Mind blowing concepts@quant

MBA CAT 2024
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Going by consensus a new thread on concepts and articles on mathematics by Quant pagals…lets make it a handbook of mathematics by putting best of the cocepts here…we start with this:------> hi guys, mathematics is beautiful…

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Going by consensus a new thread on concepts and articles on mathematics by Quant pagals....lets make it a handbook of mathematics by putting best of the cocepts here.....we start with this:------>



hi guys,

mathematics is beautiful.......most of us are not able to njoy its beauty because we generally are not clear about why we are using a particular approach and that at times restricts us from further investigation.....here is one fundamental we use frequently a^b > b^a subject to the condition a2....actually if we impose the condition 3 {271/(n+1)}^(n+1)
or {n+1}^(n+1)/n^(n+1) > 271/n
or {n+!/n}^(n+1) > 271/n
or {1+1/n}^(n+1) > 271/n
now we have come to an interesting number called e or eular's number....
e is formally defined as the value of f(n)={1+1/n}^n when value of n becomes very very large or infinite....at infinite value of n, solution to this definition gives a value of e = 1/0! + 1/1! + 1/2! + 1/3! + .................upto infinite number of terms....
that gives its value as 2.71......so the value on the left hand side is very close to e since in this case we know that n should be sufficiently large.... so the value of 271/n should be close to the value of e hence maximum product will be obtained when each number is equal to 2.71 that makes a total of hundred numbers...hence maximum product is (2.71)^100......for different values of sums, answers may need more precise calculations....but it is sufficient to know this much....now i faced another question in one of the Coaching institute's tests in which satisfactory solution was not given (there are a lot many questions with unsatisfactory solutions in every coaching institute material)....question states that
a = 100^100; b = 101^99 ; c = 102^98 ,find the order of magnitude...
i used the basic result we are discussing here that
100^101 > 101^100
or (100/101)*100^100 > 101^99
or 100^100 > (100/101)*100^100 > 101^99
and can b solved further using the same rule.....

now that got me into investigation of our main problem....

lets analyse a case lets say that sum of some numbers is 10, what is the maximum product
analysing all the cases with different number of numbers.
if there are 10 numbers then corresponding maximum product is
forn=10,(10/10)^10 =1
for n=9, (10/9)^9 = 2.5811
for n=8, (10/^8 = 5.9604
for n=7, (10/7)^7 = 12.1426
for n=6, (10/6)^6 = 21.4334
for n=5, (10/5)^5 = 32
for n=4, (10/4)^4 = 39.0625
for n=3, (10/3)^3 = 37.0370
for n=2, (10/2)^2 = 25
for n=1, (10/1)^1 =10

wow....what does this investigation reveal, it confirms our thm that maximum product for a particular sum is reached when each number approaches e....
in this case max product is obtained at n=2.5........looking at this investigation ,we can also prove using the previous method (when we proved for 271) that when numbers increase from 1 to a close value to e, the corresponding maximum product increases and afterwards the corr. max product decreases(subject to the condition that sum of numbers is constant...)........

If we want to find out that which one is greater a^b or b^a; 3e ; we know from our investigation (which can also be proved) that if a b^a ...
remember this we proved for values of a,b>=3....in general the rule for real numbers should be that a^b > b^a if ebest way to learn from any article or general proof is by taking an example and xploring as u read further....

i hope it clears some basics...........
rgds
Sumit

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Guys got this file of formulae and stuff related to QA....hope it adds to the thread.....if u already have the file please revise it again.

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How to find the pythagoras triplets for any number ?

(some parts of this note have been extracted from the web)

You probably know that 32 + 42 = 52. Those three whole numbers, known as "Pythagoras Triplets", satisfy the Pythagoras Theorem, a2 + b2 = c2. Did you know there are many more such whole number triplets?

The Pythagoras Theorem, states that, for a right-angled triangle represented by three sides, a, b and c, where a & b form the right angle, and c is the hypotenuse, the equation:

a2 + b2 = c2

relates the three sides, and the inverse is also true. For example, a triangle with sides 3, 4 and 5 is right-angled, since 32+ 42= 52.

How do we find sets of integers that satisfy the equation?

Many formulas abound that allows us to find the triplets. One that I recalled from school days was to...

square an odd number, a, and calculate b=(a2-1)/2, and c=(a2+1)/2.

If a=5, b=(52-1)/2=12, c=(52+1)/2=13, which satisfies 52 + 122 = 132

However, the greatest formula devised by Brahmagupta in the year 628, according to it, it provides ALL the triplets involving a particular number greater than 2, whether it is odd or even. Interestingly, the number of triplets depends on the factors of the square of the number. A prime number will yield only one triplet, so does even numbers not divisible by 4 (4n+2). The following paragraphs describe how the formula works.

We are looking for a triplet of the form:

A2 + B2 = C2

Where A, B and C are integers (whole numbers).

Furthermore, we look for ALL possible combinations of B and C for a given value of A.

Step 1:
We look for all values of M that are factors of A2, and where A-M= a positive even number.
If A is even, the quotient A2/M must also be even to have integral answers.
If A is odd, A2/M must be odd. Since all factors of A2 are odd, therefore the quotient is always odd.

For example, for A=15, possible values of M are: 9 (225/9=25), 5 (225/5=45), 3 (225/3=75, and 1 (225/1=225).

Step 2:
For each value of M, calculate the value of B=(A2/M-M)/2
Thus B=(225/9-9)/2=8, or (225/5-5)/2=20, or (225/3-3)/2=36, or (225/1-1)/2=112

Step 3:
Similarly, calculate C=B+M, or
C=8+9=17, or 20+5=25, or 36+3=39, or 112+1=113

A by-product of the calculations is the radius of the inscribed circle that is completely inside the triangle, and that is tangential to all three sides,

R=(a-m)/2


cheers,

:)
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How to find the pythagoras triplets for any number ?

(some parts of this note have been extracted from the web)
..................

R=(a-m)/2

cheers,

:)


hey anupum,
check out this....
we know that 3 ,4 and 5 are the basic triples....
hence,
1*(3,4,5)
2*(3,4,5) = (6,8,10)
3*(3,4,5) = (9,12,15)
4*(3,4,5) = (12,16,20)
5*(3,4,5) = (15,20,25)
and so on , are all triplets which satisfy pythagorus theorem....
-sagar
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However, the greatest formula devised by Brahmagupta,
..............
:)


Good Anupam,

looking for the mathematics of this formula,
lets start with an example,
suppose that the question is that how many pyth. triplets are possible with 15 as one of the smaller sides.....lets say h is the hypo. and a is the other side...

h^2 = 15^2 + a^2
or 15^2 = h^2 - a^2
or 225 = (h-a)(h+a)

this may contain the answer why the number of triplets depends upon number of factors.....
i can assume those values of (h-a) that are the factors of 225....but how many different values of (h-a) can i assume? starting from the smallest factor 1,3,5,9
but can i assume (h-a) as 15... may be no because a companion of (h-a) i.e. (h+a)
is to be taken care of when (h-a)=15,(h+a)=15 too...that means a=0 which is not possible as a is a side of triangle...nor can i assume any other factor greater than 15 as value of (h-a) because that makes (h-a) > (h+a) that is not possible in case of sitive numbers.....means at these four values of (h-a), i may get some pyth. trips.

lets analyse, (h-a) = 1; (h+a)=225 solving h=113, a=112;
(h-a) = 3; (h+a)=75 solving h=39 , a=36;
(h-a) = 5; (h+a)=45 solving h=25 , a=20;
(h-a) = 9; (h+a)=25 solving h=17 , a=8;

we may deduce that number of natural number trips possible with an odd number x (because we have proved this for an odd number only.simple.) as one of the smaller sides are = (number of factors of x^2 - 1)/2......because the number of values of (h-a) that can be assumed are all the factors of x^2 that are less than x......i.e. next time we look at a question like this:----->

how many right triagles with integral sides has one of the smaller sides as 7007?
we will go for finding the number of factors of (7007). 7007 = 7^2*11*13
or (7007)^2 = 7^4*11^2*13^2 number of factors = 5*3*3.and according to our formula number of such rt triangles = (5*3*3-1)/2 = 44/2 = 22....

what if the side in question is even integer, now that is home work......

this raises more Q's.....we still dint calculate the trips(triples) in which this odd side is hypotenuse...home work again...

getting late, i'll post more maths about the pythtrips.....

regards
Sumit
8

If you're good with long division, here's a quick way to find pretty accurate square roots without the aid of a calculator.

Let's try 24.6.

  1. Make a guess. It can be a very bad guess. It doesn't matter. You can even guess 1. Let's try 5 since 52 is 25, which is pretty close to 24.6.
  2. Divide 24.6 by 5. 24.6 / 5 = 4.92.
  3. Now, comes the trick: Pick a new guess between 5 and 4.92 and divide it into 24.6 again. Let's try 4.95. 24.6 / 4.95 = 4.96. 4.96 is pretty close to 4.9598 which is the actual square root of 24.6.
  4. Repeat steps 2 and 3 to any desired level of accuracy. The further you go, the harder the long division becomes. But the first few cycles yield a pretty close answer.
The reason this works is because n*n = 24.6 and n = 24.6 / n. Therefore, the real square root will always be somewhere between 24.6 / n and n.

I know this is cumbersome.. but sometimes this comes as a handy rule in DI..

cheers,

:)
9

hii there....well smit gave a real gud funda..for odd nos triplet................for even nos i have tried to gte a method..................


suppose one of the smaller side is 30......900=(h-a)(h+a)

now the factors of 900==3^2 5^2 2^2 1

i)now to get integer values of h and a we have to assign even values to h-a and h+a.............

so we eliminate all the odd values....\

3*( 3*5^2*2^2)

3^2(5^2*2^2).

similarly for 5 and 5^2.....................so we get total 4...and if we consider intrchanging ..we get 8....

ii)...now we also get odd multiples by combinig...3^2...adn 5^2........i.e....total

(3*3)-1=8 factors.....................


iii)...go for 1 ..............h-a=1....and h+a=p
h-a=p, h+a=1
therefore elimnate more 2

iv) elimnate h-a=30 nad h+a=30....therefere 1 more


theferore we have to eleimnate total 8+8+2+1 from the toatal no of factors and divide by two to eleimnate possibilities above 20...

(3*3*3-19)/2=4......................


same can be done for 20\

400=5^2 2^4 1


we have to remove 5^2 *1...or 3*2=6 ...and also..h-a=20...and h+a=20//////


hence 7 have to be eleimnated...

(15-7)/2 =4....


therfore ....we have to eleimahte al the fctoirs corresponding to odd integers and add 1....eliminate them .................



i just tried..............wat do ull say.........................

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hii there....well smit gave a real gud funda..for odd nos triplet................for even nos i have tried to gte a method..................

......... just tried..............wat do ull say.........................




i guess there,s some thing wrong with your method

total no. of even factors they are 2,4,6,10,12,18,20

first of all finding the no. of odd factors.....

900=30^2=[3*5*2]^2
even factors will be contributed by power's of 2

to find odd factors we consider only 3,5 and multiples of 3 and 5

here 3^2 will contribute 2 termms i.e 3 and 9

5^2 will contribute 2 termmsi.e 5 and 25

similarly we have 4 other terms considering different combinations of 3 and 5
i.e 3*5,3*5^2,3^2*5 and 3^2*5^2
also 1 will be a factor

hence total no. of odd factors=2+2+4+1=9 and not 19 as u said
also we cannot obtain the no of even factors


heres my method

consider 3^2 *5^2 * 2^2

here 2^2 will contribute 2 factors
consider factors obtained by 2 and 3 only

max power of 3 =2
no.of factors=3......[ 3*2,3*2^2,3^2*2]

consider factors obtained by 2 and 5 only
here max power of 5=1
no.of terms=2 .......[5*2,5*2^2]

no with 2 3 and 5 only possible combination is 3*5*2=30 which we cannot take.

hence total no. of even factors
i am trying to find a common formula but am not able to do so.
hope sumit can help
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i am trying to find a common formula but am not able to do so.
hope sumit can help


guys good try......i'll post the method for even nos as well.....on leave for 2 days...i'll come back on tuesday....have fun.....
12

suppose one of the smaller side is 30......900=(h-a)(h+a)

i)now to get integer values of h and a we have to assign even values to h-a and h+a.............

i just tried..............wat do ull say.........................



u are well on track man....perfectly right (h-a) and (h+a) should both be even....thats the way.....
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A small query for math giants:

How do you derive a formula for getting the sum of all factors of a number? Please give the derivation.

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A small query for math giants:

How do you derive a formula for getting the sum of all factors of a number? Please give the derivation.



Hello Ankur,

c if the no is 15 having factors 1,3,5,15.
now we are havin 15=3^1 n 5^1 ok.
now sum of factors (3^0+3^1)(5^0+5^1) (GP).
ie 24.

similarly for N=a^m*b^n*c^p, a,b,c are prime nos m,n,p are the highest powers present
sum is product of the individual GPs.

1 more eg

24=2^3*3^1
sum of factors = 1(2^4-1)/(2-1) * 1(3^1-1)/(3-1)

15*4=60
I hope this is clear to u.
Regards,
15

how to solve this

what is the coeff. of x^987 in the expansion of
{x+x^2+x^3+x^4+.............x^1200}^234

a good question and the funda that works behind is equally good..

17
Thats a good problem. I think its a part of Multinomial Theorem. I just know the approach.....First condense the above series as a sum of GP i.e.

^234
= x^234(1-x)^(-234)(1-x^1200)
Now there was some direct formula to find out the required coefficient in case of Binomial expansion for Negative Index which i cant recall :whatthat:
Plz let us know! Thanks :D
You are right in approach. I guess the formula is Coeff of x^r in (1+x)^ -n is (-1)^r * n+r-1Cr...I am not sure..but i tried cases..it comes out right..but CAT these kind of questions never asked...as it involves Binomial theorem,,,lot of Eng stuff... What you say?
18
juvenile Says
You are right in approach. I guess the formula is Coeff of x^r in (1+x)^ -n is (-1)^r * n+r-1Cr...I am not sure..but i tried cases..it comes out right..but CAT these kind of questions never asked...as it involves Binomial theorem,,,lot of Eng stuff... What you say?

i agree.cat questions as such that you have to do a "bit of thinking" . i strongly believe solving such questions are of no use.
And note, there won't be any question in CAT with direct formulas. you have to do the thinking. One of the reasons why i have stopped solving above questions.
-Maverick
19

shortcuts
35^2= 1225 (3*4)
45^2= 2025 (4*5)
55^2= 3025 (5*6)

also if no is of form abab then its divisible by 101 and abcabc is divisible by 1001

20
Thats a good problem. I think its a part of Multinomial Theorem. I just know the approach.....First condense the above series as a sum of GP i.e.

^234
= x^234(1-x)^(-234)(1-x^1200)
Now there was some direct formula to find out the required coefficient in case of Binomial expansion for Negative Index which i cant recall :whatthat:
Plz let us know! Thanks :D

GENFUNCTIONOLOGY: -

Concept of generating functions can be applied to solve lot many problems related to permutations and combinations.

Lets understand first what generating function is

GEN. FUNCTION IS THE ALGEBRIC REPRESENTATION OF COMBINATORICS EVENTS.

Suppose the event is tossing a coin. It can be represented as (h + t) where h means getting a head and t means getting a tail. Tossing the coin twice will have a generating function
(h + t)^2 = h^2 + 2ht + t^2 and the generating function generates all the combinatory possibilities when these 2 coins are tossed. Here the exponent of each term represents the event and the coefficient represents the number of ways in which the event can happen.
In this case h^2 means getting head both times and that can happen in 1 way only. 2ht means getting a head and a tail and that can happen in 2 ways.

Another example here (This was also a question in CL MOCK-6)

There are two cubes one of which is having 5 faces red and 1 face blue the other is colored in red and blue such that when the two cubes are rolled then probability of getting same color on top face of both the dice is . How many faces on the other cube are colored red?

I represent the one cube as 5r + b and lets say the other cube is represented by xr + yb and by now u must know that this means x number of red faces and y number of blue faces.
Here x+y = 6
The product (5r+b)(xr+yb) = 5xr^2 + (5y + x) + yb^2 generates all the possibilities when u throw the two dice. Since probability that both the dice show the same face is therefore out of 36 cases, 18 cases will be favorable that means 5x + y (i.e. total cases in which both the dice show the same face) = 18. so far we have two equations and two variables which after solving give x=3 and y=3.

So generating functions can be applied to so many combinatorial problems.
Suppose the question is that in how many ways can u put 5 identical balls in 3 boxes.
We know how to solve this problem but we may not know the genfunctinology way of solving it. Here is that way

Following generating function can represent balls in a box.
(x^0 + x^1 + x^2 +x^3 + x^4 + x^5)

And for 3 boxes (x^0 + x^1 + x^2 + x^3 + x^4 + x^5)^3 {exponent 3 because there are 3 boxes each having same generating function) and coeff. of x^5 in the expansion of it will give total number of ways in which these balls can be put into 3 boxes. Why?

Answer to this why is your task to determine. I am not explaining it here because that's too trivial to be explained.

Now how to find the coefficient? We'll use an artificial trick that u may not be aware of but after using this 5478 times you'll become used to it.

(x^0 + x^1 + x^2 + x^3 + x^4 + x^5 + x^6 + )^3 is our new generating function because coefficient of x^5 in it will have same value as the previous generating function had. Again why? Because the terms that we have added to the previous genfunction (x^6, x^7, .) has no effect on coeff. of x^5 . If more whys arise then please think about that.

Now we are concerned about the coefficient of x and not the value of x. So assume that x
This function can be expanded using binomial theorem as

{1 + (-3)(-x) + (-3)(-4)(-x)^2 / 2! + + (-3)(-4)(-5)(-6)(-7)(-x)^5 /5! + .

so coeff of x^5 is 7!/(2!*5!)

Behold we got the same answer as we would have had we applied the balls and barriers that is the normal way of solving this sum. But the question is that when this problem can be solved using an easier method then why to follow this procrustean approach. The answer is we won't follow this approach to solve that problem.

Rather we'll use the easier approach whenever asked that what is the coeff of x^5 in the expansion of ( 1 + x^1 + x^2 + x^3 + x^4 + x^5 )^3 . We know that answer will be same as number of ways of putting 5 identical balls in 3 boxes

I hope that explains another method of solving the question I asked.
Please try that question and confirm the answer. There is much more about generating functions but time and especially relevance does not allow that to be posted.

Have a nice time.
Regards
SUMIT
21

a question from CAT
u^m + v^m = w^m ( u,v,w,m are natural numbers )
then which of the following is true

(1) m >= min ( u,v,w )
(2) m > max ( u,v,w )
(3) m (4) none of these

approach of a normal person would have been assuming some values and checking which option would was correct. some 50 sec. are required to solve the question that way. And some may get confused with value of m assuming that it could be any number. But one maths afficionado who knows famous fermat's last theoram would have said that m cannot be greater than 2, and the minimum pythagorean triplet ( 3,4,5 ) along with m=2 proves that answer option (3) is correct. Some 15 seconds.
Fermat's last thm. says that the equation a^x + b^x = c^x has no integral solutions for x > 2.This thm has bewildered many a gr8 mathematicians for past 300 years (approx.). Fermat wrote in the margins of some book he was reading " I have a truly remarkable proof that a^x + b^x = c^x has no solution in integers for x>2 but the space does not allow that to be written here. " But after his death the proof was not found in any of his works. Since that time the proof has eluded everyone. Only recently, using the theory of elliptical curves and infinite descent, this thm was proved but that prrof was too complex too understand and the search is still on.

What i want to say here is that a few Qs asked in CAT are based on elementry concepts ( Please dont confuse elementry with easy ). Another such concept being Pigeonhole principle.

Here is a Q that appeared in CAT 2-3 yrs ago.

there are 128 boxes. Number of oranges in a box may be a number from 125 to 144. What is the minimum number of boxes that contain same number of oranges.

(1) 5
(2) 0
(3) 6
(4) 7
(5) 8

I have changed a few options. Post ur solutions. Afterwards i'll post the theory on pigeonhole principle.

22

there are 128 boxes. Number of oranges in a box may be a number from 125 to 144. What is the minimum number of boxes that contain same number of oranges.

(1) 5
(2) 0
(3) 6
(4) 7
(5) 8

I have changed a few options. Post ur solutions. Afterwards i'll post the theory on pigeonhole principle.

125 to 144 oranges implies....20 diff ways...
128 boxes...
128/20 = 6 and 8 remainder...
therefore atleast (6+1) = 7 boxes will have the same number of oranges....
option(4)