The number of positive integers that divide (1890)⋅(130)⋅(170) and are not divisible by 45 is ________
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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=2×33×5×7130=2×5×13170=2×5×17
Multiply them together to get the ultimate form:
N=23×33×53×71×131×171Step 2: Calculate the total number of positive integers that divide N. Add 1 to each exponent and multiply:
Total Divisors=(3+1)(3+1)(3+1)(1+1)(1+1)(1+1)Total=4×4×4×2×2×2=512Step 3: Figure out how many of these are divisible by 45. Since 45=32×51, any divisor that is a multiple of 45 must have at least 32 and 51 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: 32,33)
* Powers of 5 left: 3−1=2 (so 3 choices: 51,52,53)
* The rest (2, 7, 13, 17) keep all their original choices (4, 2, 2, 2).
Banned Divisors=4×2×3×2×2×2=192Step 4: Subtract the banned divisors from the total pool.
Clean Divisors=512−192=320Final Answer: 320
The sum up to 10 terms of the series 1⋅3+5⋅7+9⋅11+... is
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The Setup: We need the sum of the first 10 terms of the series 1⋅3+5⋅7+9⋅11+.... 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.
an=1+(n−1)4=4n−3Step 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.
bn=3+(n−1)4=4n−1Step 3: Construct the general term Tn by multiplying them, and expand it into a quadratic polynomial:
Tn=(4n−3)(4n−1)=16n2−16n+3Step 4: Deploy the summation formulas for the first 10 terms.
n=1∑10Tn=16∑n2−16∑n+∑3
Using the standard formulas ∑n2=6n(n+1)(2n+1) and ∑n=2n(n+1) for n=10:
* 16×(610×11×21)=16×385=6160
* −16×(210×11)=−16×55=−880
* 10×3=30Step 5: Crunch the final numbers.
6160−880+30=5310Final Answer: 5310
It is given that the sequence {xn} satisfies x1=0,xn+1=xn+1+2(1+xn) for n=1,2,... Then x31 is _______
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The Setup: We are faced with a recursive sequence: xn+1=xn+1+21+xn with a starting state of x1=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.
xn+1+1=(xn+1)+21+xn+1Step 2: Substitute yn=xn+1 to clean up the UI.
yn+1=yn+2yn+1
Notice that the right side is a literal perfect square: (yn+1)2.
yn+1=(yn+1)2Step 3: Take the square root of both sides.
yn+1=yn+1
This means the sequence yn is just a basic Arithmetic Progression increasing by 1 every step. Since x1=0, our base case is y1=1, which means y1=1.
Step 4: Find the n-th term of this AP.
yn=1+(n−1)(1)=nStep 5: Reverse the transformation to find xn.
yn=n2⟹xn+1=n2⟹xn=n2−1Step 6: Calculate the 31st term to finish the level.
x31=312−1=961−1=960Final Answer: 960
There are 5 parallel lines on the plane. On the same plane, there are n other lines that are perpendicular to the 5 parallel lines. If the number of distinct rectangles formed by these lines is 360, what is the value of n?
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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:
(25)=25×4=10Step 2: Set up the equation for the perpendicular lines. We need 2 lines from the n squad, which gives us (2n). The total number of distinct rectangles is 360.
10×(2n)=360Step 3: Solve for n. Divide both sides by 10 to strip it down:
(2n)=362n(n−1)=36n(n−1)=72Step 4: Find the consecutive integers that multiply to 72. Since 9×8=72, we know n=9. No cap, it's that clean.
Final Answer: 9
Q5:ipmat indore 2021QA › Time & WorkMediumSA · TITA
There are two taps, T1 and T2, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T1 is opened keeping T2 closed, the water tank (initially full) becomes empty in half an hour. If both T1 and T2 are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T2 is opened while T1 is closed is
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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.
Rate of T1=301 tanks per minuteStep 2: Look at their combined synergy.
Rate of T1+Rate of T2=201 tanks per minuteStep 3: Subtract T1's rate from the combined rate to expose T2's true solo speed.
Rate of T2=201−301Step 4: Find a common denominator (which is 60) to crunch the fractions:
603−602=601
This means T2 clears 601 of the tank every minute.
Step 5: Flip the rate to find the total time. If T2 does 601 of the job per minute, it takes exactly 60 minutes to finish the whole thing solo.
Final Answer: 60
Q6:ipmat indore 2021QA › Mean, Median & ModeMediumSA · TITA
A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is
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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.
150×200=30000 total possible marksStep 2: Find the raw total marks before the revaluation. The squad averaged an 80% hit rate.
0.80×30000=24000 marksStep 3: Find the raw total marks after the revaluation bump. The new average slightly leveled up to 80.02%.
0.8002×30000=24006 marksStep 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
Q7:ipmat indore 2021LRDI › WeightsEasySA · TITA
What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan)
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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 2k−1 distinct positive values.
Step 2: Set up the inequality to cover everything up to 100 grams.
2k−1≥1002k≥101Step 3: Find the smallest integer k that satisfies this. We know our binary powers:
* 26=64 (too small)
* 27=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
If one of the lines given by the equation 2x2+axy+3y2=0 coincides with one of those given by 2x2+bxy−3y2=0 and the other lines represented by them are perpendicular then a2+b2=
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The Setup: We're dealing with homogenous equations of degree 2, which represent pairs of straight lines passing through the origin. The equations are 2x2+axy+3y2=0 and 2x2+bxy−3y2=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 m1 and m2.
Step 1: Translate the given equations to quadratic form in terms of slope (y=mx).
For the first pair, dividing by x2 gives:
3m2+am+2=0
The product of the roots (slopes) is m⋅m1=32⟹m1=3m2.
Step 2: Repeat for the second pair.
−3m2+bm+2=0
The product of slopes here is m⋅m2=−32⟹m2=−3m2.
Step 3: Use the perpendicularity condition for the non-coinciding lines. Their slopes multiply to −1:
m1⋅m2=−1(3m2)(−3m2)=−1⟹−9m24=−1⟹m2=94Step 4: Set up the algebra to find (a2+b2). We know m is a root for both equations, so both evaluate to 0:
3m2+am+2=0 (Eq. 1)
−3m2+bm+2=0 (Eq. 2)
Add them together: (a+b)m+4=0⟹a+b=−m4
Subtract Eq. 2 from Eq. 1: 6m2+(a−b)m=0⟹a−b=−6mStep 5: Calculate the final target, a2+b2. Use the identity 2(a2+b2)=(a+b)2+(a−b)2.
Square the sums:
(a+b)2=m216=4/916=36(a−b)2=36m2=36(94)=162(a2+b2)=36+16=52⟹a2+b2=26Final Answer: 26
Q9:ipmat indore 2021QA › FunctionsMediumSA · TITA
If a function f(a)=max(a,0) then the smallest integer value of x for which the equation f(x−3)+2f(x+1)=8 holds true is _______.
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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≤−1, then both x−3 and x+1 are negative or zero. The bouncer zeroes them both:
0+2(0)=0=8
That's a hard pass.
Step 2: Level up to the middle tier (−1<x≤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)=8x+1=4x=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)=83x−1=83x=9⟹x=3
Since x=3 is the only valid root across the whole domain, that's our main character.
Final Answer: 3
Q10:ipmat indore 2021QA › Set TheoryEasySA · TITA
In a class, 60% and 68% of students passed their Physics and Mathematics examinations respectively. Then at least ________ percentage of students passed both their Physics and Mathematics examinations.
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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∪M)≤100Step 2: Hit them with the standard Set Theory formula:
n(P∪M)=n(P)+n(M)−n(P∩M)Step 3: Plug in the known stats and the boundary we just established:
60+68−n(P∩M)≤100128−n(P∩M)≤100Step 4: Solve the inequality to find the minimum intersection threshold:
128−100≤n(P∩M)n(P∩M)≥28
The absolute bare minimum overlap is 28%. Period.
Final Answer: 28