Concept 1 Perfect Squares

Concept I Perfect Squares

There has been a huge surge in the number of questions about perfect squares, in almost all mocks. The basic trick to any such question is assuming the number as a perfect square of an integer k and then using techniques of completion of square and then the formula of (a^2-b^2) and solving using divisibility theory

Example I Find all natural n such that n(n+16) is a perfect square

step 1 n(n+16)=k^2

step 2 (n^2+2.8.n+8^2)-8^2=k^2

step3 (n+8+k)(n+8-k)=64

see now lhs and rhs both are integers then both of (n+8-k) and (n+8+k) are divisors of 64. But note that we add the two equations we will get 2n+16, so teh sum of two divisors should be even hence both divisors even or both odd

so n+8+k=32,16,8,4,2 and n+8-k=2,4,8,16,32

but see this n is positive hence k is positive, thus n+8+k>n+8-k
so only two options

and solving we get 2n+16=34,20
so n=9,2

Note : The source of this problem is Pomona Wisconsin mathematics talent search exam!

Practice problem!!
Find the sum of all such positive integers m’s such that m^2+25m+19 is a perfect square

Now we will extend the method to other kinds of problems
Basically what we used in the above problem is difference of square method

lets take an example
x^6=y^2+127, find the no of pairs of postive integers (x,y)

first step in this problem is recognising that 127 is a prime
then we move to
(x^3+y)(x^3-y)=127
so clearly 2x^3=128 x=4 and y=63

so one pair (4,63)

Concept 4 Prime Numbers

I was recently reading a blog by one MIT student, wherein he quotes one of his professors, ” Mathematicians are annoyingly Precise” and he further adds, ” think deeply of simple things“. So thats what we will do, be precise and rigorous in our deep thinking of simple things, that is exactly what CAT wants us to do. Lets roll!

Theorem 4.1 Prime numbers are odd, except for 2, and they have exactly two factors, the number and 1 itself.

You would be wondering, why I have started with this, but whole prime number theory is based on this only.

From here, I will formulate something, which I use excessively in problem solving !. But first lets solve an example. This question is taken from My quant problem set III, the link of which can be found,” free material for cat”.

Example 4.1 Let a,b,c,d be distinct prime numbers satisfying :

2a+3b+5c+7d=162

11a+7b+5c+4d=162

Given that abcd=k. Find the number of distinct values of k?

A)  0                   B) 1                  C)  2                   D) 3                 e) 4

How we go about this? We were told in school, that n variables need n equations, but we have n-2 here. A road-block? No, a call to think deeply. Just see how we can reduce variables or increase equations .

We subtract the two equations and get 9a+4b=3d => 4b=3(d-3a)

RHS is divisible by 3, so should be LHS and therefore b=3

put this in the initial equations, and we are sure the max value of a can be =7 ( i leave it to u to figure out how, a hint: all prime numbers are distinct, and we have used 3, we are left with the two smallest as 2 and 5 ).

Back again 3a=d-4=>d=3a+4 gives us (a,d)=(5,19),(11,37)..  but clealry the second set wont work, very large values. We found the set, just by using the constraint, all are distinct primes and 3 has been used.

so we have b=3,a=5 d=19, there is no further need to go as we need the no of values of k which will obviously be 1. But for the sake of completeness we can check c=2

Seems like a marathon, but no its a 3-4 minute problem, once you start doing what I want you to !

Now, if you have understood this concept, you should be able to get the practice problem, which is taken from one of the simcats.

Practice Problem 4.1 A boy spends Rs. 81 in buying some pens and pencils. If a pen costs Rs.7 and a pencil Rs 3, What is the ratio of pens to pencils when the maximum number of pens are purchased such that no extra money is given to the shopkeeper?

A) 3:2    B) 2:1   C) 5:4   D) 7:2   E) none of these1

The next concept which I am going to take up is Prime squares:)

Theorem 4.2 : All prime squares ( p>3) are of the form 6k+1, i.e , p^2=6k+1, for all primes p>3.

Lets try to prove this,  any three numbers  (p-1)p(p+1) will be divisible by 6. but as p is a prime greater than 3, it would neither be divisible by 2 nor 3, hence p^2-1=6k so p^2=6k+1.

Some purists will say, that as p is a prime greater than 3, then, p^2-1=24k+1, yupp I agree, but 24k+1 becomes cumbersome to handle sometimes. The proof is simple again, p is odd so both will be divisible by 2 and one by 4. also one of them by 3. hence p^2-1=24k so p^2=24k+1

But, I have always used 6k+1, may be just used to it. You may pick the one that suites you.

Kindly note, this is a necessary condition not a sufficient one, means all prime square will be of form 6k+1, but all no of 6k+1 cant be prime square

Lets handle our next example based on this.

Example 4.2 : Find the number of primes p, such that p^2+3p-1 is also a prime?

A quick check will tell 2 does not satisfy and 3 does.

now we check for higher primes

p^2+3p-1=6k+1+3p-1=3(2k+p) hence divisible by 3, not  a prime

So, only one prime p=3 . We are done here !

More to follow, do tell us how you like this, press the rating button on the left !

Concept 5 Pigeon- Hole Principle

I am delighted with the response, my concept lessons are being well received. Thank you.

Today, we will be taking something called “Combinatorics”. It is generally the nemesis of many students, especially the ones who do not understand why do we need to arrange something, and that too in some weird way. Well I have sympathy for you, but no matter how chaotic our lives be, we still like to maintain some order, and therein comes the concept of ordering, arranging, partitioning and so on. And all of it come together to make a branch of mathematics called Combinatorics.

It will be injustice to combinatorics, if I write just once lesson, so I will try to write a few more, for the moment, I will pick up one of the darling principles of Combinatorics, known as Pigeon-Hole Principle.

Theorem 5.1if n+1 pigeons fly to n holes, there must be a pigeonhole containing at least two pigeons

Well this theorem, look apparently simple and trivial, but its extremely powerful. Lets take a test of it.

Example 5.1 Let A be any set of nineteen integers chosen from the arithmetic progression 1,4, . . . ,100. Prove that there must be two distinct integers in A whose sum is 104.

Now how do we go about this? remember n and n+1. The hint is to make n+1=19. Something clicked?

see we have 34 numbers of the form 3k+1, from 1 to 100. If we do not want a sum of 104 , we will break them in the sets of 2 integers whose sum is 104

{4,100},{7,97}..{49,55} and {1}, {52}. Clearly we have 16 two element sets and 2 one element set.

So if we make a set of 19 integers, we will have to pick both the integers from atleast one of the two element sets, which will give us a sum of 104.

We are done here.

If you still have doubts, let me explain again, suppose you are four friends ( boys) and there are three girls. And each one of you like a girl out of the three. So at least one of the girls will be liked by two boys.

Lets solved a more involved example, wherein we need not prove a thing, but find a thing. Some people may be feeling cat does not want us to prove but find. Here is how we do that.

Example 5.2 Let there be n balls with Ram. he decides to colour one ball with colour 1, two balls with colour 2 and so on upto, fifty balls with colour 50. At the end of it , all n balls are used, and no ball is coloured twice. Ram then draws balls from the lot at random, without replacement. What is is the minimum number of balls that he must draw in order to gurantee drawing 10 balls of the same color?

What the hell is his problem. Why coloring and then taking out. Stupid chap. Let us help him with the math now.

see if he picks all the balls with colors which are less than 10 it will come upto (1+2+3..+9)=45.

Now for the worst case he will pick 9 balls each from rest of the balls, which is 41*9

so total is 41*9+45=41*10+4=414. ( avoid multiplying, be watchful)

now if he picks one more ball, atleast one of the set will be of 10. so we are done

he needs to draw 414+1=415 balls.

Practice Problem 5.1 A circular table has exactly 60 chairs around it. There are N people seated at this table in such a way that the next person to be seated must sit next to someone. What is the smallest value of N?

Practice Problem 5.2 We call a set “Sum-free” if no two elements of the set add upto a third element of the set. What is the maximum size of the “sum-free” subset of {1,2,3…2n-1}?

Thats all in this lesson, this is more than enough, if you need more, look into the mock papers, you will find something.

I would like to thank Mr. David Santos, as I have used his book on number system extensively to make this lesson, but I have tried to add my own flavor to it.

Concept 6 Data Sufficiency

Concept 6: DATA SUFFICIENCY

“The ultimate goal of mathematics is to eliminate any need for intelligent thought.”-A. N. Whitehead

Well, for starters try to think what’s the meaning of the above quote, it has a nice and beautiful meaning. To make things so simple that an average mind is able to see it. So lets put this into practice, with another concept lesson.

Today, we will discuss DATA SUFFICIENCY, one of the very scoring problems in cat and other mba entrance exams. The good thing about DS is we get DS both in quant and DI, and it makes for 6-8 problems in almost every paper. As there are no theorems in DS, we will take things to note

Things To Note:

1) DS problems, we need to answer if it is sufficient information to answer, means, there should be one conclusive answer.  We do not need to find the answer, just if it can be found or not?

2) Some questions ask , is this true? So if we can find that the information available is enough to prove that it is not, we are still able to answer the question, that it is not true. Hence we are able to answer the question.

3) Check for all possibilties, that is using one statement, using second, then only combine the two.

Lets take up an example.

Instructions For DS Questions

Each question is followed by two statements, X and Y. Answer each question using the following instructions:
Mark (A) If the question can be answered by using the statement X alone but not by using the statement Y alone.
Mark (B) If the question can be answered by using the statement Y alone but not by using the statement X alone.
Mark (C) If the question can be answered by using either of the statements alone.
Mark (D) If the question can be answered by using both the statements together but not by either of the statements alone.
Mark (E) If the question cannot be answered on the basis of the two statements.

Example 6.1

In a triangle ABC, D is a point on the side BC and M, N are length of perpendicular dropped on line AD from the vertices B and C respectively. Is M > N?

(X) AB > AC

(Y) BD < DC

The problem is simple, but again see this, this question asks is M>N? The answer may be yes or no, but what we are concerned with is our ability to answer it and not the actual answer.

AB>AC gives us no idea, just imagine a few figures and you will know this.

Look at the other statement it says BD<DC, if BP and CQ are the perpendiculars then, BDP and CDQ are similar

so BD/DC=BP/CQ so we know the ratio and thus are able to answer and see this teh answer came NO.

so, we dont really want the answer, but the ability to give a unique answer.

So, answer is B)

This question was taken from QQAD practice test 4, here is the official solution.

Let D’ be the midpoint of BC and let X’ and Y’ be the feet of the perpendicular from B and C to AD’ respectively => X’ = Y’. As D’ shifts to right or left, we can know which of M or N is bigger. Thus, (Y) answers

(X) doesn’t tell us anything

=> choice (B) is the right answer

Practice Problem 6.1

Is x=y?

X: (x+y)(1/x+1/y)=4?

Y: (x-50)^2=(y-50)^2

Practice Problem 6.2

What is the value of m and n?

X: n is an even number, m is an odd number, m>n

Y: mn=30

Note: Both the practice problems are old cat problems, enjoy!

Example 6.2

The distance of point P=(x,y,z) from the origin is sqrt(62) units, then find the coordinates of point P.

X: x+y+z=12

Y: x,y, and z are positive integers.

From original questions x^+y^2+z^2=62

From statement X

(x+y+z)^2=144

2(xy+yz+zx)=62

can’t say for sure

From statement Y:

positive integers

but we can easily find two pairs (1,5,6), (2,3,7). can’t find a unique solution

Option E

This question is taken from SIMCAT 9

Practice Problem 6.3

Rahim plans to draw a square JKLM with a point O on the side JK, but is not successful. Why is Rahim unable to draw the square?

X: The length of  OM is twice that of OL

Y: The length of OM is 4 cm

This problem is taken for CAT 07 paper.

Example 6.3Let p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?

(X) p(i) – p(j) is not a composite number

(Y) p(2i) + p(2j) is a composite number

One of my favourite problems !

p(i) – p(j) is not a composite number
=>i^2-j^2 is a prime as i ,j are positive integers and i >j, ( i^2-j^2) can’t be 1
=>(i+j)(i-j)= prime
so i-j=1
let p be the prime so i=(p+1)/2
j=(p-1)/2
clearly p is not 2 hence all p is odd
p(i) + p(j)=80 +(p^2+1)/2

now p^2=6k+1 ( Refer my lession on prime numbers for this )
therefore
p(i) + p(j)=80 +(p^2+1)/2
becomes
80+(6k+2)/2=81+3k=3(27+k)
so not a prime => can be answered by using X

now, p(2i) + p(2j) is a composite number
4(i^2+j^2+20) is composite

now i and j can be anything
can’t make any conclusions

=> choice (A) is the right answer

This is QQAD pratice test problem!

Thats all, do post your queries and suggestions !

Good Luck!

IIM A Interview Experience

Date 6 March

Cat percentile 99.97

We were 8 students in our panel and there were 4 such panels.

Essay

We were given 10 mins to write an essay in the area provided in the interview form. We were forewarned to stop rightaway when we are asked to stop. Essay topic was how other games were not eclipsed by Cricket and something like that.

Wrote a decent essay.. examples from olympic winners not getting padma awards etc..

All in all a good 10 mins to start with.

Interview

I was last in the panel and the panelists told us that they will take 15 mins each.

They took a break just after the 7th guy left and it started to scare me..

P1: Old prof to my left

P2: Old prof to my right. ( Mr Rajendra Patel) http://iimahd.ernet.in/faculty/facultydetails.php?id=214&farea=refarea&section=area

M: me

P2 called me in.

Pleasantries and was asked to sit.

P2: Please show me this certificate and explain what it is.

I stood up went close to P2 and showed the Gold medal certi and explained it was given to me for getting the highest marks in CBSE Board amongst all thirteen schools run by Bokaro Steel.

P1: Good, Very Good!

P2: Ok XXX , you have done electrical engg from IIT Kanpur.

M: Yes Sir!

P2: There were a few electrical enggs before you, what did they tell you?

M: Sir, they told me you were asking probability questions?

P2: Probability? What probability?

M: Sir, they were telling you gave questions about probability etc

P2: Ok.

P2: How would you explain to a layman why he pays for electricity? Electrons come to his home and then he returns it back, he doesn’t keep any why should he pay for it?

M: Sir, although he is not keeping any electrons, but energy is required to maintain this flow of electrons and the power company thus needs to be compensated.

P2: Speak in layman’s terms

M: Sir, it is like the current won’t flow unless there is potential difference.

P2: Speak in layman’s terms.

M: Sir, we can think it as water flowing, water flows on a slope similarly .

P2: Ok

P2: Ok tell me how will you explain integration to a layman. And dont tell me area under the curve ok.

M: Sir, it is basically a better form of summation when the function is discrete you can use sum, but when it is continuous it can’t be summed, so we break into parts.

P2: Now you are using discrete continuous, I have told you four times speak in layman’s terms.

M: Sir, take a Asymmetric figure, now if we need to find the area of this figure, summation would not help us, but we would need integration to do it.

P2: DO you know transmission losses form the major part of power losses?

M: Yes sir

P2: What is done to prevent these losses?

M: Electricity is transmitted a very high voltage 11 KV or 22 KV or even higher. This is done to reduce i^2r losses.

P2: Any other?

M: We can use better equipment, better transformers

P2: Is the purpose of transformer to cut power losses?

M: no sir, thats not it primary purpose but in course of transmission we need a lot of transformers and if we have efficient ones, we can reduce losses.

P2: WHat is the primary purpose of a transformer?

M: Either to step up or step down voltage depending on the need.

P1 Takes over

P1: Ok XXX, You have done electrical engg from IIT Kanpur, Can you make electricity flow without a wire

M: Sir it won’t be practical but it can be done.

P1: How?

M: Sir, if we apply voltages at sharp points, the air around the points will breakdown into ions and then start conducting. Sound and corona discharge will accompany this phenomena and this will be very inefficient way of transmitting electricity.

P1: How far?

M: very small distances around 1 mm.

P1: Ok, this sound and light thing doesn’t look practical. ANy other way?

M: Sir, basically what I did there was made air an conductor and replaced wire by air. If we have to remove wire, we can use any other conductor ( smile). May be water.

P1: Yes ( smiles)

P1: Ok do you know function of a function, can you differentiate it?( passes me a paper) We use a lot of calculus in Management.

M: Sir, it is basically defined when the independent variable is not a basic. For example y=f(x) and x=g(z) then y=f(g(z))

P1: Now differentiate it?

M: y’=f’(g(z)g’(z)

P1: What is that prime business ? (smiles) write simply

M: dy/dz=d(f(g(z)))/d(g(z))xd(g(z))/dz

P1: why is this a product and not a sum? Can you prove it?

M: yes sir.. I prove it using basic definition of derivative and the chain rule fo limits. he takes the paper and verifies it.

P1: Can you give an example?

M: yes, y=cos(sinx).

P1: Now differentiate it

M: y’=-sin(sinx) cos x

P1: Good.

P1: XXX, Can you tell me the difference between electrical force and gravitational force?

M: Electrical force can be attrcative or repulsive but gravitational force is attractive only.

P1: Ok, Do you know the nucleus, there are only protons, are they like charges?

M: yes sir.

P1: Then why don’t they come out and break the nucleus?

M: Sir, there are three forces inside the nucleus. Electrical force, gravitational force and nuclear force. The gravitational force is very small and hence neglected. But to counter the repulsive electrical force there is a nuclear force which keeps the protons inside the nucleus.

P2 takes over

P2: Ok, Where are you from?

M: Bokaro Steel City

P2:Hmmm, have you heard of the city Mclean? Can you tell me the country it is in?

M: No sir! ( I did not take a guess )

P2: Brussels?

M: Belgium?

P2: DO you know why most indian rivers originate from north and then flow into east or west?

M: Sir, the melting ice of himalayas act as the source of rivers, hence they originate from there. The rivers then flow southwards due to the slope and they move eastwards or westwards due to the terrain they face in their course.

P2: Ok can you tell me the names of all major cities on the banks of river ganga from west to east?

M: Haridwar.

P2: Before Haridwar?

M: Rishikesh.

P2: Ok Continue

M: Rishikesh, Haridwar, kanpur, allahabad, banaras, patna, sultanganj, Kolkata.

P2 looks at P1

P2: Take a toffee

I pick an alpenlibe from the bowl

and say thank you to both of them

P1: All the best.

I am out now..

There was no one left who could tell me the time but it was approx 20-25 mins.

Problem of the day 26.08.09

Five students Implex, Slam, Sanyo, dewan and nbangalorekar are wearing caps of Blue or Green color without knowing the color of his own cap. It is known that the students wearing the Blue cap always speaks the truth while the ones wearing Green always tell lies. If the students make the following statements

Implex: I see 3 blue caps and one Green
Slam: I see 4 Green caps
Sanyo: I see 1 Blue cap and 3 Green
dewan: I see 4 Blue caps

Then, which among the following (Student, Cap Color) combination is correct?

(1) (Implex, Blue) (2) (Slam, Green) (3) (dewan, Blue) (4) at least two of the foregoing (5) none of these

Problem of The day 06.08.09

1. Four digits of the number 29138576 are omitted so that the result is as large as possible. The largest omitted digit is (A) 9 (B) 8 (C) 7 (D) 6 (E) 5

Problem of the Day 30.07.09

Find the sum of the digits of the least natural number N, such that the sum of the cubes of the four smallest distinct divisors of N is 2N?

1)  9                                2) 8                             3) 7                    4) 6                      5) 10

Bonus QUestion 28.07.09

Suppose K be the number of integers n such that (2^n+1)/n^2 is also an integer.
Then K is
a) 0               b) 1               c) 2              d) 3              e) none of these

Problem of the day 28.07.09

if a<b and 12²+4²+5²+3²=a²+b² the find (a+b)?

Problem of the day 27.07.09

Find the number of quadratic polynomials ax² + bx + c such that:

a) a, b, c are distinct.

b) a, b, c ε {1, 2, 3, …2008}

c) x + 1 divides ax² + bx + c
a) 2013018            b) 2013021            c) 2014024             d) 2018040       e) none of these

Bonus Question 26.07.09

The perimeter of a right triangle is 60. The height to the hypotenuse is 12 what is the area?
(A) 75 (B) 144 (C) 150 (D) 300 (E) none of these

Problem Of the Day 26.07.09

If x² + y²= 1 and x, y are real numbers. Let p, q be the largest and smallest possible
value of x + y respectively. Then compute pq
a) 0                    b) 1/2         c) −1/2                          d) 2                           e) −2

Problem of the day 25.07.09

In 1896 lord Coin has decided to play a game. From the January 1 till December 31 every day he chooses among two match boxes an arbitrary one and placed a match from it to another box (if the chosen box was not empty). If the chosen box was empty then he placed a match from
the other box to the chosen one. What is the probability that after the December 31 the both boxes will have an equal number of matches if at the beginning each box had a) n = 400 b) n = 200 c) n = 100 matches?

Bonus Question 16.07.09

Find the number of solutions in distinct positive  integers of x^4+y^4=z^4

A) 0                 B) 1                  C) 2                         D) 3                E) More than 3