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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## Comments

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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.

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Oh, I get it! Thanks!

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[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 \(98\).

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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.

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@Swapnil Das which book you refer for solving these types of questions?and also are you giving nmtc this year?

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

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@Swapnil Das ,

In your question trigo fun, I think the answer is \(\sqrt { 3 } sin(\theta )\). Can you post the solution please.

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Thanks, I have deleted the problem and I will repost it later.

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Yeah. I had made a grave error for the second question. Here's the answer: \(34\)

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

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Please post the solution.

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Sure, let's talk in the lounge?

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2cos(theta)+sin(theta) is not equal to 2root(3)cos(theta).

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1) 99. 2) 551?

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Comment deleted Aug 16, 2015

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Can you post your Solution for the first one?

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we get 1, 99 times by multiplying various terms such that the sum of their powers is 98.

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1(1 and x^98)...

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For the last doubt i think the number of divisors of \(n\) is \(18\)

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

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let \(p\) and \(q\) the prime divisors of \(n\) and \(n=p^m.q^k\) so the number of divisors of \(n\) is \((k+1)(m+1)\).

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 \(3 \times 6=18\)

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It is 18.

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!

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