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Re: lulu% - 29-03-2012, 03:11 PMQuote:

there r 3 studnts ta3. 504. not5. cbd

bio=chem/430=120/4.

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Re: Official Quant Thread for CAT 2012 [part 2] - 29-03-2012, 04:05 PMQuote:

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Re: lulu% - 29-03-2012, 04:14 PMQuote:

Originally Posted by bs0409 How many solutions does n^2 - [n^2] = (n - [n])^2 have satisfying 1 ≤ n ≤ 5 where [n] denotes greatest integer less than or equal to n?

n^2-[n^2] = (n-[n])^2 = n^2+[n]^2-2n[n][n]^2+[n^2] = 2n[n]

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n =1, 3/2, 2, 9/4, 5/2, 11/4, 3, 19/6, 10/3, 7/2, 11/3, 23/6, 4, 33/8, 17/4, 35/8, 9/2, 37/8, 19/4, 39/8 , 5total 21 sulutionsLast edited by culdip; 29-03-2012 at 05:11 PM.

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Re: lulu% - 29-03-2012, 04:21 PMQuote:

Originally Posted by bs0409 1.How many solutions does n^2 - [n^2] = (n - [n])^2 have satisfying 1 ≤ n ≤ 5 where [n] denotes greatest integer less than or equal to n?

2.Let [n] denotes greatest integer less than or equal to n. The number of real solutions to the equation 4n^2 - 40[n] + 51 = 0 are

4n^2 - 40[n] +51=0[n]=(4n^2+51)/40....RHS can never be an integer as numerator is odd and denominator is even....so no solutions...

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Re: lulu% - 29-03-2012, 04:23 PMQuote:

Originally Posted by culdip n^2-[n^2] = (n-[n])^2 = n^2+[n]^2-2n[n][n]^2+[n^2] = 2n[n]

n = 1, 1.5, 2, 2.25, 2.5, 2.75, 3, 3.5, 4, 4.25, 4.5, 4.75, 5total 13 solutions

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Aekdum sahi..........

But how did you get the values?? is it by trial and error??

Let N be the largest integer divisible by all positive integers less than its cube root. The number of divisors of N are

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Re: lulu% - 29-03-2012, 04:24 PMQuote:

Originally Posted by bs0409 How many solutions does n^2 - [n^2] = (n - [n])^2 have satisfying 1 ≤ n ≤ 5 where [n] denotes greatest integer less than or equal to n?

I think its 21 solutions

If [n] = k, then [n²] can be k², k² + 1, ...., k² + 2k

=> n² - (k² to k² + 2k) = n² - 2kn + k²

=> n² = k, k + 1/2k, k + 2/2k, ..., k + (2k - 1)/2k

So, For k = 1, we will have two solutionsk = 2, 4 solutionsk = 3, 6 solutionsk = 4, 8 solutionsk = 5, 1 solution as n ≤ 5

But n = 2, 3, 4, 5 are counted twice

So, 21 solutions

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Re: lulu% - 29-03-2012, 04:38 PMQuote:

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Originally Posted by Debtanu12 Number of students taking up all four subjects = xNumber of students taking up any three subjects = 0Number of students taking up any two subjects = 0Number of students taking up physics only = pTherefore, number of students taking up physics = p+xTherefore, number of students taking up chemistry = 2*(p+x)Therefore, number of students taking up maths = 1.5*(p+x)Therefore, number of students taking up biology = 0.5*(p+x)

=> 5*(p+x)=120=> p+x=24

Therefore, Physics = 24Chemistry = 48Maths = 36Biology = 12Hence, maximum value of x = 12

I think you are doing it wrong 12+24+36+48=120 

In this case each student will attend only one class, which is not asked in the question

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Re: lulu% - 29-03-2012, 04:44 PM

WARNING : This is going to be a Loooooooong post .. 

SORRY, that this post is not organized...   Its nothing but just the random collection of tips and tricks that I have gathered over past few months.

Puys, below is the list of concepts/tricks/fundamentals/formulaes etc. that I have learnt from puys like you and GODS viz., Hemant saar, Nachiket bhai, Naga Saar, Angad bhai, Mn bhai, Varun saar, Vikas bhai and many

more .. 

Download these :MBA|CAT|CAT 2011|CAT 2012|CAT Online|MBA 2012|MBA Entrance Exams|CAT Test|Preparation|CAT Questionshttp://totalgadha.com/mod/forum/discuss.php?d=8392 - Find last two non-zero digits.MBA|CAT|CAT 2011|CAT 2012|CAT Online|MBA 2012|MBA Entrance Exams|CAT Test|Preparation|CAT Questions - Number of solutions of an equation.Microsoft Word - Venn Diagrams- Basics, Problems, Maxima and Minima_4 - Venn DiagramsTransformation of graphs by modulus function - Plotting graphs of quadratic equations...

Inclusion?exclusion principle - Wikipedia, the free encyclopedia - Inclusion - Exclusionhttp://www.math.cornell.edu/~putnam/inclExcl.pdf - Inclusion - Exclusionhttp://www.math.ust.hk/~mabfchen/Mat...-Exclusion.pdf - Inclusion - Exclusionhttp://www.pagalguy.com/forum/cat-an...ml#post1742160 (PG Underdogs Team ' 09) - Last minute MUST

Geometry :  

Ptolemy's theorem :

For a quadrilateral ABCD to be a closed figure , if AC and BD are diagonals.AC * BD >= AB*CD + AD*BCFor a cyclic quadrilateral the equality holds true i.eAC * BD = AB*CD + AD*BC

For any quad ABCD, side lengths a,b,c,d respt. and q=1/2(Sum of Opposite Angles) Area=((s-a)(s-b)(s-c)(s-d)-abcd*cosq)^1/2, So in case of Cyclic Quad, Area=((s-a)(s-b)(s-c)(s-d))^1/2. ????

for quadr. circumscribing a circle ...sum of opposite sides is equal

The perimeter of a right angle triangle=2(r+2R)And the area =r^2+2rR = r(r + 2R)

-> when two circles are of same radius are overlapped with center common area is 1/2 (4*pi /3 - root 3) r^2 -> Sum of the exterior angles of a polygon add up to 360 

-> In general, for n-sided polygon, No. of triangles that can be drawn by using the vertices, but not using the sides = n/3 * C(n-4,2)

If |x| +|y|=a, then the area of the region bounded by the given curve will be 2a^2, This is also valid also when |x+k|+|y+k|=a 

Q. Find the area of the region bounded by the graph |x-7|+|y-9|=12 (AIMCAT 1209)

Area bounded by the graph |x-7|+|y-9| = 12 is same as Area bounded by the graph |x|+|y| = 12 and that is =>2*12^2=288 sq units

nC0 + nC1 + nC2 + ... + nCn = 2^nto get an intersection point inside the polygon, we need to select 4 of the vertices. After choosing them we can get two intersecting diagonals in only one way. So, total intersection points = C(n, 4)

we get nC4 intersection points if polygon has n sides with n being odd. And (nC4 - (n/2)C2 + 1) if n is even as we get centre point as a common point to n/2 diagonals.

-> that if nCa=nCb then either a=b or a+b=n

When two equations a1x^2 + b1x + c1 = 0 and a2x^2 +b2x + c2 = 0 have ONE common root, then (c1a2- c2a1)^2 = (a1b2-a2b1)(b1c2-b2c1)COMMON ROOT is:

c1a2-c2a1/a1b2-a2b1

No. of co-prime pairs of factors = (a+1)(b+1)(c+1) + ab + bc + ca +3abc - 1

A perfect number is the number whose sum of all factors except itself is equal to that number itself.First 4 perfect numbers are 6, 28, 496 and 8128.

Binomial :  

(a + b)n = nC0an + nC1an − 1b + nC2an − 2b2 + nC3an − 3b3 + ... + nCnbn(1 - x)^(-r) = (1 + C(r, 1)x + C(r + 1, 2)x^2 + .... + C(n + r - 1, n)x^n + ....(x + y)^n - (x - y)^n = 2{C(n, 1)yx^(n - 1) + C(n, 3)y^3*x^(n - 3) + ..... }(1-x)^(-3) = 1 + C(3,1)*x^1 + C(4,2)*x^2 + C(5,3)*x^3 + ......

Basically if no of marks are +n for correct and -1 for incorrect, thenNumber of score which are not possible is given by (n - 1) + (n - 2) + ... + 1 = n(n - 1)/2

after crossing if they take time t1 and t2 to reach destination, then, v1/v2=root(t2/t1)

Sum of digits formed without repetition : 

If zero is not there,S = (n-1)! * (Sum of digits)*11111.. n times.

If zero '0' is there .

S = (n-2)! (Sum of all digits) [(n-1)(1111...n times) - (1111... n-1 times)]

Remainders :  

There is a theorem in quant Very ImpAny no written 6 times is divisible by 3,7,11,13,37 and the no itselfTherefor 111111 will be divisible by 3,7,11,13, 37 and 1now similarly 6 8,s will be divisible by 7.hence 48 8's will be divisible by 7

any digit repeated 3 times is divisible by 3*37

1^k + 2^k + 3^k...+ n^k is divisible by n(n+1)/2 when k is odd

-> No's repeated 3^n times are divisible by 3^n

-> a^6 - 1 is divisible by 7, where a and 7 are coprime. (Fermat;s theorem)=> 10^6 - 1 will be divisible by 7=> 999999 will be divisible

-> group 3 digits each and then alternatively subtract and add. The final number when divided by 7 will give u the remainder.123456789(789+123-456 )/7 would give u the same reminder

-> Sum of square of 12 consecutive odd numbers is S = 48k² + 572

-> 1^p+2^p...is divisible by 1+2+3+4..when p is odd

-> Odd perfect squares are always of form 8k + 1

-> Square numbers are always of the form 3n or 3n+1

-> 2^20n has last four digit as c9376 2^(20k) always has last two digits as 762^(20k + 10) always has last two digits as 24

-> When a three digit number has last two digits 24 and 74 , then their cube always ends in 24

-> Numbers ending with 49 and 99, when raised to even power end with 01. And when raised to odd power, end with 49 or 99 resp.

-> when number in base n is divided by n -1 , the divisiblity rule is sum of number should be divisible by n -1 

-> In any base 'n' last digit is always the remainder when the number is divided by n.

-> The remainder we get when we divide the sum of the digits by 9 is the same as the remainder when we divide the number by 9

Note:- If we have to find remainder on division by 1001, then form pairs of 3 from the right side, and find the difference between the sum of alternate terms.Eg: - 12345678 = 678 - 345 + 12 (mod 1001) = 345 (mod 1001)

-> Euler's number of number N is the number of numbers less than N and coprime to N.

-> Say the number is xxxxxx (that is x occurring 6 times)Now xxxxxx=1001*xxx for any value of xIn fact for a set of 3 numbers recurring twice it will be divisible by 1001.123123 = 123*1001abcabc=abc*1001Now 1001=7*13*11So the number mentioned in the first line will always be divisible by 7,11 & 13 

-> Last three digits of 7^k has a pattern.

7^4 = 4017^8 = 8017^12 = 2017^16 = 6017^20 = 0017^24 = 4017^28 = 801 and so on

-> Product of all factors= A ^(No. of factors of A)/2

-> In an AP, when u divide any term with the common difference remainder is first term!! >-> In how many ways can a number be written as a sum of ‘n’ consecutive natural numbers, where ‘n’ is greater than 1? 

no of ways= (no of odd factors -1)

e.g. n = 1000 1000= 2^3* 5^3no of ways= (no of odd factors -1)= (3+1)-1= 3

the number of ways in which N can be written as sum of n consecutive positive even integers is B', whereB' = {no of odd factors of (N/2)} - 1So, for odd N it will be zero.

the number of ways in which N can be written as sum of n consecutive positive odd integers is B'', whereB" = {number of ways in which N can be written as product of two factors} - 1 ........ (When N is odd)B" = {number of ways in which (N/4) can be written as product of two factors} - 1 ........ (When N is a multiple of 4)For N = 4k + 2, B" = 0

Largest possible value of k for which 3 raise to power 11 is expressible as sum of k consecutive positiv integersfor odd numbers , 2*3^((n-1)/2) and for even 3^(n/2)

-> divisibility rule of 999 is : starting from right hand side of number , form triplets and add all these triplets.. divide this by 999 to obtain remainder ..eg .. 123876 --> 123 + 876 = 999 .. so perfectly divisible ..

-> For a quadratic equation of the form ax^3 + bx^2 + cx + d,Sum of the roots is -b/aSum of the roots taken 2 at a time is c/aProduct of the roots is -d/a

-> Cauchy - schwarz inequality:   (a^2 + b^2)(x^2+y^2) >= (ax+by)^2(x1² + x2² + x3²)(y1² + y2² + y3²) ≥ (x1y1 + x2y2 + x3y3)²

-> x^4 + y^4 + z^4 = (x² + y² + z²)² - 2(x²y² + y²z² + z²x²)

-> Coef of a^p*b^q*c^r in (a+b+c+...+k)^n = n!/(p!*q!*r!)

-> Suppose we have to find the number of ways in which an ordered pair (a, b), where a and b are natural numbers, can be choosen such that LCM of a and b is (p^x)*(q^y)*(r^z)*... or HCF = 1 isSo, total number of such ordered pairs = (2x + 1)(2y + 1)(2z + 1)..

a and b are natural numbers. Let f(a, b) be the number of cells that the line joining (a, b) to (0, 0) cuts in the region 0 ≤ x ≤ a and 0 ≤ y ≤ b. For example f(1, 1) is 1 because the line joining (1, 1) and (0, 0) cuts just one cell. Similarly f(2, 1) is 2 and f(3, 2) = 4.

So, f(a,b) = a + b - HCF(a,b)

Baye's theorem : P(I/R)=P(R/I)*P(I)/(P(R/I)*P(I)+P(II)P(R/II)+P(R/III)P(R/III)

SQUARES : 

Any perfect square when divided by 4 leaves remainder as either 0 or 1.Any perfect square when divided by 7 leaves a remainder of 0 or 1 or 2 or 4

Squares of even numbers will be of form 4n

-> The digit sum of perfect squares are always 1,4,7 or 9.

-> Numbers of form (50k ± 12)² ends in 44

-> Numbers of form (500n ± 3 ² will end in 444

-> Number of ways a number can be expressed as diff of perfect square is : (number of factors)/2 45 = 3^2 * 5(number of factor)/2= 6/2 = 3 .

If each term of an infinite GP is equal m times the sum of all the terms that follow it, then r (common ratio) = 1/(m+1)

If p is prime then any no repeated p-1 times is divisible by p, for p > 5.

If p is prime, k^(p-1) mod p = 1 mod p

We know that with two coprime numbers a and b, the largest number that can not be formed is (ab - a - b).

m*n parallelograms => m+n+2 lines

If p! ends with k zeros, then k can take [p/5] + 1 different values for p<= 0

If a,b,c are distant from point P which is interior to triangle to vertices and d is side of triangle , then, 3(a^4+b^4+c^4+d^4)=(a^2+b^2+c^2+d^2)^2put a,b,c value and solve for d ..

Max value of asinx + bcosx = (a^2+b^2)^1/2Min value of asinx + bcos = -(a^2+b^2)^1/2

how many pairs of positive integers r there (a,b) so that their LCM is < a number > Total Cases = (2x+1)(2y+1) where x,y are powers of prime factors of the LCM of the number

-> The last non-zero digit of a number N! is LNZ(4^n)*LNZ(2n)! ..... (for N = 10,20,30....), where n=N/10 

R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]

where n = 5a + b

Example: What is the rightmost non-zero digit of 25! ?

→ R (25!) = Last Digit of [ 2^5 x R (5!) x R (0!) ]

→ R (25!) = Last Digit of [ 2 x 2 x 1 ] = 4

-> Last two non zero digits of nth multiple of 10 i.e., (10n)! is LTNZ(44^n)*LTNZ(2n!) or (10*n) = (44)^n * (2n)!

The last two non-zero digits of a number N! isZ(A)=Z(B) *Z(x/5) *Z(Unit Digit) Where, Z(A) is the last non zero digit of N's factorial.Z(B)= 4 if tens digit is odd ; 6 otherwiseZ(x/5)=Smallest Integer <= N/5.

So, for 99

Z(99)=4*Z(19)*Z(9)

Now, For Z(19) you can run the method recursively to find out it's 2Z(9)=8

So, Z(99)=4*2*8=64

The number of ways one can go from P to Q (using the shortest path) is (m+n) C m or (m+n) C n, where, m,n are number of rows and columns of the grid.

sum of the squares of the roots of a quadratic equation : (b/a)^2 - 2(c/a)sum of the cubes of the roots of a quadratic equation : (b/a)^3 - 3(c/a)(b/a)

-> For any expansion of the type (x1+x2+x3........xn)^r, the number of terms is given by (n+r-1)C(r-1)

-> For a quadratic equation, Sum of roots = - Coefficient of x^(n - 1)Product of roots = (-1)ⁿ*Constant termSum of roots is always -b/a

Product of roots is (-1)^n * k/a

... where k is constant term.

So, with quadratic equations, we get product of roots as c/a

With cubic equations we get product of roots as -d/a

In general,

ax^n + b*x^(n-1) + ...

Sum of 1 roots at a time = -b/a

Sum of 2 roots at a time = c/a

Sum of 3 roots at a time = -d/a

And so on..

number of positive integral solutions of ax+by+cz=d is equal to coefficient of x^d in(x^a + x^2a + x^3a......)(x^b+ x^2b+.........)(x^c+x^2c+......)

if x, y and z can take zero value also..then, number of solutions are coefficient of x^d in(1+x^a + x^2a + x^3a......)(1+x^b+ x^2b+.........)(1+x^c+x^2c+......)

If an equation is 1/a+1/b=1/c, then the no. of solution of (a,b), which

satisfies this will be {factors of C^2}/2 for unordered case.

-> if sides of rectangle are mutually prime, the number of squares cut by diagonal is a+b-1 ..

-> the hands are at right angles when the lag between them is 15 min.

-> For polygon with sides 'n' (n>5), the number of triangles which have no side common with any side of the side of the polygon is given by n/6 * (n-4)*(n-5)

Number of additional lines created through intersection = 1/8 * [ n! / (n-4)!] , where 'n' is the number of straight lines

Problems on Intersection of Straight lines , Circles, Formation of Points and Formation of Triangles, Quadrilaterals Etc

Basic Concept Fundas

1. If there are n number of straight lines , They intersect each other in nc2 ways

2. If there are m number of circles , They intersect each other in

2*(mc2) ways = m (m-1)= 2p2 ways

3. When n straight lines and m circles intersect each other , they intersect in

at most 2 * m * n = 2* ( no. of circles ) * ( no.of straight lines)

4. When n parallel lines intersect m straight lines , Then no. of parallelograms possible = nc2 * mc2= mn (m-1) (n-1)/4

5. There is one case when collinear and non- collinear points are given , and asked how many triangles it can formed=>

The funda for this - ( Triangles that can be formed with all points ) – ( Triangles formed with collinear points )

And the same funda is applied whenever such variations in condition occurs

Maxim number of regions given by 'n' lines = 1 + (Sum of all natural numbers till n)

So, 5 lines will give 1 + (1+2+..+5) = 16 regions

As circle will give 2*5 more regions..

Circle always gives extra regions which are equal to 2*(number of lines) **So, total 16 + 10 = 26 regions

contd. ...Last edited by pkaman; 29-03-2012 at 04:47 PM.

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Re: lulu% - 29-03-2012, 04:45 PM

... contd.

Permutations & Combinations :  

how many ways one can put 

(a) 4 different boll in 3 different beg(b) 4 identical ball in 3 different bag(C)4 different boll in 3identical beg(d)4 identical boll in 3identical beg

1) 4 identical balls in 4 identical bags: (This is easy one)5 ways:(4, 0, 0, 0), (3, 1, 0, 0), (2, 2, 0, 0), (2, 1, 1, 0) or (1, 1, 1, 1)

(2) 4 identical balls in 4 different bags: (This is also easy one)It is like this:a+b+c+d = 4where a, b, c and d are number of balls in each bag.So, 7C3 ways

(3) 4 different balls in 4 different bags: (This is also easy one)Each ball can go to one of the 4 bags. So, it has 4 choices. So, it will be 4*4*4*4.. 4 choices for each.So, 4^4.

(4) 4 different balls in 4 same bags: (This is calculation bsed..)This is based on first case (1)(4, 0, 0, 0) => 1 way only(3, 1, 0, 0) => 4C3 ways3 balls to be grouped can be chosen in 4C3 ways and remaining one in 1 way(2, 2, 0, 0) => 2 balls can be chosen in 4C2 ways.So, 6 ways. But, as we select 2, at the same time we make one more pair of remaing 2. So, at one time, we take care of 2 cases. So, we can take 4C2/2 = 3 ways.(2, 1, 1, 0) => 4C24C2 ways to chose 2 balls and then reaming will be 1 and 1.(1, 1, 1, 1) => 1 way only

Fractions : 

a/b < (a+n)/(b+n) ... when a < b

We can do a quick test like:1/2 < 99/100

a/b < (a+n)/(b+n) ... when a > b

Quick test: 3/2 > 101/100 .. 

Cases reverse for -ve numbers.. [Assumung that we do -(a+n) and -(b+n) and n being +ve number]

Cases also vary for sqaures...

Actually, it is better to create such cases on the fly by doing quick test for un-usual formats.

We should alyways keep in mind cases for a/b when a, b, n > 0 and a > b or b > a.These are mentioned above in the post.

Time, Speed, Distance :

If the two bodies are moving in opposite direction.

# They will meet first time after covering a total distance of d# For subsequent meetings they will meet after covering a total distance of 2d.

So we can generalize it to be d(2n+1)

Similarly, If the two bodies are moving in same direction.

# They will meet first time after covering a total distance of 2d# For subsequent meetings they will meet after covering a total distance of 2d.

So we can generalize it to be 2dn.

Tournament Funda

There are 16 teams and they are divided into 2 pools of 8 each. Each team in a group plays against one another on a round-robin basis. Draws in the competition are not allowed. The top four teams from each group will qualify for the next round i.e round 2. In case of teams having the same number of wins, the team with better run-rate would be ranked ahead.1. Minimum number of wins required to qualify for the next round _____?2. Minimum number of wins required to guarantee qualification in the next round _____?

Now, i don't know how many of you are aware of the following method. But 1 thing I mention in advance that this should take only 30 seconds to solve1.1 group is consisting of 8 teams. So each team will play 7 match each. Suppose each of the 8 teams were seeded and we consider the case where a higher seeded team will always win.So the number of wins for the 8 teams would be 7,6,5,4,3,2,1,0 with highest seeded team winning all and lowest seeded team losing all.For minimum number of wins we allow 3 teams to win

maximum number of matches. Of the remaining 5 teams just find out the mean of their number of wins.In this case it would be (4+3+2+1+0)/5=2.So 5 teams can end up with 2 wins each and a team with better run rate will qualify with 2 wins.

2.In this case consider the mean of first 5 higher seeded teams(7+6+5+4+3)/5=5So it may be the case that 5 teams can end up having 5 wins each. And hence 1 team will miss the second round birth. So minimum number of wins to guarantee a place would be 6.

Round Robin/League Games:

Suppose A,B,C,D,E,F play each other exactly once. Top 2 Qualify for Finals.

4 Standard Questions:Q1. Find total no. of matches.Q2. Min. no. of matches a team must win in order to have a chance of qualifying.Q3. Max. no. of matches a team can win and still not qualify.Q4. Min. no. of matches a team must win to reach next round undisputed/ guaranteed.

A1. Easiest of the lot.

A2. If top "k" teams are supposed to qualify for the next round then choose top k-1 teams and make them win as many matches as possible. Distribute the remaining points equally/ as equal as possible.

A3. If top "k" teams are supposed to qualify then choose top k+1 teams and make them beat every other team then you'll be left with matches between k+1 teams. Distribute as equally as possible.

A4. A3 + 1

Miscellaneous :  

a,b,c in AP log a,log b,log c in GP vice versa

How to find pythagorean triplets :

Kisi bhi odd number,a, ka square nikalo. Usko aadha kar do. Aadhe se 0.5 upar, aur aadha neeche do values aengi, let these be b and c. Ye do values, aur a pythogorean triplets hain. example lo ye...3 is the first side: 3^2=9.9/2=4.54.5-0.5=44.5+0.5=5So, triplet is {3,4,5}

5: 5^2=25, 25/2=12.5

12.5+0.5=1312.5-0.5=12Triplet is {5,12,13} and so on..

Agar kisi even number of smallest side banana hai, then express the number in the form x*y, where x is a power of 2, and y is an odd number. 

Find the triplet taking y as the smallest side. Eventually multiply all the three sides by x. 

Example: 28 needs to be the smallest side. 28=4*77 ke saath triplets aate hain: 24 aur 25. So, hamare triplets hain {28,96,100}.

HOPE IT HELPS !!!  

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