My country's national olympiads 1st round is to be conducted tomorrow. I will most likely post all of them here for our consumption! I expect them to be pretty easy.

@Ekene Franklin
–
No. At present I am not interested. The main goal of student student studying in class 11 and 12 in India is to get into IIT (Indian Institute Of Technology)

This is a marvelous secret and I benefited a lot from this. If you want to write any question for the problems of the week the better option is to write a physics problem rather than maths. Out of 5 problems 3 are allotted for maths and 2 for science. In the maths category there will be a much competition from many users. But in the science category there are only 4 competitors namely Rohit Gupta, Pranshu Gaba, and two staff members Blake Farrow, Danielle Scarano. So, I posted some good physics problems and 3 of them are featured now. But I won't force you to post only physics problems then there will be much competition for science. If you have good problems in maths you can post them (as 3 of my maths problems are also featured although I say that I am basically a physics man).

It will be in the main option bar. It is the place where you can find links like "profile", "stats", "search", etc. The "view" full site option will be present there.

@Manuj Rajpal
–
Ohh !! You are doing in the app. Sorry, I don't know whether this is possible in app or not. But can't you find any option like this. in the screenshot.

@Manuj Rajpal
–
In every set there will be a option at the bottom called "save". If you click that, that particular set will be saved into your sets. Some of my problems are original whereas some other are completely taken from books or the problems and the remaining are the problems which I have taken from books and modified them.

@Manuj Rajpal
–
The community members write problems and post them to community and the staff selects the best 15 questions among all the problems and they will feature them as problems of the week. There is no forum involved in this process.

Before brilliant used to have biology. But recently when the staff updated brilliant they removed biology topic due to lack of problems in that category.

Hello Everyone !!! As this note has got some immense response from young students I would like to continue this further. But the problem is that this note is getting more and more comments day by day due to which the process of loading and opening any notifications is taking more time. So, on the request of @EKENE FRANKLIN any further discussion can be continued in the new note Brilliant young students messageboard

If the mole fraction of a solute is changed from \(\dfrac14\) to \(\dfrac12\) by adding some solute in the \(800~g\) of \(H_2O\) solvent, then what will be the ratio of molarity of the two solutions ?

Simply balance the redox equation and substitute the stoichiometric coefficients of the balanced reaction into your expression.( I would have posted a solution if only my device could type LaTeX.)

I don't think uncertainty in position and momentum are relativistically equal. However, if a non-relativistic assumption is to be made, then from Heisenberg's, and taking p=mv, then uncertainty in velocity>=(h/2pi)X(1/(uncertainty in position X mass)). Sorry for the math ambiguity. My device doesn't type LaTeX.

I have an answer, but I am not totally sure it is correct. Coordination structure chemistry of silicates and silicones is my weakest point in chemistry. Perhaps I need to practise more!:)

@Ram Mohith
–
My goal in many interactions is to make them frictionless. Trivially speaking, I resist every enmity and conduct friendliness. I have even successfully transmuted people from an angry ground state, to a happy excited state!

Quantitatively speaking, momentum is the product of mass of a body and its velocity. It is denoted by \(\overline{P}\).
\[\large \overline{P} = m \overline{v}\]
But the above definition is valid only when velocity of the body is constant. If the velocity of the body is variable we cannot use this formula and we should go by calculus. Physically speaking, momentum is the rate of change of kinetic energy with respect to its velocity. So, you ca say it as kinetic energy at that instance. Here is how we get \(\overline{P} = m \overline{v}\) from kinetic energy.
\[K.E = \dfrac12 m (\overline{v})^2\]
Now, differentiate both side with respect to velocity. We get :
\[\dfrac{d}{dx} K.E = \dfrac{d}{dx} \left ( \dfrac12 m (\overline{v})^2 \right )\]
As, we know \(\dfrac{d}{dx} K.E = \overline{P}\), we get :
\[\overline{P} = \dfrac12 \cdot m \cdot 2 \overline{v}\]
So, we get :
\[\large \boxed{\color{blue} \overline{P} = m \overline{v}}\]

I think you have posted many problems in brilliant. However, when you are writng a question there will be two options, either Number or Multiple choice. You can select whatever you wish.

Thanks. The results will be uploaded online in about 1.5 hours. Meanwhile I am posting them here for our consumption. Solve some, post solutions and invite others to do same!

Problem 1:
A password consists of \(4\) distinct digits such that their sum is \(19\) and exactly two of its digits are prime, e.g. \(0397\). How many possible such passwords exist?

There is a trick. First, take the 20th triangular number. \(\displaystyle\frac{20(20+1)}{2} = 210\)
Add 1. \(210+1=211\) These numbers are known as pizza numbers and are always one more than a triangular number.

@Ekene Franklin
–
Since the divisibility rule of 5 only concerns the last digit, we have to find the last digits of the values. \(1^{124}\) is obviously 1. Since the repetition in the last digit is (for 2): 2, 4, 8, 6 then \(241 \div 4 = k (\text{Remainder}1\)) the first number is 2 (in this set). Apply this to the other powers. Summing the last digits, we get 10 and \(10 \div 5 = k (\text{Remainder}0)\)

@X X
–
Not exactly what I did, but something similar. Thanks a lot. If I may ask, have you ever participated in an olympiad before, or do you hope to participate in one?

@X X
–
Ok. Thanks. By the way, I really admire your problems. They are very cool. One of my major problems here is creating problems. My original problems are too easy(you can check them out). But solving others problems(of course not those extremely hard ones) is not a big problem for me.

@Ram Mohith
–
You're right! (Ha, ha! I'm a little jealous of you because your problems are often the first problem in the easy category, and in your contributions page, the number of people are banging up:p)

@X X
–
But I am not fully done. I expected that Number + Reciprocal will take me to 1 lakh (100 K) solvers. But it fell short of 7 K. But luckily this week too my question came in one of the problems of week (the first one) so it is confirmed that I will reach the milestone in one or two days.

@X X
–
Yes, Vamprie Fan Club. It is were I had my first individual century. After that, the next week I had got 150 upvotes for my question Rising Balloons and the next week is a memorable for me, as I had my first double century for the question Counting 0's.

I have made a hatrick in June 9 - June 29 (3 weeks) as 3 of my problem continuously came in easy part of the POTW. Now again the chance has come for me. Last week and this week too two of my problems are featured. So, if luck favors next week to I can have a problem in POTW. I am thinking very hard on which question should I post the next.

@Mohmmad Farhan
–
No. Still under construction. Before that one I have two more questions, Raising Balloons - 2 and Popping balloons which are yet to be released. After these two the sequel for My. Everest vs Mt K 2 will be released.

Thanks a lot. But no one apart from us is solving the problems. I want to check my solutions and my faulty areas. Please direct the problem solving masters of Brilliant here to help me out.

@Ekene Franklin
–
If you want to direct anyone to this place just keep the symbol @ and type the name of the person. Immediately a notification will be sent to that particular user.

I think you there is some error in spelling or you have kept a gap between the symbol and name. Once check again. If it is correct then you should get it in blue color link and not just normal text.

For example you should type it as @ _ Ekene Franklin ( _ symbol means no gap) then you will get it as @EKENE FRANKLIN

@Ekene Franklin
–
You should select from the list which will appear when you type the name. Also the name should be exactly matching with that particular user even capital letters should be taken into consideration.

My first trophy was in Grade 1(third)[english], Second in grade 2 (1st)[first scrabble competition] and third in grade 2(Best school player award)[first scrabble competiton]

I have a proof for the existence of infinite dimensions.

On a line (1 Dimension), a number can be represented by a point. On a grid (2 Dimensions), an intersection is represented for the data in a cell. In a cube (3 Dimensions), data can still be represented on a point (where straight lines intersect). Following this pattern, there are infinite dimensions (because there are \(\aleph_0\) numbers).

Exactly. Just what I did during the exam, using the logic that if a prime satisfies those conditions, then the only even prime, 2 must be in the sum and difference.

@X X
–
I think you must minimum have 5K people reached. But the main parameter I think is that the contribution page is mostly obtained either by up votes or by solvers to the question (more than 1K). When I got my contribution page I have around 120 upvotes, around 300 solvers, around 30 views for my notes.

@X X
–
@X X, I want your help. I am trying to write a new problem but I got struck up at one point. Here is my doubt :

Suppose lets say a man jumps from a building with initial velocity \(u\). After \(x\) seconds another man freely falls from the same building and from the same height. Now, we can say that the displacement of both the man's are equal as they both reach the ground simultaneously after they jump from same point, so we can equate their distances. Here is what I did :
\[ut + \dfrac12 gt^2 = \dfrac12 g(t - x)^2\]
where \(t\) is the time taken by the first person to reach the ground. The main part of my doubt is that whether we should write the time taken by second person as \((t - x)\) or \((t + x)\).

@Ekene Franklin
–
And also due to the high specific heat capacity of water. This question is a slight modification of air currents (due to which it is cool during the day and warm during night near coastal areas) and I am going to post this very soon.

Hello to everyone! I have not been around for some time now on Brilliant. However, I met on my return a delightful change in the method of posting options for multiple choice questions on Brilliant. Kudos to Calvin Lin and the truly brilliant Brilliant team!

Actually, I sort of did that. So, I made a correct answer then removed it (crossing it with the x) and then the two dummy answers above it, became the correct answers.

Hi guys. Why can't we together contribute to some wiki's. At present I am working on Heat Transfer, Types of Matrices, JEE Complex Numbers, Midpoint of a Line Segment and Section Formula. I need some help to build them so why can't you join me ad show brilliant the power of young students. If you have any such wiki's to build please share them to and I will try my best to contribute them. Thank You !!!

OH. You are saying about me reposting the problems again and again. What to do ? It's my natural habit that I want all of them to see and solve my problems. Man is made up of desires and everybody have their own desires. You cannot stop them and they go on flowing continuously.

@Ekene Franklin
–
What they can do ? They will just leave it off. If all are having problem of me reposting problems then I will stop it. Don't make this as a big issue.

Um, Ram, I think I need some clarification on your problem 'Competition in Leaf'. Transpiration occurs on the upper part, right? So cobalt (II) chloride should come in contact with more water.

Hello, I have been off from Brilliant for a month as my sports championships were coming close.

And I have term tests starting in a fortnight so I have to pack my self up. Similarly, my IGCSE ICT exam is coming closer.

However, I have a doubt in one of the questions of IGCSE Further Pure Mathematics... Can you help?

Q)
a) Show that \[ \displaystyle a^3 - (a - 1)^3 \equiv 3a^2 - 3a + 1 \\ \implies \color{blue} \text{ I know how to do this. But the next one uses this, so I have given here. }\]

@Syed Hamza Khalid
–
The first red box is just like we write r=1 down, we write r= n up (blue)
Second red box, down as r =1,r-1=0 and up(blue) , r=n therefore r-1=n-1
Yellow box, वे take the upper limit as n-1 and add n^3 in first and add 0^3 to make the lower limit as 1 in second
Green : still thinking

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestMy country's national olympiads 1st round is to be conducted tomorrow. I will most likely post all of them here for our consumption! I expect them to be pretty easy.

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Ok. Thanks and also are you writing that exam.

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It is actually meant for students two years older than I am, but I will give it a shot.

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This is a marvelous secret and I benefited a lot from this. If you want to write any question for the problems of the week the better option is to write a physics problem rather than maths. Out of 5 problems 3 are allotted for maths and 2 for science. In the maths category there will be a much competition from many users. But in the science category there are only 4 competitors namely Rohit Gupta, Pranshu Gaba, and two staff members Blake Farrow, Danielle Scarano. So, I posted some good physics problems and 3 of them are featured now. But I won't force you to post only physics problems then there will be much competition for science. If you have good problems in maths you can post them (as 3 of my maths problems are also featured although I say that I am basically a physics man).Log in to reply

Where is the show full site option??(from ur mt. Everest Q discussion)

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It will be in the main option bar. It is the place where you can find links like "profile", "stats", "search", etc. The "view" full site option will be present there.

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Here is the screenshot of where is it present. Can you find it now ?

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How do you send questions to them to select for weekly problem??Do they themselves select from your sets?

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No. Only staff will select the problems of the week. In general, the notification will be as if I have written the problem of the week.

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I personally think Brilliant should include a biology and biochemistry section. What do you think?

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Yes.

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Before brilliant used to have biology. But recently when the staff updated brilliant they removed biology topic due to lack of problems in that category.

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Actually I asked them to remove it, because the biology problems could not be seen in the community page. You can see Calvin Lin's messageboard.

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link. It's still available.

Really? Here is theLog in to reply

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yes

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Here is a link of a biology problem.

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Yes, but they cannot be found in the community page.

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Those are from a set biology is so easy-2, biology is so easy-3, biology is so easy-4, biology is so easy-5biology is so easy-6, biology is so easy-7, biology is so easy-8, biology is so easy-9, biology is so easy-10, biology is so easy-11, biology is so easy-12, biology is so easy-13, biology is so easy-14 by Mohammad Khaza.

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Hello Everyone !!! As this note has got some immense response from young students I would like to continue this further. But the problem is that this note is getting more and more comments day by day due to which the process of loading and opening any notifications is taking more time. So, on the request of @EKENE FRANKLIN any further discussion can be continued in the new note Brilliant young students messageboard

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\(\text{Doubt : 1}\)

If the mole fraction of a solute is changed from \(\dfrac14\) to \(\dfrac12\) by adding some solute in the \(800~g\) of \(H_2O\) solvent, then what will be the ratio of molarity of the two solutions ?

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\(\text{Doubt : 2}\)

In the following reaction : \[\large aNO_3^- + bCl^- + cH^+ \rightarrow dNO + eCl_2 + fH_2O\] find the value of \(\dfrac{a^2(b + c)}{d \times f}\)

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Simply balance the redox equation and substitute the stoichiometric coefficients of the balanced reaction into your expression.( I would have posted a solution if only my device could type LaTeX.)

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Ok. Try now after keeping it in \ (....) without spaces which will give you \(...\)

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Did you tried it. What is the answer you got.

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Yeah, I solved it. My answer is 7.

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\(\text{Doubt : 3}\)

Uncertainty in position and momentum are equal. Then what will be the Uncertainty in velocity ?

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I don't think uncertainty in position and momentum are relativistically equal. However, if a non-relativistic assumption is to be made, then from Heisenberg's, and taking p=mv, then uncertainty in velocity>=(h/2pi)X(1/(uncertainty in position X mass)). Sorry for the math ambiguity. My device doesn't type LaTeX.

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Yes you are correct. You got the correct answer. Regarding the latex you can keep it in \ (...\ ).

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\(\text{Doubt : 4}\)

In \((Si_4O_{12})^{8-}\), how many \(Si - O - Si\) linkages are there ?

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I have an answer, but I am not totally sure it is correct. Coordination structure chemistry of silicates and silicones is my weakest point in chemistry. Perhaps I need to practise more!:)

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What is the answer bro?:)

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It is 4.

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Thanks guys. Keep sending problems and lets discuss!

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Hi guys! For sure we all have noticed that the derivative of kinetic energy is momentum. Is there a physical explanation for this?

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Quantitatively speaking, momentum is the product of mass of a body and its velocity. It is denoted by \(\overline{P}\). \[\large \overline{P} = m \overline{v}\] But the above definition is valid only when velocity of the body is constant. If the velocity of the body is variable we cannot use this formula and we should go by calculus. Physically speaking,

momentum is the rate of change of kinetic energy with respect to its velocity. So, you ca say it askinetic energy at that instance. Here is how we get \(\overline{P} = m \overline{v}\) from kinetic energy. \[K.E = \dfrac12 m (\overline{v})^2\] Now, differentiate both side with respect to velocity. We get : \[\dfrac{d}{dx} K.E = \dfrac{d}{dx} \left ( \dfrac12 m (\overline{v})^2 \right )\] As, we know \(\dfrac{d}{dx} K.E = \overline{P}\), we get : \[\overline{P} = \dfrac12 \cdot m \cdot 2 \overline{v}\] So, we get : \[\large \boxed{\color{blue} \overline{P} = m \overline{v}}\]Log in to reply

Perfect bro!

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Explain the Bohr theory of atoms, combining both mathematics and physics in your explanations.

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Does anyone know how to post a solution of a problem that doesn't have a posted solution yet?

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If you got it correct you click the button "No solution" under Continue and write. And if you got it wrong you can't.

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Thanks a lot, bro!

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@Mohmmad Farhan why can't you come here. I think this is a best place for a student like you. You can post your quires here.

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Sorry! I am not really active these days because of exams

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Ok. No problem !!!

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How do I post numerical answer problems on Brilliant?

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I think you have posted many problems in brilliant. However, when you are writng a question there will be two options, either Number or Multiple choice. You can select whatever you wish.

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Gearing up for today's 1st round olympiads!

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Good luck!

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Thanks. The results will be uploaded online in about 1.5 hours. Meanwhile I am posting them here for our consumption. Solve some, post solutions and invite others to do same!

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Problem 1: A password consists of \(4\) distinct digits such that their sum is \(19\) and exactly two of its digits are prime, e.g. \(0397\). How many possible such passwords exist?

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2,3: (2,3,6,8)

2,5: (2,5,4,8)

2,7: (2,7,4,6),(2,7,1,9)

3,5: x

3,7: (3,7,1,8),(3,7,0,9)

5,7: (5,7,1,6)

There are \(7\times4!=168\) possibilities

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How did you get all possibilities?

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Problem 2: What is the maximum possible area of a quadrilateral of sides \(1, 4, 7, 8\)?

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I think we should use the concept of maxima and minima.

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Perhaps you should solve it. I used the maximality condition of Bretschneider's formula. My answer was \(18\).

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Looks similar to:

https://brilliant.org/problems/hexagon-octagon-in-a-rectangle/?ref_id=1538063

Maybe a harder version perhaps! :D

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Problem 3: \(20\) lines are drawn in a plane. What is the maximum number of parts into which the plane can be divided by the \(20\) straight lines?

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I think it is \(211\)

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There is a trick. First, take the 20th triangular number. \(\displaystyle\frac{20(20+1)}{2} = 210\) Add 1. \(210+1=211\) These numbers are known as pizza numbers and are always one more than a triangular number.

P.S. @EKENE FRANKLIN did you use this method?

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Great! Just what I wanted, and what I used during the examination!

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What is the remainder when \(1^{241} + 2^{241} + 3^{241} + 4^{241}\) is divided by \(5\)?

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\(0\)

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Of course that is the answer, but prove it.

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Problem 5: Find the general term of the sequence described by \(a_n = 4a_{n - 1} - 3a_{n - 2}\) given that \(a_1 = 1\) and \(a_2 = 9\).

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\(a_n=4\times3^{n-1}-3\)

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Great. This was what I got. But do you mind showing your solution so I can clarify mine?

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The first few terms of \(b_n\) is \(2,8,26,80...\) and \(b_n\) also satisfies the recursion \(b_n=4b_{n-1}-3b_{n-2}\)

Compare the two sequences: \(a_n=\frac43b_n-\frac53=4\times3^{n-1}-3\)

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@X X I think this is your first problem of the week in easy catogery.

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Number + Reciprocal will take me to 1 lakh (100 K) solvers. But it fell short of 7 K. But luckily this week too my question came in one of the problems of week (the first one) so it is confirmed that I will reach the milestone in one or two days.

But I am not fully done. I expected thatLog in to reply

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Thundering Question) is the third one.

OH !!! Your problem is featured as the second one in the problems of the week. But my first problem (Log in to reply

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It's a Fact:I have made a hatrick in June 9 - June 29 (3 weeks) as 3 of my problem continuously came in easy part of the POTW. Now again the chance has come for me. Last week and this week too two of my problems are featured. So, if luck favors next week to I can have a problem in POTW. I am thinking very hard on which question should I post the next.

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here. It is posed by B.D

How is my new problem, check itLog in to reply

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I made the cut for the 2nd round! I scored \(76.70\) out of a maximum possible 110. Hurray!

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Oh !!! Congrats.

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Thanks a lot. But no one apart from us is solving the problems. I want to check my solutions and my faulty areas. Please direct the problem solving masters of Brilliant here to help me out.

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Hi guys !!! I want to tell you some secret. I think next week would be a week of physics. Because this week I have found six problems which have equal chances in coming to problems of the week. There are Rohit Gupta's Trick Spring Balance and Spring Balance, Pranshu Gaba's Three blocks Sliding and Relative Stuntmen Motion, Danielle Scarano's Thermographic Dog and my Mt. Everest vs Mt. Godwin Austin. I think these question will be featured definitely this week if not possible then next week.

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@Calvin Lin

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I think you there is some error in spelling or you have kept a gap between the symbol and name. Once check again. If it is correct then you should get it in blue color link and not just normal text.

For example you should type it as @ _ Ekene Franklin ( _ symbol means no gap) then you will get it as @EKENE FRANKLIN

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It is not still working. Do I need to add something like a code or anything, or should I just type it?

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@Calvin Lin

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@Calvin Lin.

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@X X, @Michael Mendrin, @Chew-Seong Cheong, @Mohmmad Farhan,and the rest of all, Can you please participate in this message-board (as requested by Ekene Franklin)

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What courses would you recommend for a 4th grader like me?

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You are 4 the grade ????!!!!!

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10 years old. I told you

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I am going to be 5th grade

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My first trophy was in Grade 1(third)[english], Second in grade 2 (1st)[first scrabble competition] and third in grade 2(Best school player award)[first scrabble competiton]

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## BrilliantForLife

Dare you to toggle \(\LaTeX\)Log in to reply

I have a proof for the existence of infinite dimensions.

On a line (1 Dimension), a number can be represented by a point. On a grid (2 Dimensions), an intersection is represented for the data in a cell. In a cube (3 Dimensions), data can still be represented on a point (where straight lines intersect). Following this pattern, there are infinite dimensions (because there are \(\aleph_0\) numbers).

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Problem 6: Find all primes which are the sum of two primes and difference of two primes.

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5=7-2=3+2,this is the only possibility.

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Exactly. Just what I did during the exam, using the logic that if a prime satisfies those conditions, then the only even prime, 2 must be in the sum and difference.

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Problem 7: Simplify \(S = (cot 10^\circ - 3\sqrt{3})(cosec 20^\circ + 2cot 20^\circ)\)

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Why are you keeping square brackets. For square root you just need to keep flower brackets.

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For trigonometric functions, logarithms, limts, etc you must keep a slash before them. Like \cot will be as \(\cot\) and \log will be as \(\log\).

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Problem 8: A unit square is completely covered by \(3\) identical circles. Find the smallest possible diameter of the circles.

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Problem 9: How many \(6\) digit numbers can be formed using only odd digits?

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Is it 5^6?

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Correct. Nice stuff. What did you do?

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DO YOU SERIOUSLY HAVE TO GIVE ME 46 NOTIFICATIONS OVERNIGHT!

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Is that bad, Mohmmad? I am usually on Brilliant throughout the night.

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I HAVE EXAMS

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Don't call me Mohmmad

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Did you know that :

\[1 + \omega = \omega \ but \ \omega + 1 \neq \omega\]

Freakin' cool right?

P.S. \(\displaystyle \omega \ is \ the \ second \ lowest \ level \ of \ \infty \ after \ \aleph_0 \)

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Finally I reached it. I have reached

1 lakh (100 k)solvers in 91 problems.Log in to reply

Congratulations!

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Do you know the minimum people to reach?

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I didn't get your point !!! Are you asking about minimum people required to get contribution page.

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Suppose lets say a man jumps from a building with initial velocity \(u\). After \(x\) seconds another man freely falls from the same building and from the same height. Now, we can say that the displacement of both the man's are equal as they both reach the ground simultaneously after they jump from same point, so we can equate their distances. Here is what I did : \[ut + \dfrac12 gt^2 = \dfrac12 g(t - x)^2\] where \(t\) is the time taken by the first person to reach the ground. The main part of my doubt is that whether we should write the time taken by second person as \((t - x)\) or \((t + x)\).

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DO or DIEYou are lost in a forest on a cold midnight day. The temperature is quite low and it is bitter cold. Now you have only two chances :

You should lie on the ground and should shrink your body like a tight pack.

Go and stay in the water in a pond nearby.

There not much time left. Take your decision before cold kills you. So,

what wold you choose?Log in to reply

First option.

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But the second one is the correct choice you have to make.

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Hello to everyone! I have not been around for some time now on Brilliant. However, I met on my return a delightful change in the method of posting options for multiple choice questions on Brilliant. Kudos to Calvin Lin and the truly brilliant Brilliant team!

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Which questions would you handpick for people? (ie. Your favourite questions)

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@Calvin Lin Sir, I am very much happy. But I hope multiple correct answer format will also be soon available for the community.

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I think and hope this change is a step up to the multiple correct answer releasing.

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Actually, I sort of did that. So, I made a correct answer then removed it (crossing it with the x) and then the two dummy answers above it, became the correct answers.

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Hello guys! Which of my recent problems did you like the most?

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To be honest, this note has been quiet for a while

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Yup

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Hi guys. Why can't we together contribute to some wiki's. At present I am working on Heat Transfer, Types of Matrices, JEE Complex Numbers, Midpoint of a Line Segment and Section Formula. I need some help to build them so why can't you join me ad show brilliant the power of young students. If you have any such wiki's to build please share them to and I will try my best to contribute them. Thank You !!!

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You keep on changing the topic of your problems and they keep on resurging and I have to say: It is quite annoying

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I didn't get your point. Are you asking anything about my set.

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OH. You are saying about me reposting the problems again and again. What to do ? It's my natural habit that I want all of them to see and solve my problems. Man is made up of desires and everybody have their own desires. You cannot stop them and they go on flowing continuously.

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I will do what I can.

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Thank you.

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Hello guys. I just completed the second stage of my country's olympiads. Much more difficult.

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Good for you.

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Um, Ram, I think I need some clarification on your problem 'Competition in Leaf'. Transpiration occurs on the upper part, right? So cobalt (II) chloride should come in contact with more water.

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No. Stomata is present on the lower surface of leaf and hence more transpiration will occur from lower surface. Search in internet if you want.

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But this is the case in dicots only. In monocots, it is on both surfaces

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Hello, I have been off from Brilliant for a month as my sports championships were coming close.

And I have term tests starting in a fortnight so I have to pack my self up. Similarly, my IGCSE ICT exam is coming closer.

However, I have a doubt in one of the questions of IGCSE Further Pure Mathematics... Can you help?

Q) a) Show that \[ \displaystyle a^3 - (a - 1)^3 \equiv 3a^2 - 3a + 1 \\ \implies \color{blue} \text{ I know how to do this. But the next one uses this, so I have given here. }\]

b) Hence show that

\[ \displaystyle \sum_{r=1}^{n} { r^3 - (r - 1)^3 } = 3 \sum_{r=1}^{n} r^2 - \dfrac{1}{2} (3n^2 + n) \]

c) Show that

\[ \displaystyle \sum_{r=1}^{n} { r^3 - (r - 1)^3 } = n^3 \]

d) Hence deduce that

\[ \displaystyle \sum_{r=1}^{n} r^2 = \dfrac{1}{6} n(n+1) (2n+1) \]

Please help me from (b) onwards.

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Isn't last one n(n+1)(2n+1)/6

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Yeah he probably forgot...

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(b)

\(r^3-(r-1)^3=3r^2-3r+1\)

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)=\sum_{r=1}^n(3r^2-3r+1)\)

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)=\sum_{r=1}^n 3r^2-3\sum_{r=1}^n r+\sum_{r=1}^n 1\)

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)=\sum_{r=1}^n 3r^2-3\times\frac{n^2+n}2+n\)

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)=3\sum_{r=1}^n r^2-\frac{3n^2+n}2\)

(c)

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)\)

\(=\displaystyle\sum_{r=1}^n r^3-\sum_{r=1}^{r=n} (r-1)^3\)

\(=\displaystyle\sum_{r=1}^n r^3-\sum_{r-1=0}^{r-1=n-1} (r-1)^3\)

\(=\displaystyle\sum_{r=1}^n r^3-\sum_{r=0}^{n-1} r^3\) (Change \(r-1\) to \(r\))

\(=\displaystyle(\sum_{r=1}^{n-1} r^3+n^3)-(0^3+\sum_{r=1}^{n-1} r^3)\)

\(=n^3\)

(d)

According to (b) and (c),

\(\displaystyle\sum_{r=1}^n (r^3-(r-1)^3)=3\sum_{r=1}^n r^2-\frac{3n^2+n}2\)

\(n^3=3\displaystyle\sum_{r=1}^n r^2 -\frac{3n^2+n}2\)

\(\displaystyle n^3+\frac{3n^2+n}2=3\sum_{r=1}^n r^2\)

\(\displaystyle\frac{2n^3+3n^2+n}2=3\sum_{r=1}^n r^2\)

\(\displaystyle\frac{2n^3+3n^2+n}6=\sum_{r=1}^n r^2=\frac{n(n+1)(2n+1)}6\)

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In this how does this happen, isn't it supposed to be 1?

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Similarly, I don't understand the blue, red, green and yellow highlighted stuff

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So it will become §(3r^2 - 3r - 1) = 3§r^2 - 3§r - § 1

= 3n(n+1)(2n+1)/6- 3n(n+1)/2 - n

=n^3

Sorry i didn't have submission sign so I used §