# Solutions to “next number in sequence”: preRMO, pRMO and RMO

What is the next number in sequence?

A) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

Solution to A:

Ans is 20. The sequence is the position in the letter of the alphabet of the first letter in the numbers 1 to 12, when given in full. e.g. ONE: O=15.

B) 1, 8, 11, 18, 80, ?

Ans is 81. The sequence comprises whole numbers beginning with a vowel.

C) 1, 2, 4, 14, 21, 22, 24, 31, ?

Ans is 32, The sequence comprises whole numbers containing the letter O.

D) 4, 1, 3, 1, 2, 4, 3, ?

Ans. is 2. The sequence is as follows: there is one number between the two I’s, two numbers between the two 2’s, three numbers between the two 3’s and four numbers between the two 4’s.

E) 1, 2, 4, 7, 28, 33, 198, ?

Ans is 205.

F) 17, 8, 16, 23, 28, 38, 49, 62, ?

Answer is 70. Sum of digits in all previous numbers in the sequence.

G) 27, 216, 279, 300, ?

Ans is 307. Difference divided by 3 and added to the last number.

H) 9,7,17,79,545, ?

Answer is 4895. Each number is multiplied by its rank in the sequence, and the next number is subtracted.

I) 2,3,10,12,13, 20,?

Answer is 21. They all begin with the letter T.

J) 34, 58, 56, 60, 42, ?

Answer is 52. The numbers are the totals of the letters in the words ONE, TWO, THREE, FOUR, FIVE, SIX when A=1, B=2, C=3, etc.

Regards,

Nalin Pithwa.

# Age related questions: pRMO, preRMO training: and their solutions

**Age old questions:**

Dave is younger than Fred and older than George.

Alan is younger than Ian and older than Colin.

Ian is younger than George and older than John.

John is younger than Colin and older than Edward.

Fred is younger than Barry and older than Harry.

Harry is older than Dave.

**Who is the youngest?**

Answer: You have to try to create an “ascending” or “a descending” sequence and try to “fill in the gaps” :— answer is Edward.

Regards,

Nalin Pithwa.

# Next number in sequence: PreRMO, pRMO, RMO

What is the next number in the sequence?

a) 15, 20, 20, 6, 6, 19, 19, 5, 14, 20, 5, ?

b) 1,8,11, 18, 80, ?

c) 1,2,4,14,21,22,24,31,?

d) 4,1,3,1,2,4,3,?

e) 1,2,4,7,28,33,198,?

f) 17,8,16,23, 28, 38, 49, 62, ?

g) 27, 216, 279, 300, ?

h) 9,7,17,79,545,?

i) 2,3,10,12,13,20,?

j) 34, 58, 56, 60, 42, ?

Regards,

Nalin Pithwa.

# Miscellaneous Questions: part I: solution to chess problem by my student RI

Some blogs away I had posted several interesting, non-trivial, yet do-able-with-some-effort problems for preRMO and RMO.

A student of mine, RI has submitted the following beautiful solution to the chess problem. I am reproducing the question for convenience of the readers:

Question:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge on vertex. Thus, a square can have 8, 5 or 3 neighbours depending on its position.) Show that all the sixty four entries are in fact equal.

Answer (by RI):

Let us denote the set of all integers on the chess board by S (assume they are distinct). [Now, we can use the Well-ordering principle: every non-empty set of non-negative integers contains a least element. That is, every non-empty set S of non-negative integers contains an element a in S such that for all elements b of S}. So, also let “a” be the least element of set S here. As it is the average of the neighbouring elements, it can’t be less than each of them. But it can’t be greater than all of them also. So, all the elements of S are equal.

QED.

Three cheers for RI ðŸ™‚ ðŸ™‚ ðŸ™‚

Regards,

Nalin Pithwa

# Miscellaneous questions: part II: solutions to tutorial practice for preRMO and RMO

Refer the blog questions a few days before:

**Question 1:**

Let be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

**Answer 1:**

Without loss of generality, we may take …..call this relation (i).

Let, if possible, no three of the given numbers be the lengths of the sides of a triangle. (That is, no three satisfy the triangle inequality. Note that when we say three numbers a, b and c satisfy the triangle inequality —- it means all the following three inequalities have to hold simultaneously: , and ). We will consider triplets and . As these numbers do not form the lengths of the sides of a triangle, the sum of the smallest two numbers should not exceed the largest number, that is, . Hence, we get the following set of inequalities:

gives giving .

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

gives giving

contradicting the basic hypothesis. Hence, there exists three numbers among the given numbers which form the lengths of the sides of a triangle.

**Question 2:**

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide the collection into two parts such that each person has at most 1 enemy in his sub-collection.

**Answer 2:**

Let C denote the collection of given 1234 persons. Let be a partition of C. Let denote the total number of enemy pairs in . Let denote the total number of enemy pairs in .

Let denote the total number of enemy pairs corresponding to the partition of C. Note is an integer greater than or equal to zero. Hence, by Well-Ordering Principle, there exists a partition having the least value of .

Claim: This is “the” required partition.

Proof: If not, without loss of generality, suppose there is a person P in having at least 2 enemies in . Construct a new partition of C as follows: and . Now, . Hence, contradicting the minimality of . QED.

**Problem 3:**

A barrel contains 2n balls, numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw.

What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any is repeated?

**Answer 3:**

The total number of possible outcomes is . To find the total number of favourable outcomes we proceed as follows:

Let a be any odd integer such that and let us count the sequences having a as least element.

(i) There is only one sequence with a repeated thrice.

(ii) There are sequences of the form with . For each such sequence there are three distinct permutations possible. Hence, there are in all sequences with a repeated twice.

iii) When , for values of a satisfying , sequences of the form with are possible and the number of such sequences is . For each such sequence, there are six distinct permutations possible. Hence, there are sequences in this case.

Hence, for odd values of a between 1 and , the total counts of possibilities , , in the above cases are respectively.

.

Hence, the total number A of favourable outcomes is . Hence, the required probability is . QED>

Cheers,

Nalin Pithwa

# Miscellaneous questions: part I : solutions: tutorial practice preRMO and RMO

The following questions were presented in an earlier blog (the questions are reproduced here) along with solutions. Please compare your attempts/partial attempts too are to be compared…that is the way to learn:

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the integers in the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position.) Show that all the sixty four squares are in fact equal.

Solution 1:

Consider the smallest value among the 64 entries on the board. Since it is the average of the surrounding numbers, all those numbers must be equal to this number as it is the smallest. This gives some more squares with the smallest value. Continue in this way till all the squares are covered.

Problem 2:

Let T be the set of all triples of integers such that . For each triple in T, take the product abc. Add all these products corresponding to all triples in T. Prove that the sum is divisible by 7.

Solution 2:

For every triple in T, the triple is in T and these two are distinct as . Pairing off with for each , 7 divides .

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one is all the three. In a certain math examination 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists who are swimmers.

Solution 3:

Let S denote the set of all 25 students in the class, X the set of swimmers in S, Y the set of weight lifters in S, and Z the set of all cyclists. Since students in all get grades B and C, and six students get grades D or E, the number of students in . Now assign one point to each of the 17 cyclists, 13 swimmers and 8 weight lifters. Thus, a total of 38 points would be assigned among the students in . Note that no student can have more than 2 points as no one is all three (swimmer, cyclist and weight lifter). Then, we should have as otherwise 38 points cannot be accounted for. (For example, if there were only 18 students in the maximum number of points that could be assigned to them would be 36.) Therefore, and each student in is in exactly 2 of the sets X, Y and Z. Hence, the number of students getting grade , that is, no student gets A grade. Since there are students who are not weight lifters all these 11 students must be both swimmers and cyclists. (Similarly, there are 2 who are both swimmers and weight lifters and 6 who are both cyclists and weight lifters.)

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Solution 4:

Suppose E is wearing a white cap. Then, D is lying and hence must be wearing a white cap. Since D and E both have white caps, A is lying and hence, he must be wearing white cap. If C is speaking truth, then C must be having a black cap and B must be wearing a black cap as observed by C. But then B must observe a cap on C. Hence, B must be lying. This implies that B is wearing a white cap which is a contradiction to C’s statement.

On the other hand, if C is lying, then C must be wearing a white cap. Thus, A, C, D and E are wearing white caps which makes B’s statement true. But, then B must be wearing a black cap and this makes C statement correct.

Thus, E must be wearing a black cap. This implies that B is lying and hence, must be having a white cap. But then D is lying and hence, must be having a white cap since B and D have white caps. A is not saying the truth. Hence, A must also be wearing a white cap. These together imply that C is truthful. Hence, C must be wearing a black cap. Thus, we have the following: A: white cap; B: white cap; C:black cap; D:white cap; E: black cap.

Problem 5:

Let f be a bijective function from the set to itself. Show that there is a positive integer such that for each . Here denotes the composite function repeated M times.

Solution 5:

Let us recall the following properties of a bijective function:

a) If is a bijective function, then there is a unique bijective function such that the identity function on A. The function g is called the inverse of f and is denoted by . Thus,

b)

c) If f and g are bijections from A to A, then so are and .

d) If f, g, h are bijective functions from A to A and , then .

Apply at left to both sides to obtain .

Coming to the problem at hand, since A has n elements, we see that the there are only finitely many (in fact, n!) bijective functions from A to A as each bijective function f gives a permutation of by taking . Since f is a bijective function from A to A, so is each of the function in the sequence:

All these cannot be distinct, since there are only finitely many bijective functions from A to A. Hence, for some two distinct positive integers m and n, , say, we must have

If , we take , to obtain the result. If , multiply both sides by to get . We take to get the relation with . Note that this means for all . QED.

Problem 6:

Show that there exists a convex hexagon in the plane such that :

a) all its interior angles are equal

b) its sides are 1,2,3,4,5,6 in some order.

Solution 6:

Let ABCDEF be an equiangular hexagon with side lengths 1,2,3,4,5,6 in some order. We may assume without loss of generality that . Let .

Since the sum of all angles of a hexagon is equal to , it follows that each interior angle must be equal to . Let us take A as the origin, the positive x-axis along AB and the perpendicular at A to AB as the y-axis. We use the vector method: if the vector is denoted by we then have:

This is because these vectors are inclined to the positive x axis at angles 0, 60 degrees, 120 degrees, 180 degrees, 240 degrees, 300 degrees respectively.

Since the sum of all these six vectors is , it implies that

and

That is, ….call this I

and ….call this II

Since , in view of (II), we have

….(i)

…(ii)

…(iii)

The possibility that in (i), for instance, need not be considered separately, because we can reflect the figure about and interchange these two sets.

Case (i):

Here . Since , this is not possible.

Case (ii):

Here . This is satisfied by

Case (iii):

Here

Case (iv):

This is satisfied by .

Hence, we have (essentially) two different solutions: and . It may be verified that I and II are both satisfied by these sets of values.

Aliter: Embed the hexagon in an appropriate equilateral triangle, whose sides consist of some sides of the hexagon.

*Solutions to the remaining problems from that blog will have to be tried by the student.Â *

Cheers,

Nalin Pithwa.

# Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Regards,

Nalin Pithwa

# Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that . For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.

B: I see four white caps.

C: I see one black and three white caps.

D: I see four black caps.

Problem 5:

Let f be a bijective (one-one and onto) function from the set to itself. Show that there is a positive integer such that for each . Note that denotes the composite function repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:

a) all its interior angles are equal

b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Regards,

Nalin Pithwa

# Towards Baby Analysis: Part I: INMO, IMO and CMI Entrance

:

The set of all vertices of a given polyhedron, the set of all prime numbers less than a given number, and the set of all residents of NYC (at a given time) have a certain property in common, namely, each set has a definite number of elements which can be found in principle, if not in practice. Accordingly, these sets are all said to be .

On the other hand, the set of all positive integers, the set of all points on the line, the set of all circles in the plane, and the set of all polynomials with rational coefficients have a different property in common, namely, . Accordingly, sets of these kind are called sets.

Given two finite sets, we can always decide whether or not they have the same number of elements, and if not, we can always determine which set has more elements than the other. It is natural to ask whether the same is true of infinite sets. In other words, does it make sense to ask, for example, whether there are more circles in the plane than rational points on the line, or more functions defined in the interval [0,1] than lines in space? As will soon be apparent, questions of this kind can indeed be answered.

To compare two finite sets A and B, we can count the number of elements in each set and then compare the two numbers, but alternatively, we can try to establish a between the elements of set A and set B, that is, a correspondence such that each element in A corresponds to one and only element in B, and vice-versa. It is clear that a one-to-one correspondence between two finite sets can be set up if and only if the two sets have the same number of elements. For example, to ascertain if or not the number of students in an assembly is the same as the number of seats in the auditorium, there is no need to count the number of students and the number of seats. We need merely observe whether or not there are empty seats or students with no place to sit down. If the students can all be seated with no empty seats left, that is, if there is a one-to-one correspondence between the set of students and the set of seats, then these two sets obviously have the same number of elements. The important point here is that the first method(counting elements) works only for finite sets, while the second method(setting up a one-to-one correspondence) works for infinite sets as well as for finite sets.

:

.

The simplest infinite set is the set of all positive integers. An infinite set is called if its elements can be put into one-to-one correspondence with those of . In other words, a countable set is a set whose elements can be numbered . By an set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

:

The set of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between and of all positive integers: (0,1), (-1,2), (1,3), (-2,4), (2,5), and so on. More explicitly, we associate the non-negative integer $n \geq 0$ with the odd number , and the negative integer with the even number , that is,

, if , and

, if , and

:

The set of all positive even numbers is countable, as shown by the obvious correspondence .

:

The set 2,4,8, is countable as shown by the obvious correspondence .

latex \mathscr{Q}$ of rational numbers is countable. To see this, we first note that every rational number can be written as a fraction , with with a positive denominator. (Of course, p and q are integers). Call the sum as the “height” of the rational number . For example,

is the only rational number of height zero,

, are the only rational numbers of height 2,

, , , are the only rational numbers of height 3, and so on.

We can now arrange all rational numbers in order of increasing “height” (with the numerators increasing in each set of rational numbers of the same height). In other words, we first count the rational numbers of height 1, then those of height 2 (suitably arranged), then those of height 3(suitably arranged), and so on. In this way, we assign every rational number a unique positive integer, that is, we set up a one-to-one correspondence between the set Q of all rational numbers and the set of all positive integers.

.

.

Let set A be countable, with elements , and let set B be a subset of A. Among the elements , let be those in the set B. If the set of numbers has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence ).

We can assume that no two of the sets have any elements in common, since otherwise we could consider the sets , , , , instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of in the form of an infinite table

where the elements of the set appear in the first row, the elements of the set appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose , then , then move downwards, diagonally to “left”, picking , then move down vertically picking up , then move across towards right picking up , next pick up and so on ()as per the pattern shown:

It is clear that this procedure associates a unique number to each element in each of the sets thereby establishing a one-to-one correspondence between the union of the sets and the set of all positive integers.

Let M be an infinite set and any element of M. Being infinite, M contains an element distinct from , an element distinct from both and , and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset of the set .

Theorem 3 shows that countable sets are the “smallest” infinite sets. The question of whether there exist uncountable (infinite) sets will be considered below.

We arrived at the notion of a countable set M by considering one-to-one correspondences between set M and the set of all positive integers. More generally, we can consider one-to-one correspondences between any two sets M and N.

Two sets M and N are said to be (written ) if there is a one-to-one correspondence between the elements of M and the elements of N.

The concept of equivalence is applicable both to finite and infinite sets. Two finite sets are equivalent if and only if they have the same number of elements. We can now define a countable set as a set equivalent to the set of all positive integers. It is clear that two sets are equivalent to a third set are equivalent to each other, and in particular that any two countable sets are equivalent.

The sets of points in any two closed intervals $[a,b]$ and $[c,d]$ are equivalent; you can “see’ a one-to-one correspondence by drawing the following diagram: Step 1: draw cd as a base of a triangle. Let the third vertex of the triangle be O. Draw a line segment “ab” above the base of the triangle; where “a” lies on one side of the triangle and “b” lies on the third side of the third triangle. Note that two points p and q correspond to each other if and only if they lie on the same ray emanating from the point O in which the extensions of the line segments ac and bd intersect.

The set of all points z in the complex plane is equivalent to the set of all points z on a sphere. In fact, a one-to-one correspondence can be established by using stereographic projection. The origin is the North Pole of the sphere.

The set of all points x in the open unit interval is equivalent to the set of all points y on the whole real line. For example, the formula establishes a one-to-one correspondence between these two sets. .

The last example and the examples in Section 2 show that an infinite set is sometimes equivalent to one of its proper subsets. For example, there are “as many” positive integers as integers of arbitrary sign, there are “as many” points in the interval as on the whole real line, and so on. This fact is characteristic of all infinite sets (and can be used to define such sets) as shown by:

According to Theorem 3, every infinite set M contains a countable subset. Let this subset be and partition A into two countable subsets and .

Obviously, we can establish a one-to-one correspondence between the countable subsets A and (merely let ). This correspondence can be extended to a one-to-one correspondence between the sets and by simply assigning x itself to each element . But is a proper subset of M. .

More later, to be continued,

Regards,

Nalin Pithwa