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. 1 + 1 \times 2 + 3 \times 4 + 5 \times 6 + 7

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.

9 \times 1 - 2 = 7 \times 3 -4 = 17 \times 5 - 6 = 79 \times 7 - 8 = 545 \times 9 - 10 = 4895

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.

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

Problem 1:

Let a_{1}, a_{2}, \ldots, a_{10} 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

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

\bf{Reference: \hspace{0.1in}Introductory \hspace{0.1in} Real Analysis: \hspace{0.1in} Kolmogorov \hspace{0.1in} and \hspace{0.1in} Fomin; \hspace{0.1in}Dover \hspace{0.1in }Publications}

\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} Sets \hspace{0.1in} The \hspace{0.1in}Power \hspace{0.1in }of \hspace{0.1in }a \hspace{0.1in}Set}

\bf{Section 1}:

\bf{Finite \hspace{0.1in} and \hspace{0.1in} infinite \hspace{0.1in} sets}

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 \it{finite}.\it{Clearly \hspace{0.1in} we \hspace{0.1in}can \hspace{0.1in} be \hspace{0.1in} sure \hspace{0.1in} that \hspace{0.1in} a \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in}finite \hspace{0.1in} without \hspace{0.1in} knowing \hspace{0.1in} the \hspace{0.1in} number \hspace{0.1in} of elements \hspace{0.1in}in \hspace{0.1in}it.}

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, \it{if \hspace{0.1in } we \hspace{0.1in}remove \hspace{0.1in} one \hspace{0.1in} element \hspace{0.1in}from \hspace{0.1in}each \hspace{0.1in}set, \hspace{0.1in}then \hspace{0.1in}remove \hspace{0.1in}two \hspace{0.1in}elements, \hspace{0.1in}three \hspace{0.1in}elements, \hspace{0.1in}and \hspace{0.1in}so \hspace{0.1in}on, \hspace{0.1in}there \hspace{0.1in}will \hspace{0.1in}still \hspace{0.1in}be \hspace{0.1in}elements \hspace{0.1in}left \hspace{0.1in}in \hspace{0.1in}the \hspace{0.1in}set \hspace{0.1in}in \hspace{0.1in}each \hspace{0.1in}stage}. Accordingly, sets of these kind are called \it{infinite} 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 \it{one-\hspace{0.1in}to-\hspace{0.1in}one \hspace{0.1in}correspondence} 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.

\bf{Section 2}:

\bf{Countable \hspace{0.1in} Sets}.

The simplest infinite set is the set \mathscr{Z^{+}} of all positive integers. An infinite set is called \bf{countable} if its elements can be put into one-to-one correspondence with those of \mathscr{Z^{+}}. In other words, a countable set is a set whose elements can be numbered a_{1}, a_{2}, a_{3}, \ldots a_{n}, \ldots. By an \bf{uncountable} set we mean, of course, an infinite set which is not countable.

We now give some examples of countable sets:

\bf{Example 1}:

The set \mathscr{Z} of all integers, positive, negative, or zero is countable. In fact, we can set up the following one-to-one correspondence between \mathscr{Z} and \mathscr{Z^{+}} 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 2n+1, and the negative integer n<0 with the even number 2|n|, that is,

n \leftrightarrow (2n+1), if n \geq 0, and n \in \mathscr{Z}
n \leftrightarrow 2|n|, if n<0, and n \in \mathscr{Z}

\bf{Example 2}:

The set of all positive even numbers is countable, as shown by the obvious correspondence n \leftrightarrow 2n.

\bf{Example 3}:

The set 2,4,8,\ldots 2^{n} is countable as shown by the obvious correspondence n \leftrightarrow 2^{n}.

\bf{Example 4}:    The set latex \mathscr{Q}$ of rational numbers is countable. To see this, we first note that every rational number \alpha can be written as a fraction \frac{p}{q}, with q>0 with a positive denominator. (Of course, p and q are integers). Call the sum |p|+q as the “height” of the rational number \alpha. For example,

\frac{0}{1}=0 is the only rational number of height zero,

\frac{-1}{1}, \frac{1}{1} are the only rational numbers of height 2,

\frac{-2}{1}, \frac{-1}{2}, \frac{1}{2}, \frac{2}{1} 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 \mathscr{Z^{+}} of all positive integers.

\it{Next \hspace{0.1in}we \hspace{0.1in} prove \hspace{0.1in}some \hspace{0.1in}elementary \hspace{0.1in}theorems \hspace{0.1in}involving \hspace{0.1in}countable \hspace{0.1in}sets}

\bf{Theorem1}.

\bf{Every \hspace{0.1in} subset \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}set \hspace{0.1in}is \hspace{0.1in}countable}.

\bf{Proof}

Let set A be countable, with elements a_{1}, a_{2}, a_{3}, \ldots, and let set B be a subset of A. Among the elements a_{1}, a_{2}, a_{3}, \ldots, let a_{n_{1}}, a_{n_{2}}, a_{n_{3}}, \ldots be those in the set B. If the set of numbers n_{1}, n_{2}, n_{3}, \ldots has a largest number, then B is finite. Otherwise, B is countable (consider the one-to-one correspondence i \leftrightarrow a_{n_{i}}). \bf{QED.}

\bf{Theorem2}

\bf{The \hspace{0.1in}union \hspace{0.1in}of \hspace{0.1in}a \hspace{0.1in}finite \hspace{0.1in}or \hspace{0.1in}countable \hspace{0.1in}number \hspace{0.1in}of \hspace{0.1in}countable \hspace{0.1in}sets \hspace{0.1in}A_{1}, A_{2}, A_{3}, \ldots \hspace{0.1in}is \hspace{0.1in}itself \hspace{0.1in}countable.}

\bf{Proof}

We can assume that no two of the sets A_{1}, A_{2}, A_{3}, \ldots have any elements in common, since otherwise we could consider the sets A_{1}, A_{2}-A_{1}, A_{3}-(A_{1}\bigcup A_{2}), \ldots, instead, which are countable by Theorem 1, and have the same union as the original sets. Suppose we write the elements of A_{1}, A_{2}, A_{3}, \ldots in the form of an infinite table

\begin{array}{ccccc} a_{11} & a_{12} & a_{13} & a_{14} &\ldots \\ a_{21} &a_{22} & a_{23} & a_{24} & \ldots \\ a_{31} & a_{32} & a_{33} & a_{34} & \ldots \\ a_{41} & a_{42} & a_{43} & a_{44} & \ldots \\ \ldots & \ldots & \ldots & \ldots & \ldots \end{array}

where the elements of the set A_{1} appear in the first row, the elements of the set A_{2} appear in the second row, and so on. We now count all the elements in the above array “diagonally”; that is, first we choose a_{11}, then a_{12}, then move downwards, diagonally to “left”, picking a_{21}, then move down vertically picking up a_{31}, then move across towards right picking up a_{22}, next pick up a_{13} and so on (a_{14}, a_{23}, a_{32}, a_{41})as per the pattern shown:

\begin{array}{cccccccc}   a_{11} & \rightarrow & a_{12} &\hspace{0.1in} & a_{13} & \rightarrow a_{14} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \nearrow & \hspace{0.01in} & \swarrow & \hspace{0.1in} & \hspace{0.1in}\\     a_{21} & \hspace{0.1in} & a_{22} & \hspace{0.1in} & a_{23} \hspace{0.1in} & a_{24} & \ldots \\     \downarrow & \nearrow & \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\     a_{31} & \hspace{0.1in} & a_{32} & \hspace{0.1in} & a_{33} & \hspace{0.1in} & a_{34} & \ldots \\     \hspace{0.1in} & \swarrow & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in} & \hspace{0.1in}\\    a_{41} & \hspace{0.1in} & a_{42} &\hspace{0.1in} & a_{43} &\hspace{0.1in} &a_{44} &\ldots\\     \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} & \ldots & \hspace{0.1in} \end{array}

It is clear that this procedure associates a unique number to each element in each of the sets A_{1}, A_{2}, \ldots thereby establishing a one-to-one correspondence between the union of the sets A_{1}, A_{2}, \ldots and the set \mathscr{Z^{+}} of all positive integers. \bf{QED.}

\bf{Theorem3}

\bf{Every \hspace{0.1in}infinite \hspace{0.1in}subset \hspace{0.1in}has \hspace{0.1in}a \hspace{0.1in}countable \hspace{0.1in}subset.}

\bf{Proof}

Let M be an infinite set and a_{1} any element of M. Being infinite, M contains an element a_{2} distinct from a_{1}, an element a_{3} distinct from both a_{2} and a_{1}, and so on. Continuing this process, (which can never terminate due to “shortage” of elements, since M is infinite), we get a countable subset A= \{ a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots\} of the set M. \bf{QED.}

\bf{Remark}

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

\bf{Section3}

\bf{Equivalence \hspace{0.1in} of \hspace{0.1in} sets}

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

\bf{Definition}

Two sets M and N are said to be \bf{equivalent} (written M \sim N) 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 \mathscr{Z^{+}} 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.

\bf{Example1}

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.

\bf{Example2}

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 z \leftrightarrow \alpha can be established by using stereographic projection. The origin is the North Pole of the sphere.

\bf{Example3}

The set of all points x in the open unit interval (0,1) is equivalent to the set of all points y on the whole real line. For example, the formula y=\frac{1}{\pi}\arctan{x}+\frac{1}{2} establishes a one-to-one correspondence between these two sets. \bf{QED}.

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 (0,1) 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:

\bf{Theorem4}

\bf{Every \hspace{0.1in} infinite \hspace{0.1in} set \hspace{0.1in}is \hspace{0.1in} equivalent \hspace{0.1in} to \hspace{0.1in}one \hspace{0.1in}of \hspace{0.1in}its \hspace{0.1in}proper \hspace{0.1in}subsets.}

\bf{Proof}

According to Theorem 3, every infinite set M contains a countable subset. Let this subset be A=\{a_{1}, a_{2}, a_{3}, \ldots, a_{n}, \ldots \} and partition A into two countable subsets A_{1}=\{a_{1}, a_{3}, a_{5}, \ldots \} and A_{2}=\{a_{2}, a_{4}, a_{6}, \ldots \}.

Obviously, we can establish a one-to-one correspondence between the countable subsets A and A_{1} (merely let a_{n} \leftrightarrow a_{2n-1}). This correspondence can be extended to a one-to-one correspondence between the sets A \bigcup (M-A)=M and A_{1} \bigcup (M-A)=M-A_{2} by simply assigning x itself to each element x \in M-A. But M-A_{2} is a proper subset of M. \bf{QED}.

More later, to be continued,

Regards,
Nalin Pithwa

A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: (2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

P_{1}+P_{2}=P_{3}+P_{4}

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions P_{1}, P_{2}, P_{3}, P_{4} using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}

P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}

P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}

P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}

You will find that P_{1}=P_{4} and P_{2}=P_{4}

Hence, it is verified that the given identity P_{1}+P_{2}=P_{3}+P_{4}. QED.

Regards,
Nalin Pithwa.

Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff f^{-1}(f(A))=A for all A \subset X. Given that f: X \longrightarrow Y.

Problem 2:

Prove that a function if is onto iff f(f^{-1}(C))=C for all C \subset Y. Given that f: X \longrightarrow Y.

Problem 3:

(a) How many functions are there from a non-empty set S into \phi\?

(b) How many functions are there from \phi into an arbitrary set S?

(c) Show that the notation \{ X_{i} \}_{i \in I} implicitly involves the notion of a function.

Problem 4:

Let f: X \longrightarrow Y be a function, let A \subset X, B \subset X, C \subset Y and D \subset Y. Prove that

i) f(A \bigcap B) \subset f(A) \bigcap f(B)

ii) f^{-1}(f(A)) \supset A

iii) f(f^{-1}(C)) \subset C

Problem 5:

Let I be a non-empty set and for each i \in I, let X_{i} be a set. Prove that

(a) for any set B, we have B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})

(b) if each X_{i} is a subset of a given set S, then (\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'} where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)

(ii) (A-B) \times C=(A \times C)-(B \times C)

🙂 🙂 🙂

Nalin Pithwa

Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if A \bigcup B=A and A \bigcap B=A, then A=B.

Problem 2:

Show that in general (A-B)\bigcup B \neq A.

Problem 3:

Let A = \{ 2,4, \ldots, 2n, \ldots\} and B= \{ 3,6,\ldots, 3n, \ldots\}. Find A \bigcap B and A - B.

Problem 4:

Prove that (a) (A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)

Prove that (b) A \Delta B = (A \bigcup B)-(A \bigcap B)

Problem 5:

Prove that \bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})

Problem 6:

Let A_{n} be the set of all positive integers divisible by n. Find the sets (i) \bigcup_{n=2}^{\infty}A_{n} (ii) \bigcap_{n=2}^{\infty}A_{n}.

Problem 7:

Find (i) \bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}] (ii) \bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})

Problem 8:

Let A_{\alpha} be the set of points lying on the curve y=\frac{1}{x^{\alpha}} where (0<x<\infty). What is \bigcap_{\alpha \geq 1}A_{\alpha}?

Problem 9:

Let y=f(x) = <x> for all real x, where <x> is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval \frac{1}{4} \leq y \leq \frac{3}{4}? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let \mathscr{R} be the set of all ordered pairs on the form (a,a) with a \in M, and let aRb if and only if (a,b) \in \mathscr{R}. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

  • Reflexive and symmetric, but not transitive.
  • Reflexive, but neither symmetric nor transitive.
  • Symmetric, but neither reflexive nor transitive.
  • Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

 

 

A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that \tan {A} and \tan {B} are the roots of the quadratic equation x^{2}+px+q=0, find the value of

\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}

Solution:

Let \alpha=\tan{A} and \beta=\tan{B} be the two roots of the given quadratic equation: x^{2}+px+q=0

By Viete’s relations between roots and coefficients:

\alpha+\beta=\tan{A}+\tan{B}=-p and \alpha \beta = \tan{A}\tan{B}=q but we also know that \tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}

Now, let us call E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)} which in turn is same as

\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)

We have already determined \tan{(A+B)} in terms of p and q above.

Now, again note that \sin^{2}{\theta}+\cos^{2}{\theta}=1 which in turn gives us that \tan^{2}{\theta}+1=\sec^{2}{\theta} so we get:

\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}} so that

\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}

Hence, the given expression E becomes:

(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q), which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.