Sunday, September 28, 2008

Problem Of The Week 35

How many 3-d igit numbers are such that one of the digits is the average of the other two?

(A)   96   (B) 112   (C) 120   (D) 104   (E) 256

Problem Of The Week 34

Rakesh has the habit of always pouring his tea from the cup into the saucer before drinking it. He fills both the cup and the saucer to only 90% of their capacity (subject to the availability of tea). He also does not drink any tea which is below the 15% mark in the saucer. If he has to pour the tea from the cup into the saucer at least three times before emptying the cup (each time drinking from the saucer till it reaches the minimum level), then what is the maximum possible ratio of the volume of the cup to that of the saucer respectively? (1) 3 : 1 (2) 9 : 4 (3) 5 : 2 (4) 8 : 3  (5) None of these.

Problem Of The Week 33

The sum of base-10 logarithms of divisors of 10^n is 792. what is n?

(A) 11  (B) 12  (C) 10  (D) 13 (E) 14

Problem Of The Week 32

Four Equilateral triangles are formed taking one of their sides as the sides of the square, the third vertices of equilateral triangles being inside the square. The ratio of the area of fig formed by the third vertices of the triangles to that of the square is nearly

Problem Of The Week 31

A parking lot has 16 spaces in a row. Twelve cars arrive, each of which requires one parking space, and their drivers chose spaces at random from among the available spaces. Auntie Em then arrives in her SUV, which requires 2 adjacent spaces. What is the probability that she is able to park?

Saturday, September 27, 2008

Problem Of The Week 30

Let b be an odd square and n be any natural number. Then b=n^3+2n^2. Find b?

Problem Of The Week 29

Let x,y,z be distinct positive integers such that x+y+z=11. Find the maximum value of (xyz+xy+yz+zx)?

Thursday, September 25, 2008

Problem Of The Week 28

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let p be the probability that at least two of the three had been sitting next to each other. If p is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?

Problem Of The Week 27

A golden rectangle is a rectangle in which the ratio of the width to length is the same as that of the length to the sum of the length and width. Which of the following is also true about a golden rectangle?
I. The ratio of the length to width is the same as the ratio of the width to the difference of the length and width.
II. The product of the length and width is equal to the product of the sum of the two sides and the difference of the two sides.
III. The length has to be greater than two times the width.

(1) Only I and II (2) Only II and III
(3) Only I and III (4) All the three statements

(5) None of these

Problem Of The Week 26

A three digit number is such that the sum of the digit in the hundred’s place and the ten’s place is 1 more than the digit in the unit’s place. It is also given that the digit in the ten’s place exceeds the square of the digit in the hundred’s place by 1, and that the square of the digit in the units place diminished by 7 is the same as the sum of the squares of the other two digits. What is the number?

(1) 346                 (2) 256                 (3) 458                 (4) 526       (5) None of These.

Wednesday, September 24, 2008

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.

Problem Of The Week 25

Let X and Y be distinct 3 digit palindromes such that X>Y. How many pairs (X,Y) exist such that X-Y is also a 3 digit palindrome ?

Tuesday, September 23, 2008

Problem Of The Week 24

Let x,y be positive integers such that 19x+90+8y=1998. Let a be the max value of x and b  the max of y.

Find a+b.

Problem Of The week 23

Let f(n) be the number of 1s in all numbers from 1to n. then f(1)=1, f(8 )=1 and f(10)=2 and f(11)=4. Find f(10^100)?

A) 10^99+1 B) 10^100+1 C) 10^101+1 D) 10^102+1 E) None of these.

Problem Of The Week 22

Find the sum of all prime numbers below 100 which divide 3^32-2^32?

A)110          B)  115                   C)127                    D) 132 E) None of these

Problem Of The Week 21

if ||x-2|-1|<6 and ||y-1|-2|<9 where x and y are integers, then find the range of x-2y?


Tipster: When the equations look too complex, option hunting is a good method

Learn By Example I

Today we will discuss a problem, I am not giving it as an exercise as I want to discuss it, rather than discuss a whole new concept.

How many digits cannot be the unit's digit of the product of three 3-digit numbers whose sum is 989

When we read the problem, it looks like what the hell is it asking? But, it is not that difficult a problem if we go by a method

let the three  3 digit numbers be xyz, abc and pqr

then what the question essentially says is z+c+r=9 or 19

now we have to check if d is the last digit of the product of xyz, abc and pqr, then what are the values d cannot take.

it can be easily checked that it works for 0 and 2 so d can be o or 2

for d=1, (z,c,r)=(9,3,3) or(9,9,1) and others

9+9+1=19 satisfies hence d can be 1

d=3, (z,c,r)=(3,1,1),(3,7,3),(9,7,1) but none of which sum to 9 or 19

hence d cant be 3

d=4 (1,1,4) (1,2,2) (7*3*4),(4*4*4),(8*8*1)(9*8*2) and others

d=4 works

d=5 is obvious (5*1*3)

d=6 is (1*1*6) ,(1*2*3) (4*4*1)

which works

hence d=6 works too

d=7  works for 7*1*1

d=8 (4*2*1) (6*8*1) (6*2*4)(2*2*2) (2*3*8)

d=8 does not work

d=9 will work for (9*7*3)

so d=3,8 does not work

Hence 2 digits cant be unit digits !

Saturday, September 20, 2008

Problem Of The Week 20

A classroom has 4 tables, each of which has 4 chairs. There are 16 students in the class, 4 of whom are friends. If the teacher assigns the seats randomly to the students, what is the probability that the 4 friends will be sitting together at a table?

Tipster: To every probability question, there are two parts, one to find the total possible roster and the other, to find the ones which suit our case( or the ones which not). We can't just find one and leave the other.

Problem Of The Week 19

New Problem: A fresh one, I created it this morning, trying to find the most adequate data!

In a triangle ABC, altitude AD=6 is drawn to cut BC at D. From D, altitude DE=3 is drawn to cut AC at E. If it is know that AB =12. Find the ratio of the area of ABC to area of DEC?

A) 4:1 B) 16:1 C) 25:1 D) 5:1 E) Cannot be determined

Friday, September 19, 2008

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 :) !

Thursday, September 18, 2008

Miscellaneous Problem Package I

1) If |2-3/x|<=1/3 and |3-x/y|<=1/6 and ,x,y are positive reals. Which of the follwoing is a possible value of

x^2/(2x-y) ?

A) sqrt(3)-1  B) sqrt(2)  C) sqrt(1/2)   D) sqrt(3)/2   E)  sqrt(3/2)

2)There are two numbers A and B. A can be expressed as a product of 13 and a two-digit
prime number and B can be expressed as a product of 17 and a two-digit prime number. If
the unit’s digit of the product of A and B is 7, then how many distinct products of A and B
are possible?
A . 48              B. 55               C. 80                    D. 110            E. 120


3) 10 beads and two similar diamond pendants are required to form a diamond necklace. If the beads are same, how many different diamond necklaces can be formed?

A)   1    B   5   C)   6  D) 8  E 10

4) Assume the given sum of the series, 7.5 + 15.5 + 10.5 + 13.0 + 13.5 + 10.5....+ x – y + z is
20634, (x, y, z > 0) then what is the value of the expression (2x + 3y + 5z)?


A) 1084     B) 1284     C) 1464   D) 1684  E) None of These.

5) Let P be the product of all natural numbers between 45 and 293 that have an odd number of
factors. Find the highest power of 12 in P.


A)  6   B)  8  C) 10   D) 12  E) none of these

6)Assume r, s and t be three distinct integers between 0 and 10. if 1/r+1/s-3/t=2/(5r)
then find the
number of distinct values of (r + s – t).


A) 1  B)  2  C)   4   D) 3  E) none of these

7 How many integers less than 300 are relatively prime to either 10 or 18?
A)  140     B) 141    C) 142                   D) 139              E) 138


Source: OLD mocks( 2007 and before)

Problem Of The Week 18

Let p,q be prime numbers and n be a natural numbers. Find the number of ordered triplets (p,q,n) such that

1/p+1/q+1/(pq)=1/n




A) 0       B)   1   C)  2  D)  4   E) none of These

Tipster: All prime( >3) squares are of the form  6k+1 where k is a positive integer. Note this is a necessary condition not a sufficent one

Wednesday, September 17, 2008

Problem Of The Week 17

Two points are selected randomly on the surface of a sphere of radius R. What is the expected distance between them, along the surface of the sphere?

Problem is simple if you think logically, else you will have issues. This question is taken from teh CAT quant Blog by Suresh, the link of which you can see on the right!

Tipster:  The sum of the two legs of a right angled triangle is equal to sum of diameters of incircle and circumcircle :)

Concept 3 Circle and Triangles ( Part 1)

Geometry as a section wa spretty popular in CAT till 2004, consiting of 1/3rd of the problems and people used to think if they could handle it well they are clear with quant cutoff. And really that used to be the case. Cat has changed the trend reducing geometry every year since and last year in CAT 2007, there was not a single problem from geometry.  But, we can for sure be ready for nice geometry problems, so that if they come, we are up and ready for it.

Geometry has some major theorems. One should be clear about them, the ones on similarity of triangles, congruency of triangles, pythagoras, area and volume formula. Kindly refer to a text book for revising such concepts, I would recommend Quantum Cat By Arihant Publishers. Lets roll then !

The major theorems which we always need are :

Theorem 3.1 Pythagoras Theorem : a^2+b^2=c^2 where a,b,c are sides of a right angled triangle.

Clearly, C is the largest side, we call it hypotenuse.

The triplets of real numbers (a,b,c) which satisfy the above theorem is called pythagorean triplets. They are of real interest in all kinds of work.

Example 3.1 The length of one of the legs of a right triangle exceeds the length of the other leg by 10 cm but is smaller than that of the hypotenuse by 10 cm. Find the hypotenuse.

The obvious solution is  (a-10)^2+a^2=(a+10)^2 ( I have jumped a step)

solving we have a^2-20a=20a =>a=40 ( a can't be zero, its side of a triangle)

hypo is a+10=50

P.s : we have avoided the cumbersome assumption of sides as a,a+10 and a+20

Tipster clue: See this , the smallest integer pythagorean triplet is (3,4,5) so all numbers of the form (3k,4k,5k) will be pythagorean!

Practice Problem 3.1 FInd the sum of the lengths of the sides of a right angled triangle if the Circumradius=15 and inradius=6

Theorem 3.2  Sin law

a/Sin A=b/SinB =c/SIn C=2R   where a,b,c are sides opposite <A , <B and <C respectively and R is circumradius of Triangle ABC.

Very useful theorem, though we have entered the domain of trigonometry, but trigonometry, plane geometry and coordinate geometry are very important for each other to co exist.

Theorem 3.3 Cosine law

a^2=b^2+c^2-2bcCos A ( the notations remain the same as Theorem 3.2). The theorem can be similarly used for other angles too.

Practice Problem 3.2 Find the angle between the diagonal of a rectangle with perimeter 2p and area (3/16)p^2

Example 3.2 Find the length of the base of an isosceles triangle with area S and vertical angle A.

How do we start with this, we can offcourse going to need some basic geometry knowledge. let me tell you all of it. First the vertical angle of an isosceles triangle is the angle between the two equal sides( unless otherwise mentioned). The Perpendicular dropped on the unequal side from the opposite vertex, bisects the vertical angle as well as bisects the side. It means if we have a triangle ABC with AC=AB and AD perpendicular to BC then BD=BC and <BAD=<CAD.

The last thing we need is that area of a triangle is (1/2)bcsinA or (1/2)b^2Sin A for an isosceles triangle as b=c

now given (1/2) b^2sin A=S.......(1)

Now as AD bisects the vertical angle and then use BD=bsin(A/2)

hence BC=2BD=2bSin(A/2)

we can put the value of b from (1) and we are done !

Practice Problem 3.3  Find the largest angle of a triangle in which the altitude and the median drawn from the same vertex divide the angle at the vertex into three equal parts

Lets do a more involved example. This came in IMS SimCat 9. Nice and easy problem, but it might scare you for a moment if you look at the figure they drew. So I am not giving it :)

Example 3.3 In Triangle ABC, AD,BE and CF are the medians which intersect at G. ABCH is trapezium with AH=5units , and BC=10units  and Area( Tr BHC)=35 Sq units. Find the ratio of Area( BDFG): Area( ABCH). ( note we have  H and C on same side of B :) )

Here we again need to know this. The three medians divide the triangle into three triangle of equal area . Also they divide it into three quadrilaterals of equal area. So Area( BDFG)= (1/3)Area(ABC)

Next comes, the traingles drawn on the same base and between same parallel lines have equal area. Hence Area(ABC)=Area(BHC)=35 as we know the base BC, we know the altitude D= 2Area/base=70/10=7

Area of trapezium =(1/2)altitude( sum of paralle sides)= (1/2)7(10+5)

so our ratio is (35/3)/(15.7/2)=2/9

We are done :)

Monday, September 15, 2008

Problems Package ( Number Theory I)

1) If set A ={ 26, 29, 34, ...., n^2+25,...}
set B = {13501, 13504, ...., n^2+13500,....}
how many common elements do the two sets have?


2) What is the smallest positive integer with 6 positive odd integer divisors and 12 positive even integer divisors?

3) 4. Compute the number of squares between 4^9 and 9^4.

4) Given any positive integer n, Fidn the sum of all possible last digits of n^5-n?

Sunday, September 14, 2008

MockaMania : Mocks on 14th Sept

Ims Simcat 9( Some good Problems from Quant)

The paper had cat2007 pattern, only difference was it was +2, and -0.5

1) If a^k has k^4 divisors, where k is a natural number, then which of the following is true?

I a=k=1   II a^k>=210   III a^k>=2^k; k>=2

A) Only I     B)   only II    C)  Only III    D) I or II E) I or III

2) f(n)=2g(n)+f(-n); for all non-zero integers n

g(n)= n*g(n-1) for all n>0 and g(0)=1

then Find g(-10)+g(-9)+....+g(0)+...g(9)+ g(10)

A) 10!+1    B)2*10!+1   C) 2*10!    D) 1 E) None of these

3) Distinct two digit numbers are written one after the other to form a six-digit number. How  many six-digit numbers thus formed have four consecutive 1s in them?

A) 90    B) 64    C)  65    D) 56  E )72

4) x=3m-1 and y=5n-1 where m,n,x,y are natural numbers less than 16. Find the number of pairs (m,n) satisfying the equations x^2=2y^2-7

A) 0   B) 1  C) 2   D)  3  E ) 4

5) FInd the number of real solutions of the system of equations

y=|x-1|+|x-2| and y+1=x(3-x)


A) 0   B)   1 C) 2   D ) 3   E) 4

6) A natural number ( greater than one) is called squareful  number if its prime factorisation contains at least one square. How many squareful numbers below 101 are there

A) 63  B) 61  C) 39  D) 67 E) 41

7) How many pair of consecutive natural numbers less than 51 are squareful numbers as defined above

A) 2   B)  3 C)  4   D)  5 E ) 6

8) A binary number is called tri-one if it has exactly three 1s. If all tri-ones are arranged in ascending order, what is the rank of the least 8 digit tri-one number?

A) 35 B) 32 C) 34 D) 40 E) 36

9) How many tri-ones as defined above , less than  110 in decimal, when converted to decimal is divsible by 5 in base 10?

A) 5  B)  6 C) 7   D) 8 E )10

updated!

Problem Of The Week 16

Points M and N are taken on the hypotenuse of a right triangle ABC so that BC=BM and AC=AN.. Find <MCN

A relatively easy problem, still try to do it. You will benefit from it

Saturday, September 13, 2008

Problem Of The Week 15

a, b, c, and d are the solutions of x^4+7x^2-3x+2=0. Find a polynomial with integer coefficients whose roots are (a+b+c)/d^2, (a+b+d)/c^2, (a+c+d)/b^2, and (b+c+d)/a^2. Find the sum of roots of this equation

A very simple problem, one should take about 1-1.5 mins to solve this. If you take more, revise equations please!!

Friday, September 12, 2008

Problem Of The Week 14

Let X=[x] +{x} where [x] and {x} are respectively the integer and fractional parts of a real number x. Find the number of real number which exist such that they satisfy the following conditions:

a) 0<=x<=100

b) {x^2}={x}




A) 9000                  B) 9900                     C) 9990                    D) 9901                 E) 9991

Problem Of The Week 13

Suppose we have 3 real numbers x,y and z such that x^3=3y-2, y^3=3z-2 and z^3=3x-2 . Find the number of triplets of (x,y,z) which exist?




A) 0     B) 1                         C)  2          D) 3                E) None of these

Problem Of The Week 12

Let us have a series of numbers 16,1156,111556, .... . The series is formed by k ones, (k-1) fives and one 6 for k>0. Then each number of the series is a

A) Prime

B) Perfect Square

C) Prime Square

D) Atleast one of the foregoing

E) None of the foregoing

Thursday, September 11, 2008

Conundrum 2 : Middle Three Digits

A five-digit perfect square in the form of 5abc6 has a thousands digit a, hundreds digit b, and tens digit c.  If a is less than or equal to b and b is less than or equal to c, what is the sum of a + b + c?

Problem Of The Week 11

Triangle ABC is right angled at C. m<ABC=60 and AB=10. Let P be a point randomly chosen inside triangle ABC, such that BP extended meets CA at D. What is the probability that BD>5root(2)?

Problem Of The Week 10

Find the length of the shortest path from (0,0) to (12,16) in the x-y plane which does not go inside the curve x^2-12x+y^2-16y+75=0

Problem Of The Week 9

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.
2) The marksman must then break the lowest remaining target in the chosen column.


If the rules are followed, in how many different orders can the eight targets be broken?

Solutions to Power Play 1

The Answer key is as

1) C  2)  E  3) B  4) A 5) E  6) D 7) B 8) E 9) C  10) E

Question 1 is easy just use a-1/a= 1 and 1/a-a=1 will give two roots sum them

Question 2  10^j-1o^i=10^i(10^(j-1)-1)

1001=7.11.13=10^3+1

so clearly 10^3+1 divides 10^6-1 and therefore 10^6k-1

hence j-i=6k, k is a natural number

applying other constraints we get option e

Question 3) 1+2+3..30=30.31/2=31.15=465

[465/2]=232 now suppose a subset A of S doe not have sum more than 232 then A' must have sum more than 232 hence 1/2 of the subsets of S will have sum more than 232

so 2^30/2=2^29

question 4 use the concept of reflection we will get min distance sum as 5root(2)

Question 5) Function is not correctly defined as 0 is not a natural number

so option e)

Question 6)

let the radius be r
and the point of tangency be P and Q and triangle be ABC. P lies on AB and Q on BC

let AP = m and BQ = n

m^2 = 15^2-x^2
n^2 = 20^2-x^2
m = 9 and n = 16 x =12 arcPQ = 6pi

question 7) see n numbers product is n and sum is zero

if n is odd then sum can't be zero

similalry check for other cases it will easily come out n =4k

question 8 toughest problem of the test

let g(n)=p(n)-n

then g(17)=-7=g(24)

let a,b be integers such that p(n)=n+3

then a-17 divides g(a)-g(17)=3-(-7)=10

similarly for 24

hence we find a-17 and a-24 both divide 10 this means k=a-17 and k+7 both divide 10

this means k=-5 or-2

a=19 b=22

hence ab=418

question 9 can be easily done

question 10) tricky enough problem

look for a series and solve it you will get E