This is my message board. I am moderator. My job is to keep the community clean and engaged. Any types of questions can be asked here. Any irrelevant comment , note or problem(report the problem and leave the link here) can be conveyed through here or slack.

\[\huge \color{red}{\text{Happy}} \quad \huge \color{blue}{\text{Brillianting !}}\]

## Comments

Sort by:

TopNewestRajdeep, it has come to my attention that you are aggressively @mentioning several members numerous times a day. Please refrain from doing so, especially if you notice that they have not responded to any of your past mentions.

In general, unless you have a strong reason, please refrain from @ mentioning any person more than once a day. If you continually bombard people with such requests, they will (similar to the boy who cried wolf) most likely start to ignore you. – Calvin Lin Staff · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

sorryLog in to reply

@Vighnesh Raut The solution to your question \(\int{\frac{dx}{1 + \tan^{4}{x}}}\) is given below.

Solution : \[\int{\frac{dx}{1+ tan^4 {x}}} \\ \text{Substitute tan x = t . Then the integral will become } \\ \int{\frac{dt}{(1 + t^2)(1 + t^4 )}} \\ \text{ Separating using partial fractions we get} \\ \int{(\frac{1}{2(1 +t^2)} + \frac{2t - \sqrt{2}}{4\sqrt{2}(-t^2 + \sqrt{2}t - 1)} + \frac{2t + \sqrt{2}}{4\sqrt{2}(t^2 + \sqrt{2}t + 1)})} \\ \text{Integrating we get} \\ \frac{1}{2}\arctan{t} - \frac{1}{4\sqrt{2}}\log{(t^2 - \sqrt{2}t + 1)} + \frac{1}{4\sqrt{2}}\log{(t^2 + \sqrt{2}t + 1}) \\ \text{We can easily substitute back in t = tan x and then simplify.}\] – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Log in to reply

@Azhaghu Roopesh M . Please try my 100 followers problem! – Mehul Arora · 2 years ago

Fully agreed!Log in to reply

Sorry for misprint here \(\phi = \varphi\)

Now If body is in equilibrium. Resultant = 0

So that means \(mg\sin{\theta} = T\cos{\phi}\)

Plugging in values and solving we get

\[T = \frac{10 \times 9.8 \times \sin{45^{\circ}}}{\cos{25^{\circ}}}\] If Doubt comment otherwise reply that you understood. – Rajdeep Dhingra · 2 years ago

Log in to reply

– Rishabh Tripathi · 2 years ago

ohh.. thank you.. I got it.. :D :DLog in to reply

– Rajdeep Dhingra · 2 years ago

Your welcome!Log in to reply

unable to download fiitjee big bang sample papers. anybody who knows where to find them? – Himanshi Sethi · 6 months ago

Log in to reply

@Harsh Shrivastava Please post your JOMC Questions here . – Rajdeep Dhingra · 1 year, 1 month ago

Log in to reply

In the above solution mass 1 = 1 kg , mass 2 = 2 kg , mass 3 = 3 kg. This will make calculation easier. You can try and generalize it for variables . If any doubt persists then do ask. – Rajdeep Dhingra · 1 year, 3 months ago

Log in to reply

@Rajdeep Dhingra – Anik Mandal · 1 year, 3 months ago

How do we get that block m1 moves upwards? If we did m1 downwards we get a negative value?Log in to reply

@Anik Mandal – Rajdeep Dhingra · 1 year, 3 months ago

Yes.Log in to reply

Log in to reply

– Anik Mandal · 1 year, 3 months ago

Yes thanks for your co-operationLog in to reply

Log in to reply

– Swapnil Das · 1 year, 9 months ago

Well I would be happy if you post problems on Newton's sums.Log in to reply

Try my new questions Find the Truth 7 and Dirty Algebra. – Archit Boobna · 1 year, 10 months ago

Log in to reply

Rajdeep try my question "I think factorization may help" – Archit Boobna · 1 year, 11 months ago

Log in to reply

– Rajdeep Dhingra · 1 year, 11 months ago

I have already done it.Log in to reply

– Archit Boobna · 1 year, 11 months ago

Do "Dirty Maxima" thenLog in to reply

Here's the link to the staff list , in case you missed out on my comment on Calvin sir's Message Board . – Azhaghu Roopesh M · 2 years ago

Log in to reply

Just help me out in this: -

ABandCDare 2 straight lines meeting atOandXYis another straight line. Show that in general 2 points can be found inXYwhich are equidistant fromABandCD. When is there only one such point ?See this – Rajdeep Dhingra · 2 years ago

Log in to reply

Also I see that Shashwat has explained it to you already . – Azhaghu Roopesh M · 2 years ago

Log in to reply

Guys I have a problem in questions like this one.

If \(\displaystyle f \left( \dfrac{x+y}{2} \right) = \dfrac{f(x) + f(y)}{2}. \)

\(\text{Find } f(2) \text{ given } f(0) = 1 \text{ and } f'(0) = -1. \)

Also give me links to many such problems so I can practice them. Any book or wiki or PDF also will work.

– Rajdeep Dhingra · 2 years ago

Log in to reply

this . – Azhaghu Roopesh M · 2 years ago

You might want to seeLog in to reply

Also why do you want to become a mod ? Just curious . – Azhaghu Roopesh M · 2 years ago

Log in to reply

@Azhaghu Roopesh M , You can Solve the problem now... I have reposted it – Mehul Arora · 2 years ago

Log in to reply

– Rajdeep Dhingra · 2 years ago

It was easy but good. When I tried the same problem with the wrong answer I wondered where had I went wrong.I thought of disputing it but didn't do it as the changed problem had come.Log in to reply

– Mehul Arora · 2 years ago

Yeah! sorry for the inconvenience caused.Log in to reply

P.S - If I get a request to become one well I won't deny it. \(\ddot\smile\) – Rajdeep Dhingra · 2 years ago

Log in to reply

– Azhaghu Roopesh M · 2 years ago

Frankly speaking , Megh has a better chance of becoming a mod . Look at the number of his followers !Log in to reply

– Rajdeep Dhingra · 2 years ago

Yes, But I think you'll also be chosen if megh is.Log in to reply

– Azhaghu Roopesh M · 2 years ago

Really ? We'll see .Log in to reply

Could you help the guy out who posted a question on the RMO/INMO board. The \(\omega\) one. – Rajdeep Dhingra · 2 years ago

Log in to reply

@Prasun Biswas @Rajdeep Dhingra help me in starting that question . Please help. – Mehul Arora · 2 years ago

Log in to reply

@Prasun Biswas Help him. I am exhausted. Not because of you but other things. I typed almost 50 comments each no less than 200 - 300 words. – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

rajdeep, I got my answer as 2013/2008 why did I go wrong?Log in to reply

\(P(1)=P(2)=P(3)=P(4)=P(5)=1\)

\(\Rightarrow P(x) = A(x-1) (x-2) (x-3) (x-4) (x-5)+1\)

Therefore,

\(\dfrac {P(2015)-1}{P(2014)-1} \)

\(= \dfrac {A(2015-1) (2015-2) (2015-3) (2015-4) (2015-5)+1-1}{A(2014-1) (2014-2) (2014-3) (2014-4) (2014-5)+1-1} \)

\(= \dfrac {A(2014) (2013) (2012) (2011) (2010)}{A(2013) (2012) (2011) (2010) (2009)} = \dfrac {2014}{2009} \)

\(\Rightarrow a = 2014\), \(b = 2009\) and \(a+b = 2014+2009 = \boxed{4023}\) – Rajdeep Dhingra · 2 years ago

Log in to reply

@Chew-Seong Cheong Sir's solution... XD – Mehul Arora · 2 years ago

Hahaha.. I truly understood how exhausted u are... This isLog in to reply

– Rajdeep Dhingra · 2 years ago

Well yes, I didn't want to type this whole stuff out. So I did copy his.Log in to reply

– Mehul Arora · 2 years ago

One last question, Is the polnomial always having +1 in the end?Log in to reply

P.S - I will post 3 problems tomorrow on this. Then you can check your newly learnt skill. \(\ddot\smile\) – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

Thanks!! In those problems can You add a solution to 1 problem which helps me to understand how the polynomial expansion will end i.e. with which constant?Log in to reply

See , A polynomial is anything in the form \(a_0 x^n + a_1 x^{n-1} + ......... + a_{n-1} x + a_n\).Provided that only \(a_0 \neq 0\). Rest anything is possible.There no particular constant. – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

Where can I read more about this?? Is there a brilliant wiki? or something like aops or wolfram|alpha?Log in to reply

– Rajdeep Dhingra · 2 years ago

Well the brilliant wiki is empty. I learnt it from class 9 NCERT.Log in to reply

– Mehul Arora · 2 years ago

k.. I have that!! Which Chapter?Log in to reply

link.

Sorry class 10th. Here's theoffline – Rajdeep Dhingra · 2 years ago

Log in to reply

@Rajdeep Dhingra :* – Mehul Arora · 2 years ago

Hell man.... I got that as 4022 !!! Though I finally understood all this.... Thank you so much for your helpLog in to reply

@Rajdeep Dhingra Can You please help me with this ? – Mehul Arora · 2 years ago

Log in to reply

Extra hint: Try forming a new polynomial whose roots are the given \(x\)'s. – Prasun Biswas · 2 years ago

Log in to reply

– Rajdeep Dhingra · 2 years ago

\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\]Log in to reply

\[\ln L=\lim_{x\to\infty}\frac{1}{x}\cdot \sum_{i=1}^x f\left(\frac{i}{x}\right)\]

This is a Riemann sum and can be evaluated by using limit sum definition of definite integral with \(f(x)=\ln x\). We proceed as follows:

\[\ln L=\int\limits_0^1\ln x\,dx\]

Using IBP, we have,

\[\ln L=\left[x\ln x-x\right]_0^1\implies \ln L=(1\times \ln 1-1)-\lim_{x\to 0^+}(x\ln x-x)\]

The limit \(\displaystyle\lim_{x\to 0^+}(x\ln x)\) can be evaluated to be \(0\) using L'Hopital's rule. Thus,

\[\ln L=(0-1)-(0-0)=(-1)\implies \boxed{L=e^{-1}}\] – Prasun Biswas · 2 years ago

Log in to reply

You are watching four people play bridge, where a hand begins by dealing each player 13 cards.

a. After a hand is dealt, you ask a player, “Do you have at least one Ace?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

b. On a later hand, you ask another player, “Do you have the Ace of Spades?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

See when we apply bayes theorem I can't understand how to evaluate P(A) and P(B).The probabilities I asked you.So, instead just solve the whole question please. – Rajdeep Dhingra · 2 years ago

Log in to reply

– Rajdeep Dhingra · 2 years ago

Hey bro . Did you solve that probability problem I had asked ?Log in to reply

– Prasun Biswas · 2 years ago

Nope, I got busy with some NT and inequality problems. I'll look into it later. I was actually gonna solve it in the afternoon but then I had to sleep since I stayed up late last night.Log in to reply

– Rajdeep Dhingra · 2 years ago

Have you tried solving it ?Log in to reply

– Mehul Arora · 2 years ago

I have a really low level of knowledge in Functions and all so can you please help me out? with this and the basics and all....Log in to reply

Its gonna take time to type but if you have patience I will help you. – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

Sure.. Im readyLog in to reply

– Rajdeep Dhingra · 2 years ago

Should I tell you the next step ?Log in to reply

– Mehul Arora · 2 years ago

yeah.. plzzzLog in to reply

So for values up to 5 the function is just \(\large f(n) = n + 1\)

Now we can set up are polynomial

\[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1\]

Now try the question. I am sure you'll be able to do it. If not comment.(I am going offline will be online in half an hour) – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

Ohk... then why is the answer asked in the form of m/n?Log in to reply

– Rajdeep Dhingra · 2 years ago

Because the constant is in the form of \(\dfrac{a}{b}\)Log in to reply

– Mehul Arora · 2 years ago

not getting it :( maybe I will be able to solve more of such problems after this oneLog in to reply

Now we have to find \(C \text{ from } f(x)\).

They have given us the value of the function at \(f(8)\) which is \(f(8) = 8-1 = x-1\).

So now we have \[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1 \\ \large f(8) = C(8-1)(8-2)(8-3)(8-4)(8-5) + 8 + 1 \\ \text{Plugging in value of the function at 8 , we get} \\ 7 = C\dfrac{7!}{2!} + 9 \\ \text{Finding value of C we get} \\ C = (-2)\dfrac{2!}{7!} \\ \text{Now our polynomial changes to n} \\ \large f(x) = (-2)(\dfrac{2!}{7!})(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x +1 \\ \text{Plugging 9 we get} \\ \large f(9) = \dfrac{14}{3} \]

Comment if doubt, otherwise reply that you understood. Here is a problem same as this . Try this to clear concepts – Rajdeep Dhingra · 2 years ago

Log in to reply

– Mehul Arora · 2 years ago

how should I start on this one?Log in to reply

– Rajdeep Dhingra · 2 years ago

Done it ?Log in to reply

– Mehul Arora · 2 years ago

on it .wait pls :)Log in to reply

ndegree polynomial can be expressed in the form \[P(x) = C(x - a_1)(x - a_2)(x - a_3).........(x - a_n) + k\].Try the problem now. If unable to solve I will help you. – Rajdeep Dhingra · 2 years ago

Log in to reply

Will you please tell me how did you get thiscorrect? – Gautam Sharma · 2 years ago

Log in to reply

– Gautam Sharma · 2 years ago

But the person who has posted it has written himself that this question is incorrect .Is the question correct?Log in to reply

here is the solution:

– Rajdeep Dhingra · 2 years agoLog in to reply

– Gautam Sharma · 2 years ago

Yeah i did the same but i didn't want to waste my chance and points thats why i asked.Thanks.Log in to reply

Log in to reply

– Kaustubh Miglani · 1 year ago

A square always ends in 21 41 61 01 or 81 ok Any no is in form of 10k+1....................................10k+9 Hope u can square them nowLog in to reply

Nishi Sofat Arushi Sofat

Please ask Rahul Sir to solve this.\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\] – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Please Find \[\int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx}\] Please help @Ronak Agarwal @Prasun Biswas @megh choksi – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

\( \displaystyle I=\frac { 1 }{ 2 } \int { (2\sec ^{ 2 }{ \theta } -1)\sqrt { \csc { \theta } } d\theta } \)

\(\displaystyle I = \frac { 1 }{ 2 } (\int { (2\sec ^{ 2 }{ \theta } \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } ) \)

Applying product rule to first integral we have :

\( \displaystyle u = \sqrt { \csc { \theta } } , v'=\sec ^{ 2 }{ \theta } \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } -\int { 2\tan { \theta } \frac { -\csc { \theta } \cot { \theta } }{ 2\sqrt { \csc { \theta } } } } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } +\int { \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle \Rightarrow I = \tan { \theta } \sqrt { \csc { \theta } } +C \)

Reverting back to our substitution we have :

\( I = \dfrac { x }{ \sqrt { 1-{ x }^{ 4 } } } +C \) – Ronak Agarwal · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Today, I also found a new way. \[ \int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx} \\ \text{Dividing the inner things by (x^2) we get} \\ \int{\frac{\frac{1 + x^4}{x^3}}{(\frac{ 1 - x^4}{x^2})^{3/2}}dx} \\ \int{\frac{\frac{1}{x^3} + x}{(\frac{1}{x^2} + x^2)^{3/2}}dx} \\ \text{Now we have a derivative, So we substitute the down expression. Integrating and substituting back we get} \\ \frac{1}{\sqrt{\frac{1}{x^2} - x^2}} + C \\ \frac{x}{\sqrt{1 - x^4}} + C \]Log in to reply

Hi Guys, I read your conversation and was fascinated to see your knowledge?

Could you solve one of my questions? Ques1. Prove that \(n^{4}\) + \(4^{n}\) is composite for all integer value of \(n\) \(\gt\) \(1\)

Please help.@Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Satvik Golechha – Micheal Faraday · 2 years, 1 month ago

Log in to reply

Here's the solution to your problem. – Prasun Biswas · 2 years, 1 month ago

Log in to reply

out !

Prasun Have you checked itWill you participate ? Check the edit in it. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

@PRIYANSHU MISHRA For all n \(\in\) Even . It can be easily proved that it composite(As it is a multiple of \(2\) ). For the odd I Have an approach but is not complete. Still working on that .

Here is my approach for odd.

now the cyclicity is \(\boxed{10}\) . So we find remainders for n = 1 to 10.

Here I am getting problems.Solving this would take time . Could you simplify the mod expression.@Calvin Lin @Satvik Golechha @Peter Taylor @Prasun Biswas @Ronak Agarwal Please help. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Interesting Question. On it.Log in to reply

@Prasun Biswas @Ronak Agarwal @Calvin Lin @Satvik Golechha

I have a doubt in the wiki on Chinese Remainder Theorem

I didn't understand how did this theorem is applied. In the example please tell this step.Which values we multipied in this?

Image

Log in to reply

@Prasun Biswas @Ronak Agarwal @Satvik Golechha @satvik pandey @Calvin Lin @Sreejato Bhattacharya Please help – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Prasun Biswas · 2 years, 1 month ago

The sum is, \(x\equiv \displaystyle \sum_i a_iy_iz_i\) where \(z_i\equiv y_i^{-1}\pmod{n_i}\) for the \(i^{\textrm{th}}\) congruence of the system and the other symbols have their usual meanings as given in that wiki. Read it carefully and try the example problems yourself to get a better grasp on how to utilize CRT to solve a system of congruencies. A level 5 like you would've no problem in doing that. Cheers! :)Log in to reply

Just read the theorem and went on to try the example. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

@Prasun Biswas @Ronak Agarwal @Calvin Lin Help. I solved it using basic prime factorization. But I bet there has got to be a shorter solution. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

If a,b,c and d are any 4 positive real numbers, then prove that \[\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4\]

@Prasun Biswas @Ronak Agarwal @Satvik Golechha @Calvin Lin – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Consider the following sequences: \(\{a,b,c,d\}\) and \(\left \{\dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}\right \}\) of positive reals and without loss of generality, assume \(a\geq b\geq c\geq d\). Then, we also have \(\dfrac{1}{d}\geq \dfrac{1}{c}\geq \dfrac{1}{b}\geq \dfrac{1}{a}\)

Using the Rearrangement Inequality, we have,

\[\text{Directed Sum} \ge \text{Random Sum} \ge \text{Reverse Sum}\\ \implies \frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}=1+1+1+1=4\]

\[\therefore \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4\]

and equality holds iff \(a=b=c=d\) – Prasun Biswas · 2 years, 1 month ago

Log in to reply

@Prasun Biswas Nice solution.

In this Rearrangement Inequality can we change the random sum any way we like ? Should it be the Product of the squence in any order?

Like in this case \(Directed Sum \ge \frac{a}{c} + \frac{b}{a} + \frac{d}{b} + \frac{c}{d} \ge Reverse Sum \)

Help @Ronak Agarwal @Calvin Lin – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Prasun Biswas · 2 years, 1 month ago

Yes, for the random sum, we choose any one of the permutations of the second sequence. There's a Brilliant wiki on the topic. You can read it if you want.Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

Simply apply \(A.M \ge G.M. \)Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

How did you expand (a+b+c+d)^4Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

No,No apply \(A.M. \ge H.M.\) to quantities \(\frac{a}{b},\frac{b}{c},\frac{c}{d},\frac{d}{a} \) to directly get the result.Log in to reply

@Ronak Agarwal Please give a solution to the problem Logarithms, Integrals,Trigonometry

First give a hint.Like what to set as a variable to use Leibniz's integral rule. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

Here's a hint consider : \(F(n,m) = \displaystyle \int _{ 0 }^{ \pi /2 }{ \sin ^{ n }{ x } \cos ^{ m }{ x } dx } \)Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

So that means \(F(n,m) = \frac{1}{2}B[\frac{1}{2}(n+1),\frac{1}{2}(m+1)]\). Now what ?Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

Differentiate both sides of the expression first with respect to \(m\) and then with respect to \(n\) and then put \(m=n=0\)Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

How to differentiate Beta functionLog in to reply

@Ronak Agarwal How is this related to your question ? – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

No further hints.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

I have tried the problem and got it incorrect and all 3 tries gone. Please help now.After i have read it you can delete it.Please give more hints.Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

I am putting the whole solution in my solutions discussion, I believe that you will not like my solution since it involves gamma function, anyway you see the solution to any extent you want.Log in to reply

\(F(n)\) is the nth Fibonacci number.Here are some values :

\(F(0) = 0 , F(1) = 1 , F(2) = 1 , F(3) = 2 , \dots \dots , F(n) = F(n-1) + F(n-2)\) – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

@Rajdeep Dhingra , Its not a good idea to ask problems for which no close form exist. This is quite famous, and called "Reciprocal Fibonacci Constant" . There is NO CLOSE form to this summation. You earlier asked , \( \zeta(3) \), which also has no close form, inspite of knowing that. Trying to waste anyone time who is helping you is not a good thing. – Shivang Jindal · 2 years, 1 month ago

Log in to reply

I don't remember the problems name (as i did it a month ago) but I remember the problem . Also there was no solution to it but had quite a many solvers.

The problem was \(2.2.2^{\frac{1}{2}}.2^{\frac{1}{3}}.2^{\frac{1}{5}}\dots \dots\) here the power is fibonacci numbers. hence to evaluate this we needed "Reciprocal Fibonacci Constant".

Sorry @Ronak Agarwal I didn't know it.

P.S- \(\zeta{(3)}\) was just a joke. Please don't take it seriously and please keep helping me Ronak agarwal. I won't reapeat the \(\zeta(3)\) thing. Thank you for helping me now in the future.

I have a doubt that I will ask you later Ronak Agarwal. I have proceeded with the solution till some extent but could you help me out with it. Hopefully will post in 3-4 hours. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

I am working on it.Log in to reply

Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least 2 of them will always be of the same size. Prove that there are at least 3 balls which lie in the same box, have the same colour and the same size. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Are you a moderator? Have you added the solution to your problem ?Log in to reply

@Rajdeep Dhingra – Ronak Agarwal · 2 years, 1 month ago

I have added solution to my problem. You can check it outLog in to reply

– Ronak Agarwal · 2 years, 1 month ago

Yes.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Yes to both ?Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

To only first question.Log in to reply

@Ronak Agarwal Its not working.

Please help @Calvin Lin @Satvik Golechha @Sreejato Bhattacharya – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Thank youLog in to reply

@Ronak Agarwal Have you tried my problem – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Easy Integral

You should try this@Rajdeep Dhingra – Ronak Agarwal · 2 years, 1 month ago

Log in to reply

@Ronak Agarwal Try this. \(\text{ Harmonic }\) at its best

Please Contribute a Solution.@Prasun Biswas @Calvin Lin – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

@Ronak Agarwal I solved it using hyperbolic tangent integral. How did you solve it? – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

Yep, that was too easy.Log in to reply

– Siddharth Bhatt · 2 years, 1 month ago

Simple use of Rearrangemet inequality.Log in to reply

\(\text{ Harmonic }\) at its best – Rajdeep Dhingra · 2 years, 1 month ago

Could you give a algebraic solution to thisLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Which question ?Log in to reply

@Ronak Agarwal @Satvik Golechha @Prasun Biswas @megh choksi @Calvin Lin

Do you know from where I can learn about 'Mod' ,the remainder operator, and it's properties.

Is there a wiki. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Number Theory - Modular Arithmetic.

Yes, there are lots of wikis on the Modular Arithmetic. These are all listed underYou can also do a search and filter by wiki, to see the relevant pages that appear. Often, the suggested page will be the most relevant to you, or it will contain the most information that will be useful to you.

– Calvin Lin Staff · 2 years, 1 month agoLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Thank you very muchLog in to reply

@Calvin Lin Sir please change my age. It is 12 years.My DOB is 14th August 2002. Thanks! In advance – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

@Calvin Lin There is a mistake in the typing of my date of birth. It is actually 14th august 2001. Please restore my streak, account and points. Or transfer those things in this account. – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

– Calvin Lin Staff · 2 years, 1 month ago

I sent you an email to the email address that you used on the account. Please respond accordingly.Log in to reply

@Calvin Lin My DOB is 14th August 2002 not 2001

Sorry for the trouble – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

@Calvin Lin Sir My parents have replied you back. Sir , Please restore my account. – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

@Calvin Lin Sir i have sent you there Email IDs Please reach out to them Fast as i don't want to lose my streak – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

note earlier. I wanted someone to verify whether my method is correct or not. I even tagged you there, but no response. – Prasun Biswas · 2 years, 1 month ago

Can you please help me with this? I posted aLog in to reply

@Calvin Lin Please help – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Could You Guys help me?

Find the remainder when \(2^{1990}\) is divided by 1990.

@Calvin Lin @Satvik Golechha @Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Sreejato Bhattacharya – Tanveen Dhingra · 2 years, 1 month ago

Log in to reply

here – Rajdeep Dhingra · 2 years, 1 month ago

You can see the solutionLog in to reply

A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD.Prove that the crease will divide BC in the ratio 3:5.

@Calvin Lin @Sandeep Bhardwaj @jaikirat sandhu @Prasun Biswas @Ronak Agarwal @Satvik Golechha @megh choksi @Peter Taylor @Sreejato Bhattacharya @Suyeon Khim @Agnishom Chattopadhyay @Mursalin Habib

Please help. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Look at this diagram.

Yo Yo

Here \(AP\) is the crease along which paper is to be folded.Hence \(M\) becomes the reflection of \(B\)

Since \(M\) is the reflection of \(B\) hence \(BN=NM\).

Also let side of square be \(x\) then \(MC=\frac{x}{2}\)

Also \(tan(\theta)=\frac{1}{2},\Rightarrow sec(\theta)=\frac{\sqrt{5}}{2}\)

Now \(BM=\frac{\sqrt{5}x}{2}\) (By applying pythagorean theorom in \(BMC\))

Since \(N\) is the mid-point of \(BM\) hence \(BN=\frac{\sqrt{5}x}{4}\)

Finally in triangle \(BNP\) by simple trignometry we have :

\(BNsec(\theta)=BP=\frac{\sqrt{5}x}{4} \times \frac{\sqrt{5}}{2} = \frac{5x}{8}\)

\(PC=BC-BP=\frac{3x}{8}\)

Finally we are getting \(PC : PB :: 3:5\)

Hence proved. – Ronak Agarwal · 2 years, 1 month ago

Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

And by the way how did you make the figure.Log in to reply

– Ronak Agarwal · 2 years, 1 month ago

Paint.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

And how did you mark the \(\theta\) and the perpendicular signLog in to reply

– Ronak Agarwal · 2 years, 1 month ago

By drawing curve feature and oval feature of paint. You know you can draw curves in paint.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Thanks!!Log in to reply

– Satvik Golechha · 2 years, 1 month ago

I think this is a previous RMO (maybe 1991) problem, you can find a solution online.Log in to reply

Tried but couldn't find a solution.Please help @Satvik Golechha – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Calvin Lin Staff · 2 years, 1 month ago

You are much more likely to get a response by posting that as a note instead of just as a comment on your messageboard.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Thanks for the suggestion. Anyways, do you know how to solve it.Log in to reply

Archit Boobna Calvin lin liked and reshared my integration cauchy

a

b

Log in to reply

do you like S.S.T? – Adarsh Kumar · 2 years, 2 months ago

Log in to reply

Log in to reply

– Adarsh Kumar · 2 years, 2 months ago

well,same here!just like that!Log in to reply

@Archit Boobna Please tell me your FTRE marks – Rajdeep Dhingra · 2 years, 2 months ago

Log in to reply

– Archit Boobna · 2 years, 2 months ago

333Log in to reply

@Archit Boobna Is class there tomorrow ? Tomorrow's date is 14/2/2015 – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Be Careful.

Archit Boobna In your new questionTo evaluate C can the integers be same.

Like-\((-1)^{1000} = 1 \\ also \quad 1^{1000} = 1\).

Do we have take both of them as different cases or same. – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Prasun Biswas · 2 years, 1 month ago

Read the question carefully. The values of \(n\) are redundant. Only the integers which can be expressed as \(n^{100}~,~n\in \mathbb{Z}\) are to be taken into consideration. You cannot express \((-1)\) as \(n^{100}\) for any \(n\in \mathbb{Z}\). Also, you count a single number once, so there are no different cases to begin with. You just consider the number of positive integers that satisfy the given conditions of C.Log in to reply

@Prasun Biswas So you mean the value of C is \(\boxed{1}\) I really do feel that the answer is wrong.

Also look at the other question . posted in this note – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Prasun Biswas · 2 years, 1 month ago

Nope, there is nothing wrong with that problem. I can assure you that. It's a simple one though but it made me think a bit. It was a good problem.Log in to reply

@Prasun Biswas – Archit Boobna · 2 years, 1 month ago

ThanksLog in to reply

– Prasun Biswas · 2 years, 1 month ago

You're welcome. :)Log in to reply

P is any point inside a triangle ABC. The perimeter of the triangle ABC is \(2s\). Prove that \(s\) < \(AP + BP + CP\) < \(2s\)

Also could you give a solution for tanveen dhingra Problem Above – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

Lemma: Whenever we have a point inside a triangle, the distance of it from a vertex is always less than any one of the sides of the triangle originating from that vertex.Image

Given that \(AB+BC+AC=2s\). Using the property specified in the beginning, we have,

\[AP+BP\gt AB\quad,\quad AP+CP\gt AC\quad,\quad BP+CP\gt BC\]

Adding the three inequalities, we get,

\[2(AP+BP+CP)\gt (AB+BC+AC)=2s \implies AP+BP+CP\gt \dfrac{2s}{2}=s\]

\[\therefore AP+BP+CP\gt s\ldots (i)\]

When \(P\) is a point inside \(\Delta\)\(ABC\), we always have,

\[AP\lt AB~\lor~AP\lt AC \quad \land \quad BP\lt BC~\lor~BP\lt AB\quad \land \quad CP\lt AC~\lor~CP\lt BC\]

Adding the three inequalities casewise, we have,

\[AP+BP+CP\lt AB+BC+AC\quad \lor \quad AP+BP+CP\lt AC+AB+BC\]

Note that both the cases imply the same thing which is,

\[AP+BP+CP\lt AB+BC+AC=2s\]

\[\therefore AP+BP+CP\lt 2s\ldots (ii)\]

From \((i)\) and \((ii)\), we have,

\[s\lt AP+BP+CP\lt 2s\]

P.S - I still cannot think of a rigorous proof of the lemma I stated though other than analytically verifying the validity of that lemma. – Prasun Biswas · 2 years, 1 month ago

Log in to reply

@Prasun Biswas The Lower bound is easy.

I believe that the upper bound can only be prove by drawing a mirror image

Please also solve @tanveen dhingra Question – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Prasun Biswas · 2 years, 1 month ago

I have updated the solution to include the proof of the upper bound.Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

what is \(\land\) and \(\lor\)Log in to reply

– Prasun Biswas · 2 years, 1 month ago

\(\land\) denotes the logical "AND" operator and \(\lor\) denotes the logical "OR" operator.Log in to reply

@tanveen dhingra Question. – Rajdeep Dhingra · 2 years, 1 month ago

Could you please solveLog in to reply

– Prasun Biswas · 2 years, 1 month ago

Well, you already gave him the link to the solution of his problem, didn't you?Log in to reply

here – Rajdeep Dhingra · 2 years, 1 month ago

Prasun Biswas Try my new integration problemLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Is that solution the best and easiest.I don't think so.Could you please make it shoterLog in to reply

– Archit Boobna · 2 years, 1 month ago

Just Use BinomialLog in to reply

– Archit Boobna · 2 years, 1 month ago

By the way, i have another proof. If you will help me prove that 2^1980 mod 1990 =1 , then I can prove itLog in to reply

The question is not of proof it is of finding the remainder – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Archit Boobna · 2 years, 1 month ago

DoneLog in to reply

– Archit Boobna · 2 years, 1 month ago

I am just writing it in daumLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

It won't work.Already tried.Log in to reply

(Here "a" is \(1024({ 1024 }^{ 198 }-1)\) mod 1990

Now we can easily factorize 2^1980-1 and we will see that it is a multiple of 995=5x199, so a=0.

So x-1024 is a multiple of 1990. So x mod 1990 is 1024 – Archit Boobna · 2 years, 1 month ago

Log in to reply

@Archit Boobna How would factorization help?

\(2^{1980} - 1\) = \((2^{990}-1)(2^{990} + 1)\) = \((2^{495} - 1)(2^{495}+1)(2^{990} + 1)\)

How did this help? Even further cubic factorization isn't working. @Prasun Biswas @Calvin Lin – Rajdeep Dhingra · 2 years, 1 month ago

Log in to reply

– Archit Boobna · 2 years, 1 month ago

But my approach is halfway to the solutionLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

Yes , But how to prove \(2^{1980} - 1\) \(\mod{995}\) = \(\boxed{0}\)Log in to reply

I checked using the Chinese Remainder Theorem. The result comes out to be \(2^{1980}\equiv 996 \pmod{1990}\) – Prasun Biswas · 2 years, 1 month ago

Log in to reply

@Prasun Biswas – Archit Boobna · 2 years, 1 month ago

http://www.wolframalpha.com/input/?i=2%5E1980-1+mod+995&dataset=Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

but the answer is coming right using this assumptionLog in to reply

@Archit Boobna's method is correct (as far as I can see). – Prasun Biswas · 2 years, 1 month ago

By the way, \(2^{1980}\equiv 1\pmod{995}\), soLog in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

So you mean \(2^{1980}\) is a factor of 995Log in to reply

P.S - Check again. In his solution, he said that \(2^{1980}\equiv 1\pmod{995}\). That doesn't necessarily mean that we'll also have \(2^{1980}\equiv 1\pmod{1990}\). – Prasun Biswas · 2 years, 1 month ago

Log in to reply

– Archit Boobna · 2 years, 1 month ago

I said 2^1980-1 mod 995 =0Log in to reply

– Prasun Biswas · 2 years, 1 month ago

Oh sorry. That was a typo on my part. :PLog in to reply

– Archit Boobna · 2 years, 1 month ago

http://www.wolframalpha.com/input/?i=Factorize+x%5E1980-1Log in to reply

– Rajdeep Dhingra · 2 years, 1 month ago

But during the exam we can't use wolfram alphaLog in to reply

– Archit Boobna · 2 years, 1 month ago

No! I mean to say that 2^1980-1 can be factorised... but we are just not getting howLog in to reply

– Archit Boobna · 2 years, 1 month ago

you are onlineLog in to reply

– Archit Boobna · 2 years, 1 month ago

I am trying to solve itLog in to reply

– Archit Boobna · 2 years, 1 month ago

Rajdeep if you have any problems with my question, discuss it on the solutions page or disputes page of the problemLog in to reply