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Some doubts

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 \(n\)?

Note by Swapnil Das
1 year, 5 months ago

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The 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, 5 months ago

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@Nelson Mandela Oh, I get it! Thanks! Swapnil Das · 1 year, 5 months ago

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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 \(98\). Prasun Biswas · 1 year, 5 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, 5 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, 5 months ago

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@Harshi Singh I mostly refer internet for solving these type of problems ( Brilliant is only source). No. Swapnil Das · 1 year, 5 months ago

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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. Nelson Mandela · 1 year, 5 months ago

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@Nelson Mandela Thanks, I have deleted the problem and I will repost it later. Swapnil Das · 1 year, 5 months ago

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Yeah. I had made a grave error for the second question. Here's the answer: \(34\) Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu And the values of possible length's that the third side can take are, \(\text{33, 32, 31, 30, 29, 28, 27 . . . , 1, 0}\) Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Please post the solution. Swapnil Das · 1 year, 5 months ago

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@Swapnil Das Sure, let's talk in the lounge? Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Please check the solution. Swapnil Das · 1 year, 5 months ago

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@Swapnil Das Can you explain how you got the 5th step from the 4th one? Nelson Mandela · 1 year, 5 months ago

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@Nelson Mandela Factorization: a^2-b^2 Swapnil Das · 1 year, 5 months ago

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@Swapnil Das There you did the mistake.

2cos(theta)+sin(theta) is not equal to 2root(3)cos(theta). Nelson Mandela · 1 year, 5 months ago

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@Nelson Mandela Yeah, I didn't get that too. Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Factorize it: a^2-b^2 Swapnil Das · 1 year, 5 months ago

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@Sravanth Chebrolu Sure, but before that please try my Trigo Fun problem as Nelson Sir is telling that the answer is wrong. Swapnil Das · 1 year, 5 months ago

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@Swapnil Das The answer should be \((3-\sqrt 3)cos\theta\) Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu 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 \(\tan(\theta)\). Nelson Mandela · 1 year, 5 months ago

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@Nelson Mandela Yeah. Sravanth Chebrolu · 1 year, 5 months ago

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1) 99. 2) 551? Sravanth Chebrolu · 1 year, 5 months ago

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Comment deleted Aug 16, 2015

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@Sravanth Chebrolu Can you post your Solution for the first one? Swapnil Das · 1 year, 5 months ago

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@Swapnil Das What about the second? Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Second seems to be wrong in accordance to the options. I don't know. Swapnil Das · 1 year, 5 months ago

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@Swapnil Das Could you provide the options? Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Yes- They are 29,31.32,34 Swapnil Das · 1 year, 5 months ago

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@Swapnil Das Same as nelson's Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu Can you tell me what is meant by the "out of bracket" 1's in the first problem? Swapnil Das · 1 year, 5 months ago

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@Swapnil Das The coefficient of multiplication of two terms.

we get 1, 99 times by multiplying various terms such that the sum of their powers is 98. Nelson Mandela · 1 year, 5 months ago

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@Swapnil Das I didn't get you. Sravanth Chebrolu · 1 year, 5 months ago

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@Sravanth Chebrolu I meant what is meant by the fourth line ..

1(1 and x^98)... Swapnil Das · 1 year, 5 months ago

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@Swapnil Das You were talking about Nelson's solution. . . Sravanth Chebrolu · 1 year, 5 months ago

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For the last doubt i think the number of divisors of \(n\) is \(18\) Kaito Einstein · 1 year, 5 months ago

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@Kaito Einstein \(18\) is there in the options. Can you tell the process please? Swapnil Das · 1 year, 5 months ago

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@Swapnil Das 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 \(3 \times 6=18\) Kaito Einstein · 1 year, 5 months ago

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@Kaito Einstein Excellent explanation! Thank You. Swapnil Das · 1 year, 5 months ago

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@Swapnil Das You're welcome :) Kaito Einstein · 1 year, 5 months ago

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@Swapnil Das 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! Nelson Mandela · 1 year, 5 months ago

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