (End of Yr 2, T=2): Income rises by 10k, so earns $20,000$. Rate rises by 0.5%, so = $1.0\% = 0.01$. Time invested = $3$ years.
* Deposit 3 (End of Yr 3, T=3): Earns $30,000$. Rate = $1.5\% = 0.015$. Time invested = $2$ years.
* Deposit 4 (End of Yr 4, T=4): Earns $40,000$. Rate = $2.0\% = 0.02$. Time invested = $1$ year.
* Deposit 5 (End of Yr 5, T=5): Earns $50,000$. Rate = $2.5\% = 0.025$. Time invested = $0$ years (cashed exactly as deposited).

Step 2: Calculate interest for each independent deposit.
* Interest 1: $10000 \times 0.005 \times 4 = 200$
* Interest 2: $20000 \times 0.010 \times 3 = 600$
* Interest 3: $30000 \times 0.015 \times 2 = 900$
* Interest 4: $40000 \times 0.020 \times 1 = 800$
* Interest 5: $50000 \times 0.025 \times 0 = 0$

Step 3: Sum the total interest.

$$\text{Total Interest} = 200 + 600 + 900 + 800 + 0 = 2500$$

Final Answer: 2500

The Setup: A worker finishes a partial task at shifting rates and enlists a co-worker to complete the remaining fraction. By mapping their combined work capacity against the remaining work, we can isolate the co-worker's individual rate.

Step 1: Establish Amisha's work rates.
Amisha's standard rate allows her to finish in $20$ days, so her normal rate is $A = \frac{1}{20}$ tasks/day.
When she resumes, she works at half her original rate, making her new rate $A' = \frac{1}{40}$ tasks/day.

Step 2: Calculate work completed before the break.
Amisha works normally for $4$ days.

$$\text{Work Done} = 4 \times \left(\frac{1}{20}\right) = \frac{4}{20} = \frac{1}{5}$$

$$\text{Remaining Work} = 1 - \frac{1}{5} = \frac{4}{5}$$

Step 3: Formulate the combined work phase.
She is sick for $4$ days, so Day 9 begins the new phase. From Day 9, she and her co-worker work together for $12$ days to finish the job.
Let the co-worker's rate be $C$ tasks/day.

$$\text{Combined Rate} = A' + C = \frac{1}{40} + C$$

$$\text{Work done together} = 12 \times \left(\frac{1}{40} + C\right)$$

Set this equal to the remaining work:

$$12 \left(\frac{1}{40} + C\right) = \frac{4}{5}$$

Step 4: Solve for $C$ and convert to days.
Divide both sides by $12$:

$$\frac{1}{40} + C = \frac{4}{60} = \frac{1}{15}$$

$$C = \frac{1}{15} - \frac{1}{40}$$

Find a common denominator ($120$):

$$C = \frac{8}{120} - \frac{3}{120} = \frac{5}{120} = \frac{1}{24}$$

Since the co-worker's rate is $\frac{1}{24}$ of the task per day, it would take them exactly $24$ days to complete the task alone.

Final Answer: 24

The Setup: We are given a positive integer $p$ such that the unit digit of $p^3$ is exactly $4$. We need to determine the possible unit digits of the expression $(p+3)^3$.

Step 1: Determine the unit digit of $p$.
We evaluate the cubes of all base digits from $0$ to $9$ to see which produces a unit digit of $4$:
$0^3 = 0, 1^3 = 1, 2^3 = 8, 3^3 = 27, 4^3 = 64, 5^3 = 125, 6^3 = 216, 7^3 = 343, 8^3 = 512, 9^3 = 729$.
The only digit whose cube ends in $4$ is $4$ itself.
Therefore, the unit digit of $p$ must be $4$.