Some doubts

Please take some your time to clarify my doubts:

  • Find the coefficient of x98 { x }^{ 98 } in the expansion of(1+x+x2+x3+...+x99)2 \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 55, 1010 and 2020. 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 n2{ n }^{ 2 } has 5555 divisors and nn has only two prime divisors then what is the number of divisors of nn?

Note by Swapnil Das
4 years, 2 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 - 4 years, 2 months ago

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Oh, I get it! Thanks!

Swapnil Das - 4 years, 2 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 - 4 years, 2 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 9898 by adding two elements of the set A={0,1,2,,99}A=\{0,1,2,\ldots,99\} and order of selection of elements is significant, i.e., selecting (a,b)(a,b) is different from selecting (b,a)(b,a).

The product in the problem is just the generating function for the above combinatorial problem and the coefficient of x98x^{98} denotes the number of ways in which you can sum two elements from the set AA to get 9898.

Prasun Biswas - 4 years, 2 months ago

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For the last doubt i think the number of divisors of nn is 1818

Kaito Einstein - 4 years, 2 months ago

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1818 is there in the options. Can you tell the process please?

Swapnil Das - 4 years, 2 months ago

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let pp and qq the prime divisors of nn and n=pm.qkn=p^m.q^k so the number of divisors of nn is (k+1)(m+1)(k+1)(m+1).

and n2=p2m.q2kn^2=p^{2m}.q^{2k} so the number of divisors of n2n^2 is (2m+1)(2k+1)=55=5×11(2m+1)(2k+1)=55=5 \times 11 so k=2andm=5k=2 and m=5 which mean the number of divisors of nn is 3×6=183 \times 6=18

Kaito Einstein - 4 years, 2 months ago

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@Kaito Einstein Excellent explanation! Thank You.

Swapnil Das - 4 years, 2 months ago

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@Swapnil Das You're welcome :)

Kaito Einstein - 4 years, 2 months ago

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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!

Nelson Mandela - 4 years, 2 months ago

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1) 99. 2) 551?

Sravanth Chebrolu - 4 years, 2 months ago

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Yeah. I had made a grave error for the second question. Here's the answer: 3434

Sravanth Chebrolu - 4 years, 2 months ago

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Please post the solution.

Swapnil Das - 4 years, 2 months ago

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Sure, let's talk in the lounge?

Sravanth Chebrolu - 4 years, 2 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 - 4 years, 2 months ago

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@Swapnil Das The answer should be (33)cosθ(3-\sqrt 3)cos\theta

Sravanth Chebrolu - 4 years, 2 months ago

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@Sravanth Chebrolu Which is equal to 3sin(θ)\sqrt { 3 } sin(\theta ) as tan(θ)\tan(\theta) is equal to root(3) -1. take root(3) common and put root(3) -1 as tan(θ)\tan(\theta).

Nelson Mandela - 4 years, 2 months ago

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@Nelson Mandela Yeah.

Sravanth Chebrolu - 4 years, 2 months ago

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@Sravanth Chebrolu Please check the solution.

Swapnil Das - 4 years, 2 months ago

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@Swapnil Das Can you explain how you got the 5th step from the 4th one?

Nelson Mandela - 4 years, 2 months ago

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@Nelson Mandela Yeah, I didn't get that too.

Sravanth Chebrolu - 4 years, 2 months ago

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@Sravanth Chebrolu Factorize it: a^2-b^2

Swapnil Das - 4 years, 2 months ago

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@Nelson Mandela Factorization: a^2-b^2

Swapnil Das - 4 years, 2 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 - 4 years, 2 months ago

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

Sravanth Chebrolu - 4 years, 2 months ago

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@Swapnil Das ,

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

Nelson Mandela - 4 years, 2 months ago

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

Swapnil Das - 4 years, 2 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 - 4 years, 2 months ago

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

Swapnil Das - 4 years, 2 months ago

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