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 $n$ ?

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 - 5 years, 10 months ago

Oh, I get it! Thanks!

Swapnil Das - 5 years, 10 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.

Nelson Mandela - 5 years, 10 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 $98$ .

Prasun Biswas - 5 years, 10 months ago

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

Abdeslem Smahi - 5 years, 10 months ago

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

Swapnil Das - 5 years, 10 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 $3 \times 6=18$

Abdeslem Smahi - 5 years, 10 months ago

@Abdeslem Smahi – Excellent explanation! Thank You.

Swapnil Das - 5 years, 10 months ago

@Swapnil Das – You're welcome :)

Abdeslem Smahi - 5 years, 10 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!

Nelson Mandela - 5 years, 10 months ago

1) 99. 2) 551?

Sravanth C. - 5 years, 10 months ago

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

Sravanth C. - 5 years, 10 months ago

Please post the solution.

Swapnil Das - 5 years, 10 months ago

Sure, let's talk in the lounge?

Sravanth C. - 5 years, 10 months ago

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

Swapnil Das - 5 years, 10 months ago

@Swapnil Das – The answer should be $(3-\sqrt 3)cos\theta$

Sravanth C. - 5 years, 10 months ago

@Sravanth C. – 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 - 5 years, 10 months ago

@Nelson Mandela – Yeah.

Sravanth C. - 5 years, 10 months ago

@Sravanth C. – Please check the solution.

Swapnil Das - 5 years, 10 months ago

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

Nelson Mandela - 5 years, 10 months ago

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

Sravanth C. - 5 years, 10 months ago

@Sravanth C. – Factorize it: a^2-b^2

Swapnil Das - 5 years, 10 months ago

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

Swapnil Das - 5 years, 10 months ago

@Swapnil Das – There you did the mistake.

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

Nelson Mandela - 5 years, 10 months ago

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

Sravanth C. - 5 years, 10 months ago

@Swapnil Das ,

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

Nelson Mandela - 5 years, 10 months ago

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

Swapnil Das - 5 years, 10 months ago

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

Harshi Singh - 5 years, 10 months ago

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

Swapnil Das - 5 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$