ISI UGB PYQs | 🔥Most Important Questions | Part 02 | Anirudha sir

Hinzugefügt: 2026-05-07

[00:00:01]Hello everyone. How are you all doing? I hope everything is good. Uh let's continue our preparation with ISIN and CMIPQS.

[00:00:11]Also, if you are interested in joining our IOM course, please join with the link in the description or you watch one of the videos which I made for uh as an announcement for the batch. Right? So yeah, it's a good chance to explore uh out of box thinking in a new way. Let's go. First question on the board. Please take your sweet time.

[00:00:40]Pause the video. Don't take too much time. I hope you have paused the video.

[00:00:45]I'm really assuming you have paused the video. This is a very tricky question.

[00:00:49]It relies on one of the proofs which I did in the class of seas theorem. I hope you know what is a se what is the seaas theorem we won't go through the theorem but uh I will use the method which we used in the proof of sea theorem so just uh why is it not working Anyway, so I hope that will be cut. Um, save. So suppose I have a triangle like this and uh I have A, B and C and I I have it in in a way like this.

[00:01:56]uh these two are incident and we are calling various things various things but let's not I'm not presently interested in that let's say I am interested in something else let's say I have a CVN like this and I have these two line And this is a b c d. And this is s some d.

[00:02:28]This is c. Okay. I want you all to know that you can write a by b is equal to c by d which is equal to bd by bc.

[00:02:44]Right?

[00:02:46]This is what we used while proving seaas theorem as well. All right. Now that I want to use here. How do I use it here?

[00:02:55]Oh, what's happening?

[00:03:01]How do I use it here?

[00:03:07]Uh you see we don't have the third CVN.

[00:03:13]So let's draw one more line. Right?

[00:03:17]Suppose I have it like this.

[00:03:26]And uh one CVN here, another CVN here.

[00:03:31]They are intersecting at F. This is E.

[00:03:36]This is D. Right?

[00:03:39]You join this. See this was written to be Y. This was X. This this was Z. And this was W. area of AD FE was w right but I'll divide it into W1 and W2 so I know one of the equations that W1 + W2 is equal to W3 so this we will use while finding we want to determine what W is in terms of XY Z right so let's try to find it let's try to find what is W1 and what is W2 in terms of these and then we can add and we can get So now look at it in the perspective of C.

[00:04:26]You have this CVN and you have two points coming out like this. So if you for example forget this part uh you for example forget this part and remember only this. So you have w2 + z and you have y and x. So you see w2 + z by w1 can be written as equal to y by right. So in other words we will have w2x + z x is equal to w1 y.

[00:05:16]Right?

[00:05:17]Now if I draw it again but this time I will cut off this part.

[00:05:31]This was y. This was z. This was w2.

[00:05:36]This is W1 + X right? X + W1. Now you see from this perspective ratio of these two is equal to ratio of these two. The X + W1 by Y is equal to W2 by Z. So in other words we will have uh xz plus w1 z is equal to w2 y.

[00:06:09]So you see how these two are quite similar.

[00:06:13]We have xz and we have these two. So all right.

[00:06:20]So let's say I want uh to substitute uh we don't want to use this because w is itself another unknown. So we don't want to use that.

[00:06:34]So I have w2 y is equal to w1 z + xz and I have w1 y = w 2x + x.

[00:06:56]Right? So what do how do I use both of them?

[00:07:02]uh so so so I will use uh I'll multiply this whole thing with y so I'll have w2 y² = w1 into yz + xyz and you can use this part here.

[00:07:35]Maybe I'll use a fresh page to write.

[00:07:56]Um, okay. So first of all w1 can be written as w2x by y + x by I want to substitute this here. So I have w2 y = w 2 x y + x / y into z + x.

[00:08:30]So if I want to get w2, I will multiply y². So w2 y² is equal to w2x into z + x z² + xyz.

[00:08:50]So if you bring this this side w2 into y² - xz is equal to uh you can take xz common z + y So you have w2 is equal to xz + y by y^² - xz. Now you see that w2 has to be a positive quantity because it is an area. It is a physical area. So obviously this quantity has to be positive. So that means y² has to be greater than xz.

[00:09:29]So that is a given. And then you found W2 in W1 it will be very very similarly symmetric.

[00:09:37]So W1 will be um so Y will remain the same the denominator will remain the same but instead of uh zed you'll have X here. So it will be xz x + y / y^2 - xz right and w2 is this uh xz + y by y square - x y and w you can find with w1 + w. So I hope this clear right.

[00:10:21]So we went from uh one of the proving techniques for das theorem and that was used in solving this ISIP.

[00:10:31]So guys you have to remember that in ISI UGB you will get 2 hours for eight question right 2 hours is 120 minute within 10 minutes we have solved the first one uh you know it is actually second or third question in that paper but we solved a question and yesterday uh in one of the previous videos as well we did solve ISI questions very very quickly. So it is I hope you are able to find that it's not just that uh I know the solution that's why I'm able to do this fast even in the examinations it is possible that one can do this and that can that is possible only when you if you practice a lot of questions especially the pyq right okay this is a fascinating question so please have a look at it pause the video try it right. I I really want you to pause the video and try it for once otherwise it'll be pointless that we are doing it session now.

[00:11:43]So what one those who have actually read it you'll figure that this uses pigeon hole principle and number theory together. This is such a beautiful question. It inter wheels so many things together. number of divisor part in number theory and uh pigeon hole principle part of combinatorics. Uh so it's a beautiful question and they have given us this.

[00:12:12]Okay. So the fact we will use is that if a number is p1 power a1 p2 power a2 till some pk power a k then number of divisor is equal to a1 + 1 a2 + 1 till a k + So this is a fact which we will be using. All right. You have to state it in the examination very very clearly.

[00:12:51]Okay.

[00:12:53]Now uh you see we want what do we want? You may use uh show that there must be some perfect cube among these ni and we will be using the prime factors will be at most 11. So what are the prime factors we can have? All our prime factors has to belong to 2 3 5 7 that's it. So we have five primes to choose from. All right. And when do we get 2023? So we are given 2023's this thing uh prime factorization. All right. Very kind of them actually. And this is a good thing about ISCM examination. They do give these uh facts in the beginning and I think it is also given in uh you know this RMO kind of question papers as well. Anyway, [clears throat] so so when do I get suppose I have what what is the maximum number of distinct primes I can have in the prime factorization of a number. So if a number 20 23 I want this to be our a1 + 1 a2 + 1 and so on. Right?

[00:14:20]But that would mean that a1 + 1 into a2 + 1 a3 + 1 is the maximum thing I can get to get 7 into 17 into 17 right even if I get more those will be zero a i will be zero so a4 will be zero 0 + 1 and all so that's not helpful uh we want you know strictly greater than one factors of So suppose I I take n= to p into p power a into q power b into r power c where my p q r are belonging to this 2 3 5 7 right and then so this is case one let's say case one that we have it like this that means that uh a + 1 has to be B + 1 has to be 17. P + 1 has to be 17.

[00:15:24]So basically we have A = 6, B = 16, B = 16.

[00:15:33]This won't give us uh a perfect cube, right? because my n will look like p power 6 into q power 16 into r power 16 where all these are distinct by the way not equal distinct all right so yeah so what do I do well uh for one we can choose how many primes can we choose we have to choose three primes from this so five choose three and the moment you choose P QR you have to assign 6 16 16 in some or the other way that is possible you can interchange the powers as well right you have chosen three primes and how many ways can you assign them these powers that will be three factorial by 2 factorial because 16 and 16 is repeating it's 3 factorial by 2 factorial all right so that would be 5 C3 is 5 C2 which is 5 into 4 by 2 into 3.

[00:16:42]So this is 30.

[00:16:44]So this case one is going to give us 30 possibilities. There will be 30 numbers of this form where p QR belong to that and the powers are 6 16 and 16. There will be 30 numbers which won't be perfect cubes. All right. Okay. Next.

[00:17:04]So we'll see again case two. n is equal to p power a into q power b.

[00:17:12]So that means uh a + 1 into b + 1 is equal to 2023.

[00:17:21]So we want to divide 2023 into two factors two factors which are greater than one. So that can be 7 into 7 square which is 2 17 square 289 or it can be 7 into 17 which is I think 119 right 119 into 17 these two possibilities are there yeah so within this case I'll have two possibilities one that a= 6 b = 288 or 2 a = 118 and b = 16. Now you see if in this case if n is equal to p^ 6 into q^ 288 you see that 6 and 288 both are divisible by 3. So it will be a perfect cube.

[00:18:22]It will be a perfect cube. So it will be p² into q power something whole cube right. So how many such numbers are possible?

[00:18:36]We can we have to choose 5 say2 primes into two factorial.

[00:18:42]Right? So this gives you 20.

[00:18:45]Okay let's not talk about the perfect cubes. As of now, we have to show that at least one of the 51 numbers which are like this have uh at least one of them has to be a prime. Uh what a perfect cube. Not a prime. Okay. What about this? This cannot be a perfect cube. Not a perfect cube. Not perfect cube.

[00:19:09]Right. So again this will also be 5 C2 into 2 factorial equal to 20.

[00:19:17]Right? So 5 C2 is for the primes and you have to assign A and B to one of these primes. Suppose you chose two and three.

[00:19:26]There can be two different things that is possible. So I hope you are able to understand that. So not a perfect cube 20 not a uh not a perfect cube was 30 here. This is not perfect cube. Yeah this is not perfect cube.

[00:19:47]Yeah. So we have total 50 as of now.

[00:19:51]And case three there is only one prime.

[00:19:55]It has to be 2022.

[00:19:57]So five such primes possible. Right? So these five are and if you see 222 these are five are perfect.

[00:20:11]So what is the lesson?

[00:20:16]Total such n is equal to 30 + 20 + 20 + 5 75.

[00:20:32]Right?

[00:20:34]non-perfect cubes if you see it is will be 30 + 20 50.

[00:20:45]So if I'm choosing out of the 75 out of 75 if we choose 51 by pigeon hole principle at least at least one should be perfect cube right one should be perfect cube I hope uh my handwriting doesn't bother your understanding is that clear I hope it has been clear 20 minutes two questions that's it there we go take care tata bye-bye hope you are becoming more and more confident while solving the problem Right.