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# Some 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
2 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.

- 2 years, 2 months ago

Oh, I get it! Thanks!

- 2 years, 2 months ago

[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 $$98$$.

- 2 years, 2 months ago

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.

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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

- 2 years, 2 months ago

- 2 years, 2 months ago

Sure, let's talk in the lounge?

- 2 years, 2 months ago

- 2 years, 2 months ago

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

- 2 years, 2 months ago

Factorization: a^2-b^2

- 2 years, 2 months ago

There you did the mistake.

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

- 2 years, 2 months ago

Yeah, I didn't get that too.

- 2 years, 2 months ago

Factorize it: a^2-b^2

- 2 years, 2 months ago

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

- 2 years, 2 months ago

The answer should be $$(3-\sqrt 3)cos\theta$$

- 2 years, 2 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 $$\tan(\theta)$$.

- 2 years, 2 months ago

Yeah.

- 2 years, 2 months ago

1) 99. 2) 551?

- 2 years, 2 months ago

Comment deleted Aug 16, 2015

Can you post your Solution for the first one?

- 2 years, 2 months ago

- 2 years, 2 months ago

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

- 2 years, 2 months ago

Could you provide the options?

- 2 years, 2 months ago

Yes- They are 29,31.32,34

- 2 years, 2 months ago

Same as nelson's

- 2 years, 2 months ago

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

- 2 years, 2 months ago

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.

- 2 years, 2 months ago

I didn't get you.

- 2 years, 2 months ago

I meant what is meant by the fourth line ..

1(1 and x^98)...

- 2 years, 2 months ago

You were talking about Nelson's solution. . .

- 2 years, 2 months ago

For the last doubt i think the number of divisors of $$n$$ is $$18$$

- 2 years, 2 months ago

$$18$$ is there in the options. Can you tell the process please?

- 2 years, 2 months ago

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 $$3 \times 6=18$$

- 2 years, 2 months ago

Excellent explanation! Thank You.

- 2 years, 2 months ago

You're welcome :)

- 2 years, 2 months ago

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!

- 2 years, 2 months ago

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