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-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...
@@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.
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
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.
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.
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
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.
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!
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 !! ✌🏻
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
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
@@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)
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.
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 😄
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 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.
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.
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 😁
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
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.
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.
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
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.
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.
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.
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.
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)
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 😊😊😊😊
As soon as I saw ‘including 1 and n’ it clicked that the sum of all possible divisors of n was given, and I easily solved the question.
4 หลายเดือนก่อน
lol no kuch bhi 😂
4 หลายเดือนก่อน
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
Yes sir, bhale hee sawal ka data galat ho, but similar sawal correct data ke saath puch hee sakta hai IIT. Bahut hee sundar sawaal hai, merese toh jaldi hogaya tha, fir main sochne laga yaar isme controversial kya hai, fir maine double check kiya apna method. Uske baad video play kiya toh fir sunnke realise hua 75 toh ho hee nai sakta
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
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
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
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
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 😁 🇮🇳
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. 😊
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 😮
4:27 Kaafi random point, par maano ya na maano, mujhe idhar click kiya ki kaise karna hai ye question... vo baat alag hai, ki maine kaafi lamba tareeka use kiya, but gist iska vhi tha jo sir ne use kiya
First of all I thought that the person who made the question is talking about few of the divisors of all possible divisors of the number "n" Like 48 have : 1,2,3,4,6,8,12,16,24,48 as divisors Looking carefully we get 48+1+4+6+16 = 75 (here 1 and the number 'n' is included) But still if we put these numbers in the 1/x form and take the sum it will not end up to be 75/n 😅 So it's obviously a hypothetical situation... But still enjoyed the concept ❤
Check out Bhannat Maths On Instagram (@bhannat_maths) - bit.ly/BhannatMathsInstagram
Divisors can be positive as well as negative
Oh my god sum is zero
Question is wrong
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-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...
@@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.
@@akshithmanjunath8557 right bro i also calculated it than open comment section
Difference between Other subjects vs Maths 🔥. This subject makes you to think Differently ❤️. Thank you Aman sir (legend)
youtube.com/@Kgfclasses-py4lw
Maths aur physics bhi
Kabhi zindagi mein physics solve Kari hai ?
Physics ka achhe problems is laughing at corner 😂
@@dipanghosh8470 Physics without maths is nothing😐😑
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
No. pls chk with 4, you will get 3 divisors in both cases.
1+2+4=7 and
4+2+1=7
@@kdhd100 yes in case of perfect square x maps onto inself
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.
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 : ")))
Bahar aur ek infinite loop lagado, jo ki wo number bhi Khoj se jiska divisors ka sum 75 ho.
CS wale ho kya?
@@Shreyas_Jaiswal nahi abhi to aspirant hu😅
@@ajeetsingh7209 are mai iit cs ki baat nhi kar raha, class 11,12 wala cs ki kar raha hun.
Bahi muja to cs smj hi nahi 11 ka 12 ka dekta hu tum na to pura program likh dala
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.
Conclusion: never underestimate your teacher
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
Bhai app kese approach krte ho i mean app kese sochte ho ki aeshe karenge toh shi answer niklega
@@Dx_Era_31 ques practice is only approach.
Bhai sahab esi approaches kese build kare 🥲
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
Please correct if I m wrong
Sir you are real legend of maths ♥️
legend is terence tao, andrew wiles, manjul bhargava,etc
@@anubhavupadhyay2949
A little correction needed
Srinivas Ramanujan
@@Maths_3.1415 one more correction needed
Aryabhatta
@@Maths_3.1415 ramanujan is ultra legend bro
And the above mentioned are legends
Yes brother ❤
I feel like mathematician when I study with you sir really from❤️
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.
Nahi smjh aya😢😔
The sum of each pair isnt the 'n' but its the product. You wrote that X1+ Xk is n which is wrong
What if k is odd?
Love the way you solve black book questions , pls countinew these
Using cheat method..option b
Sir showing his class..hats off
question starts from 3:03
Thanks
He bhut hi interesting question plus Aman sir’s teaching makes it better , itni acchi teacher hone ke liye dhanyavaadh sir
Superb !!! Brilliant. I am a 50 yr old guy watching your videos.
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!
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 !! ✌🏻
Ye question kitne no be hai book me ?
Wow! Very interesting questions tha and solutions ka explanation is unique ❤❤
Sir your this video give me goosebumps.
Math is ❤️
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
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
@@sleepyfellait's correct though
Proud to be a student of vg sir
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
Beautiful way brother! Great way to use AM GM HM ineq
OpOpOp ......
Nice approach bro!🙌
lol tujhe kaise pata woh AP mai hai
@@syed3344bhai kisne bola ki AM and GM sirf AP ke numbers me lagta hai
Better than any other subject 🔥 MATHS 🔥
I was able to identify this when I was in 11th
Now I got 99.9747 percentile in JEE Mains sirrrrrr
congrats broooo
btw any book suggestions for jee 2025 aspirants like me?
I got 99.88419
I got 99.9842
@@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)
@@ksp._.79How did it go ?
Similar problem in one of the aakash's book except sum of divisors was given 72 which is possible for n=30
Or sum of divisors 73 n=45
whatttt a solution dil jeet liya sir
Mathematics at whole another level 💥🎚️
Thanks to Aman Sir ,a great mathematician 🎉
*a great mathematician,sorry me for spotting your grammatical mistake
!😂😂😂!
@@yogeshsharna132 thanks bro ,emotions don't look error😅
youtube.com/@Kgfclasses-py4lw
Mathematician matlab bhi pta hain... 😂😂😂
@@anijee3526 bhai hai vo,mene do saal padhai kari hai unse,he is a mathematician,not by degree but by knowledge.
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.
Sir , please make a complete and detailed lecture on Euler's constant
Hats off to this legend
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
@@Assbeaterniggachad Nope in mathongo crash course
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
-74,-1
Very good bro
You need medicle help
Thanks sir nyi observation dekha.....❤
I love the fact that I m going to study from the author of this book
I know you are going to drop with aman dhattarwal star batch? Correct me if I am wrong
@@alltypevideos9520 A bit wrong. I am 10th to 11th moving student. So class 11th batch liya h apni kaksha ka.
@@nano1315 see you soon buddy
Same
@@nano1315 is batch sufficient i want best batch
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
😄
Nice
Yes bro right 146
Aapne 1 author ke concept ko face kiya hai sir. Last me to mja aa gya hai sir. 🙏🙏
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).
What? is there any yellow and pink book
@@mathsjanta8725 yellow book is for algebra and pink book for coordinate geometry
Author is vikas gupta sir.
He has material ready for calculas book too and make us do questions in coaching class after mains for advance level questions.
@@Pathfinder443 Bhai ye kya hai
@@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.
Sir can you please make a series from this book as the advanced paper is near
mATHS TEACHER +STORY TELLER 🔥🔥🔥🔥🔥
Sir mene jaise ye sawal dekha mene apni back book kholi aur dekha ki mughse voh sawal ho chuka tha😊
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.
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
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 😁
Best best nice consept clear sir
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
😮 correct bro👍👏🙌
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.
Name of the book is itself CONTROVERSIAL
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.
Bro your maths lol
@@Sumit_Girhe Why? what is wrong?
Sir Please Bring more problems series from Black Book
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
Wow sir 🙏🙏🙏sir plz book reviews vdo Laos na sir
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.
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
Sir please bring video or series on mathematical approach for illustrations for jee advanced 2023..
Never underestimate you teachers.
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.
Poor maths skills
Please continue this kind of videos.
Sir you are one of my greatest inspiration
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?
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
I cracked this problem myself in 2 minutes
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.
Sir ..plz make series on The best questions from A Das Gupta book of maths
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.
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)
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 😊😊😊😊
Sir apka teaching style bohot i accha lagta hai mujhko
As soon as I saw ‘including 1 and n’ it clicked that the sum of all possible divisors of n was given, and I easily solved the question.
lol no kuch bhi 😂
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
daily watch your videos and act as short practice
Bahot achha explanation hai aap ka sir thankyou and continue more videos of like this question
Yes sir, bhale hee sawal ka data galat ho, but similar sawal correct data ke saath puch hee sakta hai IIT. Bahut hee sundar sawaal hai, merese toh jaldi hogaya tha, fir main sochne laga yaar isme controversial kya hai, fir maine double check kiya apna method. Uske baad video play kiya toh fir sunnke realise hua 75 toh ho hee nai sakta
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
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
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
From which ch is this question
Thank you sirji mei yeh khud soch paaya tha
U should bring more problems of BLACK BOOK & CENGAGE
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
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 😁 🇮🇳
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
Beautiful question 😍
Maza agya sir jii.........❤
Background music makes this video a short thriller suspense video😂😂
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. 😊
Nice approach 👍
@@saumitchandhok5730 Thank you.👍
I solved a similar problem from BmMT (Berkeley mini Math Tournament) and that was fun
Sir app maths ke god hai ❤❤❤❤❤😊
One of the best teachers who left pw
Ok ok hair coloured
You look great sir
Amazing, Sir
Sir, you are great 🙏🏾
Great video sir 👏
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 😮
Sir is legend of mathematics
4:27 Kaafi random point, par maano ya na maano, mujhe idhar click kiya ki kaise karna hai ye question... vo baat alag hai, ki maine kaafi lamba tareeka use kiya, but gist iska vhi tha jo sir ne use kiya
First of all I thought that the person who made the question is talking about few of the divisors of all possible divisors of the number "n"
Like 48 have : 1,2,3,4,6,8,12,16,24,48 as divisors
Looking carefully we get 48+1+4+6+16 = 75 (here 1 and the number 'n' is included)
But still if we put these numbers in the 1/x form and take the sum it will not end up to be 75/n 😅
So it's obviously a hypothetical situation... But still enjoyed the concept ❤
Aisa lg raha Mai koi webseries dek raha....God level teacher
As sum of divisors is odd so we can also conclude they number should be of the form 2*x
But no number of this form satisfies the condition
You are best maths teacher
Absolutely marvelous.
Sir you are great mathematician