IPMAT Indore 2021 — SA

The Setup: We need the total number of positive integer divisors for a massive product, filtering out the ones divisible by 45. Think of it like running a script to ban any accounts in your Discord server that have a specific toxic role.
Step 1: Combine the numbers and find the prime factorization of the main boss, $N=(1890)(130)(170)$.
$1890 = 2 \times 3^3 \times 5 \times 7$
$130 = 2 \times 5 \times 13$
$170 = 2 \times 5 \times 17$
Multiply them together to get the ultimate form:

$$N = 2^3 \times 3^3 \times 5^3 \times 7^1 \times 13^1 \times 17^1$$

Step 2: Calculate the total number of positive integers that divide $N$. Add 1 to each exponent and multiply:

$$\text{Total Divisors} = (3+1)(3+1)(3+1)(1+1)(1+1)(1+1)$$

$$\text{Total} = 4 \times 4 \times 4 \times 2 \times 2 \times 2 = 512$$

Step 3: Figure out how many of these are divisible by 45. Since $45=3^2 \times 5^1$, any divisor that is a multiple of 45 must have at least $3^2$ and $5^1$ locked into its prime factorization.
We lock in those required factors, leaving us with a remaining pool of choices:
* Powers of 3 left: $3-2=1$ (so 2 choices: $3^2, 3^3$)
* Powers of 5 left: $3-1=2$ (so 3 choices: $5^1, 5^2, 5^3$)
* The rest (2, 7, 13, 17) keep all their original choices (4, 2, 2, 2).

$$\text{Banned Divisors} = 4 \times 2 \times 3 \times 2 \times 2 \times 2 = 192$$

Step 4: Subtract the banned divisors from the total pool.

$$\text{Clean Divisors} = 512 - 192 = 320$$

Final Answer: 320

The Setup: We need the sum of the first 10 terms of the series $1\cdot3+5\cdot7+9\cdot11+...$. We are dealing with products of Arithmetic Progressions (APs). We need to unmask the $n$-th term so we can just blast it with summation formulas and speedrun the calculation.
Step 1: Find the pattern for the first part of each term (1, 5, 9...). This is an AP with a common difference of 4.

$$a_n = 1 + (n-1)4 = 4n - 3$$

Step 2: Find the pattern for the second part of each term (3, 7, 11...). This is another AP, also with a common difference of 4.

$$b_n = 3 + (n-1)4 = 4n - 1$$

Step 3: Construct the general term $T_n$ by multiplying them, and expand it into a quadratic polynomial:

$$T_n = (4n - 3)(4n - 1) = 16n^2 - 16n + 3$$

Step 4: Deploy the summation formulas for the first 10 terms.

$$\sum_{n=1}^{10} T_n = 16 \sum n^2 - 16 \sum n + \sum 3$$

Using the standard formulas $\sum n^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum n = \frac{n(n+1)}{2}$ for $n=10$:
* $16 \times \left(\frac{10 \times 11 \times 21}{6}\right) = 16 \times 385 = 6160$
* $-16 \times \left(\frac{10 \times 11}{2}\right) = -16 \times 55 = -880$
* $10 \times 3 = 30$
Step 5: Crunch the final numbers.

$$6160 - 880 + 30 = 5310$$

Final Answer: 5310

The Setup: We are faced with a recursive sequence: $x_{n+1}=x_n+1+2\sqrt{1+x_n}$ with a starting state of $x_1=0$. It looks cursed, but it's actually just a perfect square in disguise. We just need to transform the variables to reveal the hidden meta.
Step 1: Add 1 to both sides to complete the transformation.

$$x_{n+1} + 1 = (x_n + 1) + 2\sqrt{1+x_n} + 1$$

Step 2: Substitute $y_n=x_n+1$ to clean up the UI.

$$y_{n+1} = y_n + 2\sqrt{y_n} + 1$$

Notice that the right side is a literal perfect square: $(\sqrt{y_n} + 1)^2$.

$$y_{n+1} = (\sqrt{y_n} + 1)^2$$

Step 3: Take the square root of both sides.

$$\sqrt{y_{n+1}} = \sqrt{y_n} + 1$$

This means the sequence $\sqrt{y_n}$ is just a basic Arithmetic Progression increasing by 1 every step. Since $x_1=0$, our base case is $y_1=1$, which means $\sqrt{y_1}=1$.
Step 4: Find the $n$-th term of this AP.

$$\sqrt{y_n} = 1 + (n-1)(1) = n$$

Step 5: Reverse the transformation to find $x_n$.

$$y_n = n^2 \implies x_n + 1 = n^2 \implies x_n = n^2 - 1$$

Step 6: Calculate the 31st term to finish the level.

$$x_{31} = 31^2 - 1 = 961 - 1 = 960$$

Final Answer: 960

The Setup: We are stepping into the Combinatorics arena. There are 5 parallel lines and $n$ perpendicular lines. To build a rectangle, you need a squad of exactly 4 lines: 2 horizontal and 2 vertical. It's like picking your duo for Fortnite-you need exactly two from each distinct set to lock in the shape.
Step 1: Calculate the number of pairs we can pull from the 5 parallel lines. Choosing 2 out of 5 is basic combinations:

$$\binom{5}{2} = \frac{5 \times 4}{2} = 10$$

Step 2: Set up the equation for the perpendicular lines. We need 2 lines from the $n$ squad, which gives us $\binom{n}{2}$. The total number of distinct rectangles is 360.

$$10 \times \binom{n}{2} = 360$$

Step 3: Solve for $n$. Divide both sides by 10 to strip it down:

$$\binom{n}{2} = 36$$

$$\frac{n(n-1)}{2} = 36$$

$$n(n-1) = 72$$

Step 4: Find the consecutive integers that multiply to 72. Since $9 \times 8 = 72$, we know $n=9$. No cap, it's that clean.
Final Answer: 9

The Setup: This is a classic 'Time and Work' rate problem, just flipped to deal with emptying a tank instead of filling it. T1 clears the tank in 30 minutes solo, and T1 and T2 together clear it in 20 minutes. We treat the full tank as 1 complete unit of work and need to isolate the standalone speed of tap T2.
Step 1: Establish T1's solo drain rate.

$$\text{Rate of T1} = \frac{1}{30} \text{ tanks per minute}$$

Step 2: Look at their combined synergy.

$$\text{Rate of T1} + \text{Rate of T2} = \frac{1}{20} \text{ tanks per minute}$$

Step 3: Subtract T1's rate from the combined rate to expose T2's true solo speed.

$$\text{Rate of T2} = \frac{1}{20} - \frac{1}{30}$$

Step 4: Find a common denominator (which is 60) to crunch the fractions:

$$\frac{3}{60} - \frac{2}{60} = \frac{1}{60}$$

This means T2 clears $\frac{1}{60}$ of the tank every minute.
Step 5: Flip the rate to find the total time. If T2 does $\frac{1}{60}$ of the job per minute, it takes exactly 60 minutes to finish the whole thing solo.
Final Answer: 60

The Setup: This is a weighted average and total sum problem. 30 students are taking 5 courses, with each exam out of 200 marks. Averages are just totals spread out evenly. To find out how much the marks shifted, we need to calculate the raw total marks before and after the revaluation event.
Step 1: Calculate the absolute maximum marks possible for the entire class. 30 students taking 5 courses means there are 150 individual exams.

$$150 \times 200 = 30000 \text{ total possible marks}$$

Step 2: Find the raw total marks before the revaluation. The squad averaged an 80% hit rate.

$$0.80 \times 30000 = 24000 \text{ marks}$$

Step 3: Find the raw total marks after the revaluation bump. The new average slightly leveled up to 80.02%.

$$0.8002 \times 30000 = 24006 \text{ marks}$$

Step 4: Find the net increase. The total pool of marks literally went up by exactly 6 ($24006 - 24000$).
Step 5: Allocate the points. We know two students got revaluations and neither lost points. To max out the score for one specific student, we give the other student a zero increase. This lets one student hog all 6 newly discovered points.
Final Answer: 6

The Setup: We need to find the minimum number of weights to measure any integer from 1 to 100 grams. The catch? We can *only* place weights on the left pan. That means no subtractive combinations (like putting weights on both sides). We are strictly locked into binary combinations (base-2).
Step 1: Understand the binary scale system. Since each weight can either be used (1) or not used (0), having $k$ weights can measure at most $2^k - 1$ distinct positive values.
Step 2: Set up the inequality to cover everything up to 100 grams.

$$2^k - 1 \ge 100$$

$$2^k \ge 101$$

Step 3: Find the smallest integer $k$ that satisfies this. We know our binary powers:
* $2^6 = 64$ (too small)
* $2^7 = 128$ (just right)
Therefore, $k=7$ is the absolute minimum number of weights required. You'd literally just use the weights 1, 2, 4, 8, 16, 32, and 64 to carry the whole task.
Final Answer: 7

The Setup: We're dealing with homogenous equations of degree 2, which represent pairs of straight lines passing through the origin. The equations are $2x^2+axy+3y^2=0$ and $2x^2+bxy-3y^2=0$. One line coincides, and the other lines are perpendicular. This is pure coordinate geometry boss-fight level. Let the slope of the shared line be $m$, and the perpendicular ones be $m_1$ and $m_2$.
Step 1: Translate the given equations to quadratic form in terms of slope ($y=mx$).
For the first pair, dividing by $x^2$ gives:

$$3m^2 + am + 2 = 0$$

The product of the roots (slopes) is $m \cdot m_1 = \frac{2}{3} \implies m_1 = \frac{2}{3m}$.
Step 2: Repeat for the second pair.

$$-3m^2 + bm + 2 = 0$$

The product of slopes here is $m \cdot m_2 = -\frac{2}{3} \implies m_2 = -\frac{2}{3m}$.
Step 3: Use the perpendicularity condition for the non-coinciding lines. Their slopes multiply to $-1$:

$$m_1 \cdot m_2 = -1$$

$$\left(\frac{2}{3m}\right) \left(-\frac{2}{3m}\right) = -1 \implies -\frac{4}{9m^2} = -1 \implies m^2 = \frac{4}{9}$$

Step 4: Set up the algebra to find $(a^2+b^2)$. We know $m$ is a root for both equations, so both evaluate to 0:
$3m^2 + am + 2 = 0$ (Eq. 1)
$-3m^2 + bm + 2 = 0$ (Eq. 2)
Add them together: $(a+b)m + 4 = 0 \implies a+b = -\frac{4}{m}$
Subtract Eq. 2 from Eq. 1: $6m^2 + (a-b)m = 0 \implies a-b = -6m$
Step 5: Calculate the final target, $a^2 + b^2$. Use the identity $2(a^2+b^2) = (a+b)^2 + (a-b)^2$.
Square the sums:
$(a+b)^2 = \frac{16}{m^2} = \frac{16}{4/9} = 36$
$(a-b)^2 = 36m^2 = 36(\frac{4}{9}) = 16$

$$2(a^2+b^2) = 36 + 16 = 52 \implies a^2+b^2 = 26$$

Final Answer: 26

The Setup: This is a piecewise function check. The function $f(a)=\max(a,0)$ basically acts like a bouncer: if $a$ is negative, it gets zeroed out. We need to find the smallest integer $x$ where $f(x-3)+2f(x+1)=8$.
Step 1: Test the lowest tier (where both inputs are negative). If $x \le -1$, then both $x-3$ and $x+1$ are negative or zero. The bouncer zeroes them both:

$$0 + 2(0) = 0 \neq 8$$

That's a hard pass.
Step 2: Level up to the middle tier ($-1 < x \le 3$). Here, $x+1$ is strictly positive (gets VIP access), but $x-3$ is still negative or zero (gets zeroed). The equation becomes:

$$0 + 2(x+1) = 8$$

$$x + 1 = 4$$

$$x = 3$$

This works perfectly and fits exactly at the upper boundary of our assumed domain.
Step 3: Double-check the top tier just to be safe ($x>3$). Here, both inputs are strictly positive:

$$(x-3) + 2(x+1) = 8$$

$$3x - 1 = 8$$

$$3x = 9 \implies x = 3$$

Since $x=3$ is the only valid root across the whole domain, that's our main character.
Final Answer: 3

The Setup: We are dealing with Set Theory here, specifically overlapping sets. 60% passed Physics and 68% passed Mathematics. We want to find the absolute floor (the minimum overlap) of students who survived both exams. Think of it like trying to find the minimum number of mutuals two influencers *must* share if they both have a massive following in the same city.
Step 1: Define the sets and the absolute total. Let $n(P)=60$ and $n(M)=68$. The maximum possible size of the entire union (everyone who passed at least one subject) literally cannot exceed 100%. So:

$$n(P \cup M) \le 100$$

Step 2: Hit them with the standard Set Theory formula:

$$n(P \cup M) = n(P) + n(M) - n(P \cap M)$$

Step 3: Plug in the known stats and the boundary we just established:

$$60 + 68 - n(P \cap M) \le 100$$

$$128 - n(P \cap M) \le 100$$

Step 4: Solve the inequality to find the minimum intersection threshold:

$$128 - 100 \le n(P \cap M)$$

$$n(P \cap M) \ge 28$$

The absolute bare minimum overlap is 28%. Period.
Final Answer: 28