Finally!! a problem from the JEE Advanced.
Vložit
- čas přidán 15. 03. 2021
- Problems from this exam are often requested.
Suggest a problem: forms.gle/ea7Pw7HcKePGB4my5
Please Subscribe: czcams.com/users/michaelpennma...
Merch: teespring.com/stores/michael-...
Personal Website: www.michael-penn.net
Randolph College Math: www.randolphcollege.edu/mathem...
Randolph College Math and Science on Facebook: / randolph.science
Research Gate profile: www.researchgate.net/profile/...
Google Scholar profile: scholar.google.com/citations?...
If you are going to use an ad-blocker, considering using brave and tipping me BAT!
brave.com/sdp793
Buy textbooks here and help me out: amzn.to/31Bj9ye
Buy an amazon gift card and help me out: amzn.to/2PComAf
Books I like:
Sacred Mathematics: Japanese Temple Geometry: amzn.to/2ZIadH9
Electricity and Magnetism for Mathematicians: amzn.to/2H8ePzL
Abstract Algebra:
Judson(online): abstract.ups.edu/
Judson(print): amzn.to/2Xg92wD
Dummit and Foote: amzn.to/2zYOrok
Gallian: amzn.to/2zg4YEo
Artin: amzn.to/2LQ8l7C
Differential Forms:
Bachman: amzn.to/2z9wljH
Number Theory:
Crisman(online): math.gordon.edu/ntic/
Strayer: amzn.to/3bXwLah
Andrews: amzn.to/2zWlOZ0
Analysis:
Abbot: amzn.to/3cwYtuF
How to think about Analysis: amzn.to/2AIhwVm
Calculus:
OpenStax(online): openstax.org/subjects/math
OpenStax Vol 1: amzn.to/2zlreN8
OpenStax Vol 2: amzn.to/2TtwoxH
OpenStax Vol 3: amzn.to/3bPJ3Bn
My Filming Equipment:
Camera: amzn.to/3kx2JzE
Lense: amzn.to/2PFxPXA
Audio Recorder: amzn.to/2XLzkaZ
Microphones: amzn.to/3fJED0T
Lights: amzn.to/2XHxRT0
White Chalk: amzn.to/3ipu3Oh
Color Chalk: amzn.to/2XL6eIJ
Music from Uppbeat (free for Creators!):
uppbeat.io/t/soundroll/tropicana
License code: PC8JYLCI7LYNOMSP
For the second problem, if a, b and c are in arithmetic progression then their arithmetic mean is b. As their geometric mean is b+2, by the AM-GM inequality there arises a contradiction. There's no such issue with the first problem because in that case their geometric mean is b and their arithmetic mean is b+2, which does not contradict the AM-GM inequality.
This is one of those confusing videos where I skip to the end to make sure I got the right answer but he's solving a different problem at the end.
Lmao
this is one of those confusing videos where i skip to the end to make sure i got the right answer but he's climbing a f*cking mountain
For the variant problem, one can frame it in terms of b instead. Since (a,b,c) = (b-k,b,b+k) then the geometric mean condition is equivalent to b^3 -k^2b = (b+2)^3 i.e. -k^2b = (b+2)^3 -b^3. Since b is a natural number then LHS is strictly negative and RHS is strictly positive. Therefore it has no solution.
7:45
a * ( 2 * n^2 - n )^1/3 = n * a + 2
[( 2 * n^2 - n )^1/3 - n ] * a = 2
As "a" is a natural number, "a" must be or 2 or 1 and [(2*n^2-n)^1/3-1] must be 1 or 2.
Case 1:
( 2 * n^2 - n )^1/3 - n = 1
2 * n^2 - n = (1 + n )^3
2 * n^2 - n = n^3 + 3 * n^2 + 3 * n + 1
n^3 + n^2 + 4 * n + 1 = 0
n ≈ - 0.2627
Case 2:
( 2 * n^2 - n )^1/3 - n = 2
2 * n^2 - n = (2 + n )^3
2 * n^2 - n = n^3 + 6 * n^2 + 12 * n + 8
n^3 + 4* n^2 + 13 * n + 8 = 0
n ≈ - 0.759
In both cases, n can't be expressed as a ratio of two natural numbers, so it's no doable.
And that's a good place to stop.
Cheers from Argentina, Michael!
I would have liked it better if you would have italicised that
I enjoyed this video!! Thanks Michael!!
This will absolutely blow up. Well presented solutions to these JEEs are highly popular
00:07
JEE Advance Exam is the entrance exam to get into India's Most Renowned Engineering Institutes, called the IITs (Indian Institute of Technology). Every year lakhs of students appear for this exam, and aspire to get a seat in IIT, it is one of the toughest examinations in the world, given by High School Students. IITs offer various courses for Engineering Discipline, such as Computer Science, Metallurgy etc.
It consists of mainly 3 sections, Physics, Chemistry, Mathematics and conducted in 2 shifts of 3hr each, each year the Exam Pattern and format changes. The Screening exam of JEE Mains must be cleared to appear for JEE Advance, which over a million students appear for each year.
thanks for informing saumitra!
@@targetiitbcse1761 What's your nationality ?
Thanks for copying from Wikipedia. We totally do not have access to it and couldn’t have searched it ourselves
Anyway, it’s a joke don’t be offended lol
@@srijanbhowmick9570 Indian, it was a sarcastic reply
My favourite channel that I found in 2020. Amazing content Michael. Even though I´m not nearly competent enough to do these questions I want to learn, and you going through the questions is very useful!
Many unacademy students also watching this!!
Seems like channel is more popular in🇮🇳 than 🇺🇸 :0)
💯
Lol that is because there are much more Indians
@@fearkrypton4565 sonal OP
Well ur crct😁😁😁😁😁😁😁
I'm not unacademy student
7:58 JEE? Oh boy, this video is gonna have a lot of views
Yup! These problems are way easier and approachable than olympiads. This is what I think.
@@TechToppers agreed
These don’t require much creativity as well
@@random-td7tf They _can't_ . Because each olympiad problem has something like half an hour (at least) to solve them, while each JEE problem has something like 2 minutes.
@@random-td7tf ya... but there are better questions in this JEE exam than that shown in video......
Im so glad this channel came across my recommendation.
I think INMO and RMO problems are more aligned to the kind of competitive examination content that you usually show. JEE exam is meant for placement in undergraduate institutes in India. It comes in two stages: JEE main and JEE advanced. JEE advanced score is required for joining IITs. JEE exam is brutal because the expectation is to solve these problems in a minute or two at most.
Exactly !
Not a minute or two, there are 54 questions and 3hrs. So a little more than 3 min. Plus chemistry takes less than 90 sec usually, so you can put more time into math. But still, it's really challenging.
JEE aspirant myself.
The CMI and ISI entrance examinations also have interesting problems especially in the subjective sections
i solved this in around 2 minutes so im guessing this is one of the more easy problems in the exam.
@@lgooch it is meant to be solved in this amount of time as other problems are much harder
You explain very nicely.
Well, m must be larger than n, but m = (2n^2-n)^(1/3). If we divide m/n and evaluate at n=1 then we get that m/n = 1. A bit of calculus should show that m/n is strictly decreasing as n-> +♾ . Meaning that m
We have c = (2n - 1)a = 2na - a = 2b - a.
Hence abc = ab(2b - a) = 2b²a - ba².
The condition on geometric mean leads to abc = (b+2)^3.
Thus we have ba² - 2b²a + (b+2)^3 = 0 which is a quadratic equation in a.
On can compute the discriminant and find -8b(3b² + 6b + 4).
It is strictly negative for any natural integer b, so our problem has no solution.
That's funny. I got a = 6, b = 12 and c = 24. b/a = 2. (a+b+c)/3 = 42/3 = 14 = 12 + 2 = b + 2. Solving the expression at the bottom you get 4.
@@Tiqerboy That's the solution for the first problem, I was answering to the second one.
BOOM! U just cracked the code... I'm rooting for this video to be the most viewed ever LOL
hi god
Wait u every where Unacadmey , vedanti and here
You have found the secret way to make your channel blow up
Just remember me when you get famous, will you ?
India, the best way to make views lol
@Aditya Mishra really lol
I am great tho
Awesome!
Theres this problem from Stewart multivariable calculus: given a solid box B with dimensions L, W and H. S is the set of all points which are at most 1 unit away from some point on B. The task is to find the volume of the set S in terms of L, W and H.
We are not told if L, W or H are above or below 1. They are generalized dimensions.
Stewart's problems plus are really interesting, especially the even number problems
Ohh Professor , so nice to see you while Rock climbing 😅👍
The JEE Advanced is one of the most difficult entrance exam at the +2 level. It is for entrance to prestigious engineering institutions as IIT's. It is an extremely tough exam that tests students in Physics, Chemistry and Mathematics
Hi Michael as much as we all love JEE questions, they are not the type of questions you usually solve, like Olympiad problems. Most jee problems usually can be solved with some clever insight which makes most of them attemptable within the 3-4 mins allotted to each question, also these problems are mostly objective based. What most of these questions lack is the rigour, which I mostly like in the Olympiad questions you solve. What is more suitable to the Olympiad level questions are the Indian Statistical Institute and Chennai mathematical institute exam. Have a look at them.
As always love your videos. Keep making more. JEE videos will surely get you much more views😂😂 because there is a craze about it in India. The exam in itself is so hard because of the time constraints and immense competition.
Damn Michael, didn't even know one could climb a rock like that
Are you talking about the improvisational second problem or an actual rock ? 😆
@@mikeh283 An actual rock, I didn't think that metaphorically
This is what indian viewers like me r expecting
Anyone from vedantu math channel after abhay mahajan sir told that JEE is introducing slight number theory because they are running sort of questions?
Yes, but tougher ones.
I am not, but ok
One of easier problem of JEE ADVANCE.
@@random-td7tf nah, keep making these
‘Expecting’ makes it sound like your entitled to it. Calm down a bit... He is voluntarily making these
The last equation has two solutions only, Starting from the point am=na+2, which if it has a solution the other equation does. We can divide both sides by a getting m=n +2/a. Since m is a natural number and makes up the LHS, the RHS must be a natural number. Since n is a natural number, 2/a must be, making a = 1 or 2 exclusively
WHOA !!! at 7:58 That was not a good place to stop, Sir !!!!
Man your great!!!!!
2n-1=2 otherwise a=a+2, an impossibility).
=> n(2n-1) < n^3
=> m^3 m-n=2 hence we have no solutions.
For the last question... when we have that am = an + 2 ... we could just take that equation (mod a) to see that 2 = 0 (mod a) and therefore a is either 1 or 2 (independent of the value of n) ... Then just evaluate if (2n^2-n) = (na+2)^3 for a in {1,2} has a natural solution, but for positive values of n, the RHS is always larger than the LHS. Therefore there is no solution.
I got the same
I think you missed the a on the LHS.
@@dontsetyourlimitsyt4939 True! Even though everything remains the same
It's quite easy problem to solve...I have solved it in a different way where I used the properties of natural numbers and perfect squares to assign values to 'a' and to find the actual answer from those assigned values respectively...Value of 'a' is 6.
I personally love the question from ISI Bmath/Bstat exam 2020. Paper- UGB question-P8
The one with alternating series
Yeah
Which is that?
It's a nice problem 😀
Yeah a great problem
@@ribhuhooja3137 check out the 2020 paper on aops
Excelente video! Me hiciste acordar a Alex Honnold! 😂
notice that a, b and c could also be in geometric progression if a=nc and b=n²c. but then you will quickly find that there is no integer solution for n (which makes sense as in this case b>a>c such that the arithmetic mean can never be greater than b)
Well spotted!
The Q does not say that they are in GP in any particular order.
Move na on Left hand side
Use AM>=GM on (1,(2n-1),n)
Which shows a is -ve (assuming n to be +ve)
You get
a[ (n(2n-1))^(1/3) - n ] =2
It is an examination for the 12th Standard students to get into the IIts. Besides the tough question paper comprising Physics, Chemistry, Mathematics all together in one paper, the success rate is less than 1%.
This year, 22lakh students have applied and the total number of seats is around 16000. This is what, that makes the exam even tougher and one of the toughest in the world.
Bro 16000 includes branches like Agriculture Engineering in barely developed IITs. For a core branch in an old IIT, you'll need a rank less than 5000
HOMEWORK : Some combinatorics from 2015 Harvard-MIT Mathematics Tournament
Count the number of functions f : Z → {‘green’, ‘blue’} such that f(x) = f(x + 22) for all integers x and there does *NOT* exist an integer y with f(y) = f(y + 2) = ‘green’.
Solution
Ok first we really need to look for only f(1) to f(22)
Let b(i) be a number of way we can construct f:{1,2,3,...,i} to {green,blue} such that f(i-1)=green and c(i) be a number of way we can construct function such that f(i-1)=blue
from recurrent we can conclude that
b(i+1)=c(i) and
c(i+1)=b(i)+c(i)
c(i+1)=c(i)+c(i-1)
b(2)=2 and c(2)=2
b(3)=2 and c(3)=4
We can see that c(i)=2F(i) when F(i) is a fibonacci sequence
The way we can construct a function is b(22)+c(22)=c(23)=2F(23) #
Edit:I forgot to think about f(1)=f(22) and f(2)=f(23) so can anyone do that for me?
Edit: This is not right.f(y) and f(y+2) maybe mapped to the same blue colour as well.
There exists no integer y such that f(y)=f(y+2)=green means that f(y) and f(y+2) must be mapped to different colours for all y. But that implies f(y+2) and f(y+4) must be of different colours and hence f(y) and f(y+4) must be of the same colour. Using this argument we can say that :
f(y+4k) are all of the same colour for all integers k.
And
f(y+4k+2) are all of same colour for all integers k. That means f(y+2) and f(y+22) are of the same colur, that is different from f(y). Therefore f(y)≠f(y+22) for any integer y, but that's not possible as we are given they are equal for all integers y. Hence there exists no such function f.
@@AdityaPrabhu01 only green can't be the same you can have all f(x)=blue and it is indeed one solution
@@AdityaPrabhu01 f(y) = f(y+2) = blue is possible.
My solution is incomplete bcuz I miss some important detail but I'm sure it almost end so anyone pls do the rest for me. (I'm too lazy lol)
It is easy to check that (2n^2-n)^(1/3)
Please upload more from these exams
I used guess and check, starting with 1,2,4; 2,4,8; ... until I solved a+c=2b+6.
a+c=6+2b also suggests a=6 and c=2b, which makes the guessing easier.
Last equation a*m = a*n + 2, where a, m, n are natural, doesn't demand deep thought I think. Because it means that a*(m-n)=2 and it's possible only in two ways: a=1, m-n=2 or a=2, m-n=1. In every case we return to definition of m and substitute its value (n+2 or n+1 respectively). We come to polynomial with positive coefficients that means there can be only negative solutions. But n is natural, so there is no solution.
Well the answer is 4 it was a easy question
We could directly do it by assuming the numbers as a , ar , ar^2 and their AM is ar+2 , a+ar+ar^2/3=ar+2 which results in a(1-r)^2=6 which has only one soln for a=6 and r=0,2 but as a,b,c are natural no therfore r can't be 0 hence a+14/a+1=4
The end is an awesome present!
If you take am=an + 2, you can see that the LHS has a factor of a, so the RHS does as well, but since the RHS is congruent to 2 mod a, a must be 1 or 2 (for zero to be congruent to 2 mod a). We can then see that either of these cases for a give cubics in n which have strictly positive coefficients and so don’t have any positive (and hence natural) solutions so there are no valid choices for a and so the problem is self-contradictory; there are no a,b,c that satisfy the conditions so the quantity is meaningless.
As the first variation has no solution, may I suggest a second variation where instead of the geometric mean equaling b+2 it equals 2b.
Keep solving JEE advance problems please because the exam is coming. It will help a lot. Thank you Michael
The last part. No solution since (2•n^2 - n)< n^3 so m
It wasn't obvious to me at a glance that 2n²-n
@@gregorymorse8423 the n=1 case is trivial, I don't think Klaus needed to mention it....
@@gregorymorse8423 I watch these videos for recreation like a lot of people do, but maybe the JEE folks around here are trying to be more rigorous.
After spending some time on solving this. I got the values of a,b,c as 6,12,24 respectively. And they satisfy all the given parameters.
The value of the given expression on putting a=6 will be 4.
So the required answer is 4
Edit: I solved the question as mentioned in the thumbnail taking a,b,c to be in geometric progression.
Are you sure? If I see your values that would mean n=2 m=7, and that's no solution for 2n²-n=m³? What am I missing here?
@@MrJurgenvdp The thumbnail of the video mentions that a,b,c are in geometric progression. But in the video on the board a,b,c are mentioned as arithmetic progression.
Since I am an idiot I solved the question as mentioned in the thumbnail , thinking a,b,c to be in geometric progression.
a, b c are in geometric progression, but NOT arithmetic progression. It talks about taking the arithmetic mean which I understand to be (a + b + c)/3
@@TiqerboyBoth the cases are considered in the video. When a,b,c are in G.P and A.M is b+2. And when a,b,c are in A.P and G.M is b+2.
I solved for the former one.
I stopped watching once he got the answer I got. I did see for the second half, he reversed the problem (inspired by JEE Advanced 2014) but I didn't try to solve that one.
Here it is aah
a lot to wait
JEE Advanced although hardly sees such format questions to solve but they are always interesting ones ( number theory )
Solution - b² = ac ( from the G.P. fact)
a = 2b + 6 -c ( simplified A.P. fact)
Plug it above gives
b² = 2bc + 6c - c²
(b-c)² = 6c
Hence c must be of form 6k²
and thus b is of form 6k(k + 1)
and a is of form 6k²(k+1)²
Edit : Please note that this question was solved from just by looking at thumbnail and that it's has swapped Arithmetic Mean and G.P
It's pretty easy to to show that (b-n)*b*(b+n)
We can see immediately that a divides 2 from the last equation(s), so a=1 or a=2.
However substituting a=1 gives us a cubic in n with no natural number solutions as can be verified by the rational roots theorem or simply taking the derivative (which is always positive) and computing test values.
Similarly a=2 yields no solutions for n.
So we've shown that this problem is an absurd construction.
JEE is basically SAT but on every possible steroid you could think of(that’s what makes it fun tho)
Btw be ready for a lot of salty Indian viewers coming to your channel just to complain about jee
Also instead off jee which is covered by hundreds of channels, I’d recommend checking out the ISI and CMI entrance exams
Bruh dude you can't compare JEE and SAT at all. Let's be honest, the SAT math section is like grade 9 math in india.
@@ritam8767 Agreed but JEE is an easy exam just the same.
@@ngc-fo5te u are high today ?
@@ngc-fo5te sure, if you look purely at the level of the questions objectively out of all the topics in physics, chemistry and mathematics that exist, it’s an easy exam, but there are 2 factors which make it difficult-
1) it’s an exam attempted by high school students who have to learn all that material in barely 2 years
2) the time limit is very less for the level and rigour of calculations involved in many of the questions
It looks like there are no solutions. From am=an+2, we get m>n. But then we have n^3 < m^3 < 2n^2, which implies m=n=1. But then am does not equal an+2.
Hi Michael !! There is a non-computational/number theory way to find the smallest natural number n such that the number of digits of n! is greater than n ? Love your content !!
1. Digits(n) = Floor(log_base10(n))+1
2. Log(n!) = Log(1) + log(2) + ... + log(n)
3. The sum of those logs can be calculated with an integral from 1 to n of log_base10(x), which is n*(ln(n)-1)+1)/ln(10)
4. Using the sum above for your inequality, you immediately get that the inequality surely holds for n>10*e. If you fiddle with the inequality a bit, you'll get something like n*ln(10e/n)=25, so the answer is 25.
Word of caution: I haven't done real math in 20 years, don't take my word for it.
That was a nice problem, but I was more impressed with that wicked overhanging layback at the end. All of your hobbies seem to involve chalk!
I'm more afraid of rock climbing than of math now, thanks to you!
JEE Advance is an exam for the admission in Indian Institute of Technology (IIT)
Very ordinary and simple question Alas!
I think the only way n*(2n-1) can be a perfect cube is for n=1. But then the equation of the geometric mean has no solution as it becomes 0=2.
In fact, n and 2n-1 are always coprime: if a prime p divides n, then 2n-1 leaves the rest -1 modulo p.
EDIT: so in order for n*(2n-1) to be a perfect square means that n has to be a perfect cube itself and 2n-1 has to be a perfect cube as well.
a+c=2(b+3)
ac=b^2
c=b*b/a
So, a + b*b/a=2b+6
a*a+b*b=2ab+6a
(a-b)^2 = 6a
a=6*n*n. then b=2*6*n*n. and c=4*6*n*n. [n belongs to N]
a,2a,4a
where
a+4a=2(2a+3)=4a+6
a=6
Ans.4
The second problem has no solutions because the arithmetic mean of three numbers is always greater than or equal to the geometric mean. But b > b + 2 is an impossibility.
The problem was very interesting - loved it.
In the coverage of the second version of the problem was a bit of let down - though. Why start on the problem if it was not going to be fully solved? I suppose the solution was much more involved.
Could you try a Cambridge STEP 3 paper?
wow i did this in 2 minutes. im preparing for competition math and this question is much easier than the ones i see on your channel, or im just improving.
I seriously expected you to say after "obviously 4" is "and this is a good place to stop". Done.
I solved it pretty much the way you did except I actually solved for 'n' in terms of a using the quadratic formula. Straightforward arguments lead to n = 2 as the only possible solution.
In India we have JEE mains for regular engineering colleges. Approximately 10% of top scorers of JEE mains are allowed to give JEE advanced exam. Based on ranking of JEE advanced, students get admission in colleges of the category called IIT.
Approximately 1.2 million students attend JEE mains every year.
I'm just wondering if there are other solutions involving different positionings of c in the geometric progression. (c,a,b or a,c,b)
For am = na+2, a must divide 2, since a divides n*a. Thus a = 1 or 2. for a = 1, (2n^2-n)^1/3 = n+2, or 2n^2-n = n^3+6n^2+12n+8, or n^3+4n^2+13n+8 = 0, which has no natural number roots (can be verified with the cubic formula or by brute force with polynomial long division, but that's too much for a comment), so a =/= 1. So for the alternate case you have a = 2, where 2(2n^2-n)^1/3 = 2n+2, or (2n^2-n)^1/3 = n+1, or 2n^2-n = n^3+3n^2+3n+1, which is to say n^3+n^2+4n+1=0, which also has no natural number roots (again can be verified with the cubic formula or polynomial long division, which in this case is, I believe, easier, since the last term is 1, meaning n = +/- 1 would have to be a root, I'm pretty sure, if it had natural roots, which can be checked as n = 1 would result in 7, and n = -1 would result in -3; I'm not 100% on the polynomial long division aspect, so if it's wrong, let me know, but either way you can use the cubic equation to confirm there are no natural roots.)
Edit: just realized that for the switched form, n doesn't need to be a natural number just a positive rational number. Either way, the roots of both equations are negative, and so cannot be a quotient of two natural numbers. I don't feel like rewriting the whole comment to account for that, so here's this addendum.
You could see that m-n and a are a pair of factors of 2 then pair that with the cubic equation and solve.
only solutions that are integers are 1 and 0, for both values (for the formula with the variables m and n)
but both give no solution
Given your observation about m-n being a factor of 2 (which must also be positive since a is positive), putting either m-n=1 or m-n=2 into 2n^2-n=m^3 gives a cubic polynomial equation in n with all positive coefficients. So clearly there are no roots for any positive n, without having to solve the cubic.
Thanks for doin the work i didnt feel it lol
No solution to the second one. factoring the last expression yields either m = n+ 1 or m= n+2, since m and n are naturals. plugging either into m^3 = 2n^2 -n then solving for n yields no natural solutions for n, due to descartes rule of signs causing the expressions in n to have no positive roots, therefore no natural solutions for n.
I guess the variant case has no solutions because:
Let a=b-r and c=b+r (r is the common difference from the arithmetic progression)
Their GM will be [(b-r)*b*(b+r)]^(1/3), and that is equal to b+2. Cubing both sides one gets:
b^3- b*r^2 = b^3 + 6b^2 + 12b + 8
6b^2 + 12b + 8 = -b*r^2 < 0
6*(b+1)^2 + 2 < 0, which is impossible for real b, let alone natural b.
Right from the bat you can see that a could only be 1, 6, or 13 if we seek positive integer values of that rational function....(just divide it)
Jee exams are taken after you pass std 12. The exams have 2 phases mains and advanced. It is a joint exam for b.tech, b.arch and other degrees but b.tech is most aspired by Indian students. It is the most important entrance exam for these degrees but we have other exams too like wbjee, Kerela pet, etc. The subjects for b.tech are phy, chem and maths.
The mains are taken in 2 phases since a few years. This year, it is being held in 4, due to uncertain academic year, due to the pandemic.
The mains exam has 75 questions, 3 hours. 25 phy, 25 chem, 25 mathematics.
In each subject we have 20 MCQ while 5 integer type answers. Each question holds 4 marks. There is a -1 marking for wrong answer in mcq. No minus marking in integers.
Now, the best institutes are IITs and they decided to hold a separate exam for themselves called advanced. It is taken in a single phase. The marking and pattern is not fixed. To appear in advanced, you must be among top 250k in mains but your mains marks are not considered for admission. So basically, if you just pass the mains and do good advanced, it would pose no threats. The marks of mains, however are considered for other colleges, where the marks of advanced are of no use.
*Solution of modified (2nd) problem in the video* . (a, b, c) = (b-d, b, b+d) and abc = GM^3 = (b+2)^3. After solving d^2 = 4 and b = -2. So (a, b, c) = (0, -2, -4) or (-4, -2, 0). Therefore, if a=0 then (a^2+a-14)/(a-1) = (0+0-14)/1 = -14 . If a=-4 then (a^2+a-14)/(a+1) = (16-4-14)/-3 = 2/3 .
Negative numbers are not Natural
This exam is just an exam to enter one of the prestigious engineering colleges in the country, basically like an entrance test and has two parts and the JEE advanced is the level 2 and math isn't the only section in the exam but physics chemistry as well. Topics in math are mostly calculus, trig,algebra and graph theory but only little of number theory
Please let this one be the last, there's literally hundreds of channels covering jee topics.
And a personal request- Can you please start a series of videos covering topics of number theory, vector calculus, etc.
I think he has vector calc done too, partly atleast. He has a very coherent number theory playlist
What’s ur workout schedule 😍😍😍
I found a problem that you might be interested in ! It's the geometry problem from the 2019 polish math olympiad, i found it to be extremely hard.
7:50 am = an + 2 => a=1,2
There are no integer solutions to (2n^2-n) = (n+b)^3 when b=1,2 (as n divides LHS so must divide b as well).
Dropping the b/a = n condition:
a,b,c = a, n+a, 2n+a
abc = (n+a)((n+a)^2-n^2) = (n+a)^3 - (n+a)(n^2) < (n+a)^3
abc = (n+a+2)^3 > (n+a)^3
Hence this has no solutions over the naturals. (Moreover, it has no solutions over the positive reals).
Yep
Second question has no solutions almost trivially. For any set of positive numbers (e.g. a, b, and c), the geometric mean of the set is less than or equal to the arithmetic mean of the set. But as a, b, and c are in an arithmetic progression, their arithmetic mean is b, so the geometric mean is less than or equal to b and definitely NOT b+2.
i'd like to see the follow up. coming from someone that doesn't have time or enough practice to solve it on my own.
JEE Advanced is an entrance exam to get into IITs. It is quite possible that even if you have all the concepts at your fingertips you may not be able to attempt all questions on time. It's a complete horror when you see such questions to be solved in under 2 minutes.
So we can reduce n(2n-1) = m^3 having a solution to an argument of coprimes.
If n is not 1, then these are coprime, and so, n = x^3, and we would need y^3 = 2x^3 - 1 to be a perfect cube.
But we know the difference between two cubes to be x^3 - y^3 = (x-y)(x^2+xy+y^2)
So, if 2x^3 - 1 is a square, then it would need to satisfy that.
But y^3 - x^3 = (2x^3 - 1) - x^3 = x^3 - 1 = (x-1)(x^2+x+1)
Since x and y must be coprime, if I'm not mistaken, for x not equal y, this would imply that the factors of x-1 are the same prime factors as x-y.
Therefore, we would need y = 1 or x = 0.
However, since we're in the natural numbers, we would simply get x = y = 1 as the only solution to that, making it that no progression satisfies the desired property.
Let's say instead that the geometric mean of b-d, b, and b+d is b - 2, instead of b + 2. Then
b(b-d)(b+d) = (b-2)^3
b^3 - bd^2 = b^3 - 6b^2 + 12b - 8
6b^2 - (12+d^2)b + 8 = 0
b = (12 + d^2 + sqrt(d^4 + 24d^2 - 48))/12
Through trial and error, I found that b = 2, d = 2 works.
Unfortunately, b-d = 0 and thus can't be a solution to the problem as restated.
The second problem doesn't have a solution. We need to solve the equation a(m-n) = 2, so it's very easy to see that m-n needs to be positive for any a,b,c to be solutions, but plugging in n=1 yields m=1 and m-n is a decreasing function in n for n > 1. Maybe if we loosen the constraints to let a,b,c and n be any integers it might work out
Nope, even then there is no solution. If we allow a,b,c and n to be negative or positive, we can manually check that m-n is not in \{-2,-1,1,2\} for any positive n (if n > 6 then m-n < -2). It's trivial to show that n cannot be 0. Thus, if there is any solution, n must be negative. Consider (for n > 0), a = a, b= -na, c = (-1-2n)a. Then we can reduce this solving for m + n in \{-2,-1,1,2\} for some n > 0, where m = cbrt(n(2n+1)). Again, we can easily verify that m+n is increasing in n for n > 0 and that m+n > 2 if n = 1.
am is divisible by a. na is divisible by a. then 2 must be divisible by a. So, a is equal either 1 or 2. If a=1, then m=n+2 => n^3+n^2+4n+8=0, no solution in integers. If a=2, then m=n+1 => n^3+n^2+4n+1=0, no solution in integers. Ans: no solutions
Technically it didn't say that the geometric progression is in the order a,b,c. b/a being a positive integer gives that a
You get the ordering from the fact that b/a is an integer.
@@MichaelPennMath Ok, I get that b/a being a positive integer gives that b>a but I don't see what that says about c. For example if we had a=2,c=4,b=8 we have b/a=8/2=4 and c is between a and b.
Consider the quadratic equation 2n^2 - n - m^3 = 0, where n is the variable. Then the discriminant D = (-1)^2 - 4*2*m^3 = 1 + 8m^3. However, the only way n can be natural is for 1 + 8m^3 to be a perfect square. Notice, that 8m^3 is a perfect cube. Hence, a perfect cube and a perfect square differ by 1, which can only happen if the perfect cube is 8 and the perfect square is 9. Thus, m = 1 and therefore n = 1. But n = 1 is not a solution of the initial problem. Due to this fact we can claim, that no solutions exist.
What about if n is a perfect square and c=an^{1/2} you still have a geometric progression but the order is deifferent
There is no solution (in the naturals) unless you allow the shift of b to be 0, rather than 2. Then it's trivial, all 1's.
This is due to the fact that the only time the difference between a square and a cube is 1 is when the square is 9 and the cube 8.
In your second question, you cannot have the geometric mean to be larger than arithmetic mean for a natural numbered r.
Can I give you such more problems???
You didn't slip while climbing. Would you try it without ropes and safety?
This was rather routine
Jee advanced is a 6 hr test broken in two sets with an hr of brake in between , it constitutes physics chemistry and maths and is given by nearly 250,000 students in hopes of securing a seat from 10,000 available in the IIT's
250,000? Try 1,000,000.
How did you know that b and a are one after the other in the geometric series?
I think it’s implied in the statement “a,b,c are in sequence”. I suppose you could try another variant on the problem where c is in the middle or something similar but Michael’s method is what the written problem wants.
I cracked this exam back in 2014, not so good memories 🙏
I wanna know what happens with the geometric mean variant :(((
No solutions
JEE Exam is Engineering entrance exam conducted in two phases
1.Main - for admission to NITs(or more engineering colleges) and eligiblity test for Advanced
2.Advanced- More challenging than Main.For admission to 23 IITs in India.
This Dune T-shirt, respect