Controversial Problem From Black Book For JEE Mains & Advanced

This is a tricky sequence and series question from the Black Book for JEE Mains and Advanced.
This question is conceptually wrong but you'll learn a lot from the solution to this question and can use it in JEE Maths.
You can also expect this or similar JEE Maths questions in JEE Advanced exam.
The question reads:-
Let X1, X2, X3,......, Xk be the divisors of positive integer 'n' ( Including 1 and n ). if X1 + X2 + X3 + ........ + Xk = 75 k/∑/i=1 (1/X1) is equal to :
a. 75/k
b. 75/n
c. 1/n
d. 1/75
To answer why this maths question from Black Book is wrong, the sum of divisors of any positive integer is not 75 which is given in the question.
If we ignore this and consider a hypothetical case, we can solve this question.
We can use properties of divisors and find the answer to this sequence and series question from Black Book for JEE Mains and Advanced.
Can you solve this JEE Advanced level question from the Black Book?
Give it an attempt to solve yourself.
If you are not able to solve it, stay with us during the course of this video and you'll learn the solution and important concepts.
Stay tuned with @BHANNATMATHS
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[@BHANNATMATHS]

Check out Bhannat Maths On Instagram (@bhannat_maths) - bit.ly/BhannatMathsInstagram

[@itseasy563]

Divisors can be positive as well as negative Oh my god sum is zero Question is wrong

[@akshithmanjunath8557]

Sir if we consider divisors to be positive and negative then we get 0 question would be wrong. but if we consider positive divisors I found 2 numbers for with sum is 75 (excluding 1 and n) they are 48 and 92. I think they made a typo error and said including instead of excluding.

[@ars-II]

Sir, how did u see immediately that no such number can exist whose sum of divisors is 75(before using the computer programme that is)?

[@kevinbanner5161]

​@@ars-IIHe is a mathematics teacher since a long time,He already done and seen the questions like this,Therefore he know that this is an some error...

[@k0u0s0h0a0g0r0a0]

@@ars-II proof is simple. Check my direct response. Once you are into numerical domain(such as programming or teaching) such things are easy to visualise.

[@immaturesuman]

Difference between Other subjects vs Maths 🔥. This subject makes you to think Differently ❤️. Thank you Aman sir (legend)

[@tuffaniblogs]

youtube.com/@Kgfclasses-py4lw

[@parameshwarhazra2725]

Maths aur physics bhi

[@kaustubhwadekar2204]

Kabhi zindagi mein physics solve Kari hai ?

[@dipanghosh8470]

Physics ka achhe problems is laughing at corner 😂

[@kax1230]

@@dipanghosh8470 Physics without maths is nothing😐😑

[@souhridyapatra0532]

I think the author wanted to teach us another way to think and solve the problem in a smart and better way. If the sum would exists then many of the students would work hard and get out with the problem. But the author didn't want that. So, he gave the question in this way.

[@N-methyl1phenylpropan-2-amine]

The author should've given an arbitrarily large sum instead (millions) to prevent brute force attempts, kinda like how they do it in olympiad-level problems at times. If this was a real exam question it could've been challenged.

[@manikgoswami4390]

You are absolutely right @sauhridya

[@pseudobixth]

I agree with you even at first I was also trying to solve it by finding n directly

[@RohitKumar-ll7pr]

Exactly you got it

[@Bath4848]

No.. actually this man in video still needs to work hard on his concepts..divisor & factors ka difference nhi pta isko..every factor is divisor but not all divisors are factor

[@chitraagarwal8259]

Even if we leave out "75" and assume the divisors sum to some positive integer S, there is one more concept missing for your solution. Your solution works in situations where K is even. If k was odd (this will happen in the specific case of n being a square number) and the set of xi to xk includes the square root of n just once, then you won't be able to use the mapping of x to n/x

[@kdhd100]

No. pls chk with 4, you will get 3 divisors in both cases. 1+2+4=7 and 4+2+1=7

[@sakshamsingh1778]

​@@kdhd100 yes in case of perfect square x maps onto inself

[@tirth6450]

Imagine n = 529 with divisors = 529, 23 and 1 (odd). Now we can map out the same thing as (529/529), (529/23), (529/1) = 529, 23 and 1 which is same as 529's divisors. So it works out just fine even if its a square number if i'm correct.

[@M_W_S2023]

I feel like mathematician when I study with you sir really from❤️

[@AnujG-or6rr]

Sir you are real legend of maths ♥️

[@anubhavupadhyay2949]

legend is terence tao, andrew wiles, manjul bhargava,etc

[@Maths_3.1415]

​@@anubhavupadhyay2949 A little correction needed Srinivas Ramanujan

[@allbookshorts1025]

@@Maths_3.1415 one more correction needed Aryabhatta

[@avigorai1644]

@@Maths_3.1415 ramanujan is ultra legend bro And the above mentioned are legends

[@0krishnakumarmehta]

Yes brother ❤

[@bhargavsai8014]

Solved it in less than 1 minute sir!here is my approach: Let us define a quadratic equation whose leading coefficient is unity and has roots x1 and x2,where x1,x2 are always choosen in such a way that their product gives the number n itself So our quadratic would be of the form x^2-(x1+x2)x+n=0 Now apply a transformation,t=(1/x) to the above Equation We get new equation as (nt^2-(x1+x2)t+1)=0,notice that this new equation has roots has (1/x1 and (1/x2) So sum of roots =(x1+x2)/n Doing this for all such possible pairs (x1,x2),we finally arrive to our result,that is (1/x1+1/x2 ......1/xk)=(x1+x2+x3.....xk)/n= 75/n

[@Dx_Era_31]

Bhai app kese approach krte ho i mean app kese sochte ho ki aeshe karenge toh shi answer niklega

[@ravish_gupta]

@@Dx_Era_31 ques practice is only approach.

[@2_13dekh-_-]

Bhai sahab esi approaches kese build kare 🥲

[@invincible9240]

I think so your approach has an inherent fallacy ,not sure though .how can ur approach guarantee that u r considering every divisor of number n

[@invincible9240]

Please correct if I m wrong

[@atharvgamingcafe]

Thinking of those students jo is sawal ko aise solve kar rahe honge ki konse number ke divisors ka sum 75 hai😂

[@Sumit_Girhe]

Shayad isiliye us book mai imaginary situation di ho kyu ki us metnod se asaan hai

[@ajeetsingh7209]

Python program for sum of divisors of any positive integer : def sum_divisors(num): factors=[] for i in range(1,num+1): if num%i==0: factors.append(i) sum=0 for i in factors: sum+=i print(sum) sum_divisors(int(input("Enter a number : ")))

[@Shreyas_Jaiswal]

Bahar aur ek infinite loop lagado, jo ki wo number bhi Khoj se jiska divisors ka sum 75 ho.

[@Shreyas_Jaiswal]

CS wale ho kya?

[@ajeetsingh7209]

@@Shreyas_Jaiswal nahi abhi to aspirant hu😅

[@Shreyas_Jaiswal]

@@ajeetsingh7209 are mai iit cs ki baat nhi kar raha, class 11,12 wala cs ki kar raha hun.

[@birukumar3551]

Bahi muja to cs smj hi nahi 11 ka 12 ka dekta hu tum na to pura program likh dala

[@dhananjaytudu6864]

Wow! Very interesting questions tha and solutions ka explanation is unique ❤❤

[@devkaushikyt]

Conclusion: never underestimate your teacher

[@shaloksharma6668]

Sir, the way you narrated this question itself gave some clues to the answers. The only difference is the way to enjoy a question was not taught in school or tutions but only the way to strees over marks of a question. I hope sir most students find your channel as soon as possible.

[@namansheth6345]

Love the way you solve black book questions , pls countinew these

[@arvindchirame1388]

I was able to identify this when I was in 11th Now I got 99.9747 percentile in JEE Mains sirrrrrr

[@adarsh4134]

congrats broooo btw any book suggestions for jee 2025 aspirants like me?

[@atharv8708]

I got 99.88419

[@sleepyfella]

I got 99.9842

[@ksp._.79]

@@adarsh4134 Physics: Cengage, Physics Galaxy, Shashi Bhushan Tiwari Maths: Blackbook, Any PYQs book(Personally I prefer 45 year PYQs by disha publications) Chemistry: Physical: N Awasthi, Organic: M.S. Chouhan, Himanshu Pandey Inorganic: Just notes and Any institute Module. Btw I am a Jee aspirant myself (2024)

[@tfayushmc2947]

​@@ksp._.79How did it go ?

[@shrinidhi3]

Using cheat method..option b Sir showing his class..hats off

[@MANIRAM-ct6bp]

question starts from 3:03

[@arenacoder]

Thanks

[@saurabhkumarsuman4217]

We know that each divisor of n can be paired with another divisor of n such that their product is n. For example, if n = 12, the divisors are 1, 2, 3, 4, 6, and 12, and we can pair them up as follows: (1, 12), (2, 6), and (3, 4). Notice that each pair has a product of n = 12. Using this fact, we can see that the sum X1 + X2 + X3 + ... + Xk can be written as: (X1 + Xk) + (X2 + X(k-1)) + (X3 + X(k-2)) + ... + (Xk/2 + X(k/2+1)) Each of these pairs has a sum of n, so the above expression simplifies to: n + n + n + ... + n (k/2 times) Therefore, we have: X1 + X2 + X3 + ... + Xk = kn/2 Now, we want to find k/∑/i=1 (1/X1). Using the fact that the sum of the reciprocals of the divisors of n is equal to σ(n)/n (where σ(n) is the sum of the divisors of n), we have: k/∑/i=1 (1/X1) = k/σ(n) We are given that X1 + X2 + X3 + ... + Xk = 75, so we have: 75 = kn/2 Solving for k, we get: k = 150/n Substituting into k/σ(n), we have: k/σ(n) = (150/n)/σ(n) = 150/(nσ(n)) Recall that the function σ(n) is multiplicative, meaning that if m and n are relatively prime, then σ(mn) = σ(m)σ(n). Therefore, if n is a prime power (i.e., n = p^k for some prime p and positive integer k), then we have: σ(n) = 1 + p + p^2 + ... + p^k which is a geometric series with sum (p^(k+1) - 1)/(p - 1). In this case, we can simplify the expression for k/σ(n) as follows: k/σ(n) = 150/(nσ(n)) = 150/[n(1 + p + p^2 + ... + p^k)] = 150/[np^(k+1) - n/(p-1)] = 150/[n(p^(k+1) - 1)/(p-1)] We can see that the denominator is a multiple of p-1, so we can simplify further: k/σ(n) = 150/[n(p^(k+1) - 1)/(p-1)] = 150(p-1)/[n(p^(k+1) - 1)] If n is not a prime power (i.e., n has more than one prime factor), then we can use the fact that σ(n) is multiplicative to express σ(n) in terms of the prime factorization of n, and then simplify the expression for k/σ(n) in a similar way. In any case, we can see that the answer is not one of the choices given, so we cannot determine the value of k/σ(n) without more information.

[@nafis_ul_abid]

Nahi smjh aya😢😔

[@steadytuna8753]

The sum of each pair isnt the 'n' but its the product. You wrote that X1+ Xk is n which is wrong

[@Rajis938]

What if k is odd?

[@vedantideshmukh7936]

Bhaiyaa i am accepting ur challenge for next 21 days .. for sure .. i will try to do my studies without any distraction for max 2 hours !! 👍🏻 And thank you for always making us motivated !! ✌🏻

[@AksgatSinha-tr2hd]

Ye question kitne no be hai book me ?

[@vaibhavselvakumar4924]

He bhut hi interesting question plus Aman sir’s teaching makes it better , itni acchi teacher hone ke liye dhanyavaadh sir

[@k0u0s0h0a0g0r0a0]

This is very easy to validate. 1. If no is prime, it should be 74. 2. If HCF is 2, 1 + 2 + x/2 + x = 75 gives x = 48 3. If hcf=3, same logic. 4. If hcf=4, 1+2+4+x+x/2+x/4 You have to check only till floor(sqrt(75)), which is 8.

[@lsrahulmondal]

Sir your this video give me goosebumps. Math is ❤️

[@nishkarshsaxena510]

sir i just gave 10th class boards and at the first sight i knew i couldnt solve this question because of lack of knowledge, so i proceeded with an example to solve this. i took ''15'' and applied everything mentioned. sum of its divisors is 24. then 1/d1 + 1/d2.... = 1 + 1/3 + 1/5 + 1/15 = 24/15. then i noticed that it is equal to sum/number, so i just guessed the answer to that question should be 75/n

[@sleepyfella]

I used 73 as a number and sum = 74 (as it's a prime) and answer was 1+ 1/73 = 74/73 so marked option 3, did this in my head and it worked lol

[@harshkumar7913]

​@@sleepyfellait's correct though

[@nano1315]

I love the fact that I m going to study from the author of this book

[@alltypevideos9520]

I know you are going to drop with aman dhattarwal star batch? Correct me if I am wrong

[@nano1315]

@@alltypevideos9520 A bit wrong. I am 10th to 11th moving student. So class 11th batch liya h apni kaksha ka.

[@rare0525]

​@@nano1315 see you soon buddy

[@AimersAimers-ef4em]

Same

[@learner3341]

​@@nano1315 is batch sufficient i want best batch

[@hungrywolf6124]

Similar problem in one of the aakash's book except sum of divisors was given 72 which is possible for n=30

[@ts9dream]

Or sum of divisors 73 n=45

[@TuhinSChatterjee]

Sir, I have a little doubt regarding this question.... They have asked "divisors of a positive integer n" but they didn't mention positive divisors which we generally think......at the same time they have asked to include 1 and n. It apparently looks like including all the +ve and -ve divisors. If so, then for each +ve divisor xi of n, there exists a -ve divisor (-xi) of n. Then -1 and -n are also divisors of n. In that case, the sum of all divisors must be 0. But, it is given as 75. So, to show honour to the question of the authentic book, we may consider the divisors as all +ve divisors of n including 1 and n (as specifically mentioned about 1 and n), and all -ve divisors of n excluding -1 and -n. Then, the sum of such divisors = 75 => 1 + n + 0 (since all others cancel out) = 75 => n = 74. Then, the sum of the reciprocals of those divisors would be = 1/1 + 1/74 + 0 (all other reciprocals cancel out) = 75/74 = 75/n. Sir, I think, it may be another approach to think about/explain the key concepts behind the question differently. Thank you. Your comments are most welcome.😊 Further, in this approach, we don't have to consider the problem to be only theoretically true, it is practically true as well. So, there remains no controversy. 😊

[@saumitchandhok5730]

Nice approach 👍

[@TuhinSChatterjee]

@@saumitchandhok5730 Thank you.👍

[@nonstopforever6065]

This is how I solved: let n= a × b where n is positive integer and a and b are it's divisor, if we think about possible values of a and make a set, we find that it is equal to possible values of set b . we can write a= n/b and then apply summation both sides , sum(a) =75= n × sum(1/b) hence sum(1/b) = 75/n 😄

[@saumitchandhok5730]

Nice

[@AnandKumar-cz4pe]

Yes bro right 146

[@reyhanhozaifa4306]

Sir , please make a complete and detailed lecture on Euler's constant

[@AyushSingh-sz8eq]

I actually had this ques in my DPP I thought it was easy and was able to figure it out but thanks for the outlook I didn't thinkk about it that much

[@AyushSingh-sz8eq]

@@Assbeaterniggachad Nope in mathongo crash course

[@rishirajbaul3727]

I think I can show that 75 is never the sum of the divisors of a number. Let S be the sum of divisors function. This function has an interesting property: if m, n are co-prime, S(mn) = S(m)S(n)! Suppose S(N) = 75. Then N cannot be prime (else 74 is prime). So N is a prime power or N = rs such that r,s are co-prime and S(N) = S(r)S(s). If N is a prime power, N is a power of 2, 3, 5 or 7 and these can be checked easily. It follows that {S(r), S(s)} is either {5,15} or {3, 25} (as sets). We can check directly that the sum of divisors of a number is never 5, so the first possibility is immediately ruled out. If the second possibility is what happens, then we must have S(r) = 25 or S(s) = 25. By a similar reasoning as above, this reduces to the existence of a number whose sum of divisors is 5. I appeared for JEE in 2019. Really appreciate your attention to proper rigour in mathematics!

[@angadbharsawde-gl4kw]

As it is not given whether divisors are positive or negative, we can consider n=74 and divisors as 1, 2, 37, 74, -2, and -37

[@derek.divyansh]

-74,-1

[@abhirupkundu8525]

Very good bro

[@Sumit_Girhe]

You need medicle help

[@vinayak9828]

One more approach to solve the problem Let the expression with sigma whose value is to be found out be E. Now AM 》= Hm So 75/k >= k/E So we get E 》= k^2/75 Now we know that since x1 x2 x3 are divisors of n So X1 = 75/n So obviously ,E 》= 75^2/75n So we get E 》= 75/n

[@rohanzaveri5244]

Beautiful way brother! Great way to use AM GM HM ineq

[@PiyushKumar-lv4sj]

OpOpOp ......

[@sardaar_sanveer_singh]

Nice approach bro!🙌

[@syed3344]

lol tujhe kaise pata woh AP mai hai

[@aalekhjain2682]

​@@syed3344bhai kisne bola ki AM and GM sirf AP ke numbers me lagta hai

[@movieworld8288]

You don't need to check it bu using the codes You can check it easily using the formula of sum of positive divisors of n that there is no positive integer n for which sum of positive divisors of n is 75.

[@vikasseth9544]

Superb !!! Brilliant. I am a 50 yr old guy watching your videos.

[@trulysatyam8231]

Thanks to Aman Sir ,a great mathematician 🎉

[@yogeshsharna132]

*a great mathematician,sorry me for spotting your grammatical mistake !😂😂😂!

[@trulysatyam8231]

@@yogeshsharna132 thanks bro ,emotions don't look error😅

[@tuffaniblogs]

youtube.com/@Kgfclasses-py4lw

[@anijee3526]

Mathematician matlab bhi pta hain... 😂😂😂

[@Pathfinder443]

​@@anijee3526 bhai hai vo,mene do saal padhai kari hai unse,he is a mathematician,not by degree but by knowledge.

[@priyanshagrawal780]

sir please make videos for JEE ADVANCE 2023 or atleast 10 chapters . your way of explaining is just next level no one can match it , pls sir

[@cascabellah]

Mathematics at whole another level 💥🎚️

[@ayushgautam996]

Proud to be a student of vg sir

[@IntensifiedGamingOfficial]

Name of the book is itself CONTROVERSIAL

[@insomniac276]

Sir can you please make a series from this book as the advanced paper is near

[@lavneetjanagal]

Take common denominator, it is simply equal to n. On the numerator you have obviously sum of x_i (note that given is the sum of "all" the divisors). So answer is 75/n. It could be typo because divisors of 49 sum up to 57.

[@Sumit_Girhe]

Bro your maths lol

[@lavneetjanagal]

@@Sumit_Girhe Why? what is wrong?

[@Pathfinder443]

My teacher for maths was vikas gupta sir and pankaj Joshi sir who are author of this book😊.I have studied from them in offline mode and they have already finished this book and now focusing on their other book(yellow and pink book).

[@mathsjanta8725]

What? is there any yellow and pink book

[@Pathfinder443]

@@mathsjanta8725 yellow book is for algebra and pink book for coordinate geometry Author is vikas gupta sir.

[@Pathfinder443]

He has material ready for calculas book too and make us do questions in coaching class after mains for advance level questions.

[@mathsjanta8725]

@@Pathfinder443 Bhai ye kya hai

[@Pathfinder443]

@@mathsjanta8725 this is black book (these names black/yellow/pink are given by students) Black book is written by pankaj Joshi sir and vikas gupta sir together and contains advance level ques of all chaps of jee.

[@kararriddho-mw5ew]

this is a pretty easy question to be honest....all u need to is gaussian pairing...but as a foreigner I m really glad that IIT aspirants solve this kind of cute problems unlike the redundant and memorization based admission. systems of our country

[@dakshkatiyar3962]

From which ch is this question

[@shauryagupta3644]

Sir could you pls make a video to show your other approach also, so we can understand how you found that the sum cannot be 75?

[@sunnykhan3309]

Wow sir 🙏🙏🙏sir plz book reviews vdo Laos na sir

[@redff1952]

Better than any other subject 🔥 MATHS 🔥

[@mehulpunia6174]

Hats off to this legend

[@krishanukoushik4435]

Sir , plz make one video about the sum of divisors not greater than 75 🙏🏻 I always watch your videos Your videos make us to think and increase our love towards maths and passion towards solving 😁 🇮🇳

[@imagination594]

Sir ..plz make series on The best questions from A Das Gupta book of maths

[@timirbiswas6785]

I could solve it instantly but I can never imagine that there is no natural numbers whose divisors sum up to 75....and I know there are many like me. This person thinks much beyond than common teachers and students do. I am a math teacher but he is much ahead in his thinking skill. I could solve it quickly just using reciprocity properties of divisors.

[@tarunsoni2241]

Sir ji question is correct because "Let" is written in first,which means either the scenario exist or not,you have to consider the condition

[@Dhruv-w1g]

@Bhannat sir , aman sir. ek doubt tha how did u reach the conclusion that such a number wasnt possible. i mean u mentioned a method, what was that. btw my approach was this x1.xk = n , so does x2.x(k-1) and so on. if we were to grp in addition (x1+xk)+ (x2+x(k-1)) and so on= 75 and then divide by n , (x1+xk)/ x1.xk= 1/x1+ 1/xk similarly for all giving us the required answer

[@crazyabc4365]

Sir Please Bring more problems series from Black Book

[@life_is_beatiful1883]

This problem can be solved in a little different way too. 1xn=n, x1 x xk =n and so on.. therefore (1/x1 + 1/xk) = (x1 + xk)/n.. (1/x2 + 1/x{k-1}) = (x2 + x{k-1})/n and so on So 1/x1 + 1/x2 +.... = 75/n

[@harshkumar7913]

😮 correct bro👍👏🙌

[@adarshmishra4332]

Let's take n = 75 75 / 1 = 75 75 / 2 = 37.5 75 / 3 = 25 75 / 4 = 18.75 ... And so on are divisors of n. If I sum 3+4+5+6+7+8+9+10+11+12 = 75 and all these numbers can be (in this case : are) divisors of n giving Submission 75.

[@Sumit_Girhe]

Poor maths skills

[@Rajis938]

I have finished my sequence and series chapter and im in class 11 Watching this video on full screen the way of teaching is just brilliant the amount of intrest this video contain is just brilliant Sir the editor also have done his job best with audio effects 😁

[@lifeafter498a9]

For this n should be less than 75. k should be less than 4. By trial and error 1x 74 = 1x 2 x 37 => 1,2,37,74 adds to give 114. 1x 73 is close 1+73= 74 1x 3 x 13 = 51 => 1+3+13+51= 68 We can verify this by checking factors of all numbers from 1 through 74

[@krishvora4432]

Please continue this kind of videos.

[@sarfarazsarkar7551]

mATHS TEACHER +STORY TELLER 🔥🔥🔥🔥🔥

[@varunlolage7452]

Sir please bring video or series on mathematical approach for illustrations for jee advanced 2023..

[@prithwiraj453]

The real task is to prove that no natural number can have its sum of divisors = 75

[@IIT_YTT]

I have proof.

[@anshsrivastava7744]

​@@IIT_YTT what is proof

[@IIT_YTT]

@@anshsrivastava7744 bro proof pure 2 page ka hai... Me aapko ye prompt de raha hun ChatGPT me likhna.. Prompt = "Prove that there is no integer exists whose divisors sum is 75"

[@parellax]

@@anshsrivastava7744 To prove that there is no natural number "n" whose sum of divisors is equal to 75, we can use proof by contradiction. Assume for the sake of contradiction that there exists a natural number "n" whose sum of divisors is equal to 75. Let's consider the prime factorization of "n". We can write "n" as a product of its prime factors raised to their respective powers, i.e., n = p1^a1 * p2^a2 * p3^a3 * ... * pk^ak where p1, p2, p3, ..., pk are distinct prime numbers and a1, a2, a3, ..., ak are positive integers representing their respective powers. The sum of divisors of "n" can be calculated using the formula: Sum of divisors of n = (p1^(a1+1) - 1)/(p1 - 1) * (p2^(a2+1) - 1)/(p2 - 1) * (p3^(a3+1) - 1)/(p3 - 1) * ... * (pk^(ak+1) - 1)/(pk - 1) Since we assumed that the sum of divisors of "n" is equal to 75, we can write the equation as: (p1^(a1+1) - 1)/(p1 - 1) * (p2^(a2+1) - 1)/(p2 - 1) * (p3^(a3+1) - 1)/(p3 - 1) * ... * (pk^(ak+1) - 1)/(pk - 1) = 75 Now, let's analyze the possible values of the left-hand side of the equation: If any of the prime factors pi is equal to 2, then the numerator (pi^(ai+1) - 1) will be an odd number, and the denominator (pi - 1) will be an even number. Therefore, the left-hand side of the equation will be an odd number divided by an even number, which is not an integer. But 75 is an integer, so this case is not possible. If any of the prime factors pi is greater than 2, then the numerator (pi^(ai+1) - 1) will be an odd number, and the denominator (pi - 1) will also be an odd number. Therefore, the left-hand side of the equation will be an odd number divided by an odd number, which is an integer. However, the prime factorization of 75 is 3^1 * 5^2, which means that all the prime factors are odd. So, this case is also not possible. Since both cases lead to contradictions, we can conclude that there is no natural number "n" whose sum of divisors is equal to 75. Therefore, the statement is proven false, and there is no such "n" that satisfies the given condition.

[@prithwiraj453]

@@IIT_YTT actually there is a formula for sum of divisors of any number in Number Theory

[@mathswithvinodsir3483]

The problem is tricky. It simply lists some of the divisors of n, including 1 & itself. For example, 1, 4, 14, 56 may be the four divisors of 56 (out of total 8). 1+4+14+56=75 and 1/1 + 1/4 + 1/14 + 1/56 = (56+14+4+1)/56 = 75/56. It my be noted that this question has more than one solution. Also, with the given options, detailed solution is not required; since the other three options can be easily eliminated.

[@SuryaDas-fr5gl]

Sir you are one of my greatest inspiration

[@Karamjeet_11]

Thanks sir nyi observation dekha.....❤

[@bollyfan1330]

Sir sahi kahaa aapne. Aise bhi solve kar sakate ho. Maine n ka value determine karne koh socha. Agar aisa socho kay divisors ka sum ek odd number kabh hoga. Samjho n ek odd number hain, toh (1+n) ek even number hoga baki kay joh factors of n hain woh sabh pair mein aajate hain aur woh sabhi odd hain, yani kay inka sum even hoga, unless n ek perfect square of odd number hain. Agar n even numbere hain, toh (1+n) ek odd number hoga, agar iss number kay koi ek aur odd prime factor hain, toh uska corresponding factor even hoga, yani kay uss pair ka sum odd nikalega. Agar doh odd prime factors hain (say a & b), toh teen pairs banate hain (a, n/a), (b, n/b), (ab, n/ab), inn sabhi kay sum odd hee honge, jabhi (1, n) consider karoge toh sum even ho jayega. Agar teen odd factors lelo, toh factors banenge (1, n), (a, n/a), (b, n/b), (c, n/c), (ab, n/ab), (bc, n/bc), (ca, n/ca), (abc, n/abc). Inn sabhi ka sum even nikalega. Agar 2^x ek factor hain n ka, toh sabhi factor pairs jinme ek number 2^x se divisible nahi hain, unme dono factors even honge, meaning unka sum bhi even hee hoga. Only ek case bachata hain jabhi odd number ka square ek factor ho n kah aur odd number kah cube factor na ho. n < 75 pata hain. Possible n values joh kay odd squares hain, woh hain { 1, 9, 25, 49 }. Inkay factor sums hain { 1, 13, 31, 57 } respectively. Possible n values joh even hain, jinka odd square ek factor hain, woh hain { 2x9, 4x9, 8x9, 2x25 } { 18, 36, 72, 50 }. Inkay factor sums hain { 39, 91, 195, 93 } respectively. Iska matalabh hain kay factor sum kabhi 75 nahi ho sakata.

[@ninetailedamv7983]

daily watch your videos and act as short practice

[@yogeshghadge8399]

Sir I respect your solution and way of approach.. but I believe that due to this only the question says.. "Let" and "If" in starting of the question itself...

[@focus_on_what_you_want.]

U should bring more problems of BLACK BOOK & CENGAGE

[@ashifekbal7755]

By example, let n =6 Then divisors are 1,2,3,6 Now 1+2+3+6=12 Now 1/1 +1/2 +1/3 +1/6= (6+3+2+1) /6 So 75/n(given no) Answer hoga75/k nahi hoga Since we don't know that series is in ascending or decending

[@keiken_ag]

Background music makes this video a short thriller suspense video😂😂

[@user-tp8ku5me4s]

This question isn't wrong as it is not mentioned that we have have to take " all" divisors .Also you took n/x which can be fractional but in question it is given that it should be positive integer only.😅😅

[@IIT_YTT]

Bro, value of n/x is a positive integer in this question. Ex. We can write 6 as 12/2.

[@abhirupkundu8525]

​@@IIT_YTT correct bro, if n is 12, and obviously divisors of any number are natural (we don't take decimals as divisors). That is why, n/x(x is any divisor of the number), will never give fractional value.

[@PenStuff_PS]

In that case the approach of reciprocity property of divisors won't apply as for you interchanging the sum to sum of inverses won't be equal

[@Anonymous515_Gaming]

Usually I skip maths questions simply because how overwhelming they seem(and sometimes are), and I ignored this video for sooo long simply because of the title, But sir's TWO LINES made me solve this question in, like, 2 mins or so Thanks sir

to sir ne banaya hai apko fuddu 😂

The question is correct, the question doesn't state "distinct" divisors, that means the divisors can be same, and k is also variable. So x1 x2...xk can all be equal to 1 and k will be equal to 75. So the question is correct Also, Even if the divisors were distinct then still this does not hold because 'k' is a variable, does not mean k = no. of divisors, so the solution is anyways incorrect, this solution only works when distinct divisors and k = no.of divisors, but no number has this property implying this assumption "distinct divisors and k = no.of divisors" is wrong. Proof by contradiction

[@4kofficial24]

Also there is one mistake in question. 'There is no saying about ALL divisor.'

[@IIT_YTT]

Bro question theek se read karo usme Kth term ka Matlab h all possible divisors.

[@4kofficial24]

Uska matalb all nahi hai

[@IIT_YTT]

@@4kofficial24 bro aapke basics bahut weak hai Sach me...

[@dakshsingh5891]

use the fact that X(k-i+1)*X(i)=n always, replace each term and directly use the eqn given we get a one line answer. 75/n. really easy if you know the trick else we can always find these properties of divisors in minutes in the exam.

[@vishnukondattu]

Infact sum of divisors of any positive integer n will be always zero. Because if x is a divisor of n then minus(x) will also be a divisor. Similarly if y is a divisor of n then minus(y) will also be a divisor. So sum of divisors will be x+(-x)+y+(-y)+--- = 0. If question mentions positive divisors, then it will be greater than zero.But it doesn't

[@Anonymous-lw5lw]

I got 75/n as answer in first go and used same concept..... Wow. .. and then I wondered what wud happen if n is perfect square

[@akhilwastaken]

Sir please decrease the length of the video to 5 to 8 min which can cover more topics and questions 🙏 😊

[@abhirupkundu8525]

No you can't complete topics in so less time kid

[@ajitkumarverma2403]

Aapne 1 author ke concept ko face kiya hai sir. Last me to mja aa gya hai sir. 🙏🙏

[@ShrijanKumar-pk8ye]

Sir mene jaise ye sawal dekha mene apni back book kholi aur dekha ki mughse voh sawal ho chuka tha😊

[@honestadministrator]

Here in x_m = n/ x_1 . So 1/x_1 + 1/ x_m = (x_1 + x_m) /n Similarly 1/x_2 + 1/ x_(m-1) = (x_2 + x_(m-1) ) /n In case n be a square number 1/x_((m+1) /2) = x_((m+1) /2) /n Summing up 1/x_1 + 1/x_2 + .. +1/ x_m = (x_1 + x_2 +... + x_m) /n = (sum of divisors)/n

[@debjanibanerjee5113]

Let n = x1×a1 where a1 is the quotient. given that x1+x2+x3+......+xk =75 so it can also be written as n/a1+n/a2+n/a3+......+n/ak= 75. So 1/a1+1/a2+1/a3+....+1/ak= 75/n. Now the thing is if we take the divisors of any number say 8 that are 1,2,4,8 from here the quotients will be 8,4,2,1 respectively. Here I can say that the sum of reciprocals of the divisors is equal to the sum of reciprocals of its quotients, this thing should be valid for all numbers so by using this concept I can write 1/a1+1/a2+1/a3+......+1/ak=1/x1+1/x2+1/x3+.....+1/xk=75/n. Is this solution correct sir

[@jayeshgawade230]

Sir please share the list of sums of the divisors of the natural numbers that you have made using computer programming. Please 🥺🙏

[@DwaipayanDutta]

Hi Jayesh - The video sir has made was awesome and I stumbled upon this comment so as a solution this is the list of sum of divisors of first 100 numbers [1 to 100] [1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 31, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 63, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 124, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 127, 84, 144, 68, 126, 96, 144, 72, 195, 74, 114, 124, 140, 96, 168, 80, 186, 121, 126, 84, 224, 108, 132, 120, 180, 90, 234, 112, 168, 128, 144, 120, 252, 98, 171, 156, 217] # Program Code to do this in python and use this link to check (Please change for i in range(1, 101) to get sum of divisors for more numbers " www.programiz.com/python-programming/online-compiler/" def sum_divisors(n): divisors = [] for i in range(1, n+1): if n % i == 0: divisors.append(i) return sum(divisors) sums_of_divisors = [] for i in range(1, 101): sums_of_divisors.append(sum_divisors(i)) print(sums_of_divisors) An excel function to do this task as well : =SUMPRODUCT((ROW(INDIRECT("1:"&A1))

[@deepsikhasarma1969]

@@DwaipayanDutta Thank you :)

[@atharvsharma7648]

12:32 sir kindly tell the approach as well, that made you realise that there is no such nat. number which satisfies such a condition, great insight!

[@user-pe9lt3bb6h]

Here is the sum of divisors for each number from 1 to 75 We consider only till 74-75 because after that the summ will exceed 75. 1. **1**: 1 2. **2**: 3 3. **3**: 4 4. **4**: 7 5. **5**: 6 6. **6**: 12 7. **7**: 8 8. **8**: 15 9. **9**: 13 10. **10**: 18 11. **11**: 12 12. **12**: 28 13. **13**: 14 14. **14**: 24 15. **15**: 24 16. **16**: 31 17. **17**: 18 18. **18**: 39 19. **19**: 20 20. **20**: 42 21. **21**: 32 22. **22**: 36 23. **23**: 24 24. **24**: 60 25. **25**: 31 26. **26**: 42 27. **27**: 40 28. **28**: 56 29. **29**: 30 30. **30**: 72 31. **31**: 32 32. **32**: 63 33. **33**: 48 34. **34**: 54 35. **35**: 48 36. **36**: 91 37. **37**: 38 38. **38**: 60 39. **39**: 56 40. **40**: 90 41. **41**: 42 42. **42**: 96 43. **43**: 44 44. **44**: 84 45. **45**: 78 46. **46**: 72 47. **47**: 48 48. **48**: 124 49. **49**: 57 50. **50**: 93 51. **51**: 72 52. **52**: 98 53. **53**: 54 54. **54**: 120 55. **55**: 72 56. **56**: 120 57. **57**: 80 58. **58**: 90 59. **59**: 60 60. **60**: 168 61. **61**: 62 62. **62**: 96 63. **63**: 104 64. **64**: 127 65. **65**: 84 66. **66**: 144 67. **67**: 68 68. **68**: 126 69. **69**: 96 70. **70**: 144 71. **71**: 72 72. **72**: 195 73. **73**: 74 74. **74**: 114 75. **75**: 124

[@scienceschool1762]

Bahot achha explanation hai aap ka sir thankyou and continue more videos of like this question

[@adityahon4362]

He himself needs 16 - 20 mins to solve question which in coaching classes we solve in 3 to 4 mins😂😂😂

[@nirmalkandwal5135]

He's explaining too...

[@Sumit_Girhe]

Acha uncle

[@JEEAdvAspirant-tb4qk]

Thank you sirji mei yeh khud soch paaya tha

[@ShivamMishra-yi6ds]

75(k-1) /n hoga answer

[@drishtibharadwaj9486]

Sir me isi ki preparation kr rha hu kl me vedantu ka lec dekh rha tha isi question problems wala number theory ka usme karaya tha sir ne same question ye isi ka pyq hai ye sir mene dekhte hi answer de Diya tha iska 😊😊😊😊

[@arghamaji8234]

Ok ok hair coloured You look great sir

[@rakeshrazz4103]

The number is integer but divisor can be a decimal like 65 = 130 and 1/2 also a factor because you just used fraction

[@Lucifer-pj8vc]

I also have Same doubt

[@pratikugalmugale9982]

Ese to koi bhi number prime number nahi ho sakta😂

[@IIT_YTT]

Bro Sach batana kis class me ho aap. Aap ke basics bahut weak h yrr

[@ramlaaal]

aap ki ye video dekh ke mujhe bhi aisa ek program banane ka man kiya jo given number tak ka sum of divisor de . aur mene ye banaya python program R = int(input("enter range:")) for i in range(1, R + 1): s = 0 for j in range(1, i + 1): if (i % j == 0): s = s + j print(i, "=", s)

[@shum_zx]

Never underestimate you teachers.

[@SHUBHAMKUMAR-km3tt]

Par question fhir bhi sahi hai sir 🙏 🙏🙏🙏, qki Jo sum Diya hai vo sabhi divisors ka nahi hai sir , '' 1 aur us number" n" ko include nahi Kiya Gaya hai , 🙏🙏 For example 6=1,2,3,6 Agar Mai 1and 6 ko include na Karu to 2+3=5 ayega Jo ki kisi bhi sankhya ke divisors ka sum nahi hota ......., jabki 6 ke sabhi divisors ka sum 12 hoga 🙏🙏🙏

[@xerosity]

The question is not wrong simply The sentence "be the divisors of " does not tell you if all the divisors are taken or if any repeated divisors are there 😊😊 So according to me and blackbook the Question is right !!!!!🎉🎉

[@kevinbanner5161]

Sir please make so many videos on Black book beacause it is a best book for thinking and aply the formula in the question.Black book ke solution ko samajna bohot tough hai is liye aap thode bohut video bana dege to help mil jayegiplease sir.

[@priyadadheech]

I am in class 10th and i solved this problem without paper and pen

[@amarjeetsah5431]

Please sir or anyone who understood tell me how did the answer 75÷n become because sir has multiplied N in the left side and also divided by 1÷N so, therefore in the right side we will do the same so N will cancel each other so only 75 will left... How sir has written 75÷n??????

[@harshagrawal2448]

I can't believe my eyes . YT is spying on me just in the morning i was solving this ques and got struck i resisted my temptation to search this ques online but still yt recommended me your video 😮

[@rajpal2453]

Sir, you are great 🙏🏾

[@sohumsharma2892]

Wow sir, but how did u get to the conclusion that there exists no natural number with sum of divisors = 75