Please take some your time to clarify my doubts:

Find the coefficient of \( { x }^{ 98 }\) in the expansion of\( \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 }+...+{ x }^{ 99 } \right) ^{2}\)

The sides of a quadrilateral are all positive integers and three consecutive sides are \(5\), \(10\) and \(20\). How many possible values are there for the fourth side?

What is the maximum number of intersecting points formed with four circles and two straight lines?

If a natural number \({ n }^{ 2 }\) has \(55\) divisors and \(n\) has only two prime divisors then what is the number of divisors of \(n\)?

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TopNewestThe first one,

= (1+x+x^2+x^3+..........+x^99)(1+x+x^2+x^3+..........+x^99).

to find coefficient of x^98 is,

1(by multipling 1 and x^98)+1(by multipling x and x^97)+1(by multipling x^2 and x^96)+..................+1(by multipling x^98 and 1) = 99.

Please correct me if I am wrong. – Nelson Mandela · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

Oh, I get it! Thanks!Log in to reply

[This is not a solution, just a comment]Note that your first question has a nice combinatorial interpretation:

Find the number of ways in which you can get a sum of \(98\) by adding two elements of the set \(A=\{0,1,2,\ldots,99\}\) and order of selection of elements is significant, i.e., selecting \((a,b)\) is different from selecting \((b,a)\).The product in the problem is just the generating function for the above combinatorial problem and the coefficient of \(x^{98}\) denotes the number of ways in which you can sum two elements from the set \(A\) to get \(98\). – Prasun Biswas · 1 year, 7 months ago

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The third one,

2 straight lines intersect at maximum 2C2 points which is 1. A line and a circle intersect at 2 points. So, 2 lines and 4 circles intersect at 2C1 x 4C1 x 2 points = 12. 2 circles intersect at two points. So, 4 circles intersect at 4C2 x 2 = 12 ways.

So, maximum number of intersections is 1+12+12 = 25.

Correct me if I am wrong. – Nelson Mandela · 1 year, 7 months ago

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@Swapnil Das which book you refer for solving these types of questions?and also are you giving nmtc this year? – Harshi Singh · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

I mostly refer internet for solving these type of problems ( Brilliant is only source). No.Log in to reply

@Swapnil Das ,

In your question trigo fun, I think the answer is \(\sqrt { 3 } sin(\theta )\). Can you post the solution please. – Nelson Mandela · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

Thanks, I have deleted the problem and I will repost it later.Log in to reply

Yeah. I had made a grave error for the second question. Here's the answer: \(34\) – Sravanth Chebrolu · 1 year, 7 months ago

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– Sravanth Chebrolu · 1 year, 7 months ago

And the values of possible length's that the third side can take are, \(\text{33, 32, 31, 30, 29, 28, 27 . . . , 1, 0}\)Log in to reply

– Swapnil Das · 1 year, 7 months ago

Please post the solution.Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

Sure, let's talk in the lounge?Log in to reply

– Swapnil Das · 1 year, 7 months ago

Please check the solution.Log in to reply

– Nelson Mandela · 1 year, 7 months ago

Can you explain how you got the 5th step from the 4th one?Log in to reply

– Swapnil Das · 1 year, 7 months ago

Factorization: a^2-b^2Log in to reply

2cos(theta)+sin(theta) is not equal to 2root(3)cos(theta). – Nelson Mandela · 1 year, 7 months ago

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– Sravanth Chebrolu · 1 year, 7 months ago

Yeah, I didn't get that too.Log in to reply

– Swapnil Das · 1 year, 7 months ago

Factorize it: a^2-b^2Log in to reply

– Swapnil Das · 1 year, 7 months ago

Sure, but before that please try my Trigo Fun problem as Nelson Sir is telling that the answer is wrong.Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

The answer should be \((3-\sqrt 3)cos\theta\)Log in to reply

– Nelson Mandela · 1 year, 7 months ago

Which is equal to \(\sqrt { 3 } sin(\theta )\) as \(\tan(\theta)\) is equal to root(3) -1. take root(3) common and put root(3) -1 as \(\tan(\theta)\).Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

Yeah.Log in to reply

1) 99. 2) 551? – Sravanth Chebrolu · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

Can you post your Solution for the first one?Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

What about the second?Log in to reply

– Swapnil Das · 1 year, 7 months ago

Second seems to be wrong in accordance to the options. I don't know.Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

Could you provide the options?Log in to reply

– Swapnil Das · 1 year, 7 months ago

Yes- They are 29,31.32,34Log in to reply

– Sravanth Chebrolu · 1 year, 7 months ago

Same as nelson'sLog in to reply

– Swapnil Das · 1 year, 7 months ago

Can you tell me what is meant by the "out of bracket" 1's in the first problem?Log in to reply

we get 1, 99 times by multiplying various terms such that the sum of their powers is 98. – Nelson Mandela · 1 year, 7 months ago

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– Sravanth Chebrolu · 1 year, 7 months ago

I didn't get you.Log in to reply

1(1 and x^98)... – Swapnil Das · 1 year, 7 months ago

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– Sravanth Chebrolu · 1 year, 7 months ago

You were talking about Nelson's solution. . .Log in to reply

For the last doubt i think the number of divisors of \(n\) is \(18\) – Kaito Einstein · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

\(18\) is there in the options. Can you tell the process please?Log in to reply

and \(n^2=p^{2m}.q^{2k}\) so the number of divisors of \(n^2\) is \((2m+1)(2k+1)=55=5 \times 11\) so \(k=2 and m=5\) which mean the number of divisors of \(n\) is \(3 \times 6=18\) – Kaito Einstein · 1 year, 7 months ago

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– Swapnil Das · 1 year, 7 months ago

Excellent explanation! Thank You.Log in to reply

– Kaito Einstein · 1 year, 7 months ago

You're welcome :)Log in to reply

n^2 has 55 factors which is 5 x 11.

which is (a+1)(b+1) where n^2 = x^a * y^b where a and b are powers of two primes x and y.

Hence, a and b are 4 and 10.

And n = x^2 * y^5 or x^5 *y^2.

Now the number of factors of n is (2+1)(5+1) = 18.

Hope that helps! – Nelson Mandela · 1 year, 7 months ago

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