Tutorial problems for RMO 2019 : combinatorics continued

1) In how many ways can 5 men and 5 women be seated in a round table if no two women may be seated side by side?

2) Six generals propose locking a safe containing top secret with a number of different locks. Each general will be given keys to certain of these locks. How many locks are required and how many keys must each general have so that, unless at least four generals are present, the safe cannot be opened?

3) How many integers between 1000 and 9999 inclusive have distinct digits? Of these, how many are even numbers? How many consist entirely of odd digits?

4) In how many ways can 9 distinct objects be placed in 5 distinct boxes in such a way that 3 of these boxes would be occupied and 2 would be empty?

5) In how many permutations of the word AUROBIND do the vowels appear in the alphabetical order?

6) There is an unlimited supply of weights of integral numbers of grams. Using n or fewer weights, find the number of ways in which a weight of m grams can be obtained. Prove that there is a bijection of the set of all such ways on the set of increasing words of length (n-1) or (m+1) ordered letters.

7) How many distinct solutions are there of x+y+z+w=10 (a) in positive integers and (b) in non-negative integers?

8) A train with n passengers aboard makes m stops. In how many ways can the passengers distribute themselves among these m stops as alighting passengers? if we are concerned only with the number of alighting passengers at each stop, how would the answer be modified?

9) There are 16 books on a bookshelf. In how many ways can 6 of these books be selected if a selection must not include two neighbouring books?

10) Show that there are {(n=5)} \choose 5 distinct throws of a throw with n non-distinct dice.

11) Given n indistinguishable objects and n additional distinct objects —- also distinct from the earlier n objects — in how many ways can we choose n out of the 2n objects?

12) Establish the following relations:
12a) B_{n+1}=\sum_{k=0}^{n}(B_{k}){n \choose k}
12b) \sum_{k}{p \choose k}{q \choose {n-k}}={{p+q} \choose n}
12c) S_{n+1}^{m} = \sum_{k=0}^{n}{n \choose k}S_{k}^{m-1}
12d) n^{p}=\sum_{k=0}^{n}{n \choose k}k! (S_{p}^{k})

13) Prove the following identity for all real numbers x:
x^{n}= \sum_{k=1}^{n}S_{n}^{k}[x]_{k}

14) Express x^{4} in terms of {x \choose 4}, {x \choose 3}, …by using the S_{n}^{k}‘s. Express {x \choose 4} in terms of x^{4}, x^{3}, …by using the s_{n}^{k}‘s.

15) A circular loop is divided into p parts, p prime. In how many ways can we paint the loop with n colours if we do not distinguish between patterns which differ only by a rotation of the loop? Deduce Fermat’s Little theorem: n^{p}-n is divisible by p if p is prime.

16) In problem 15, prove that n^{p}-n is also divisible by 2p if p \neq 2. Where is the hypothesis that p is prime used in Problem 15 or in this problem?

17) How many equivalence relations are possible on an n-set?

18) The complete homogeneous symmetric function of n variables \alpha_{1}, \alpha_{2}, \ldots, \alpha_{n} of degree r is defined as h_{r}(\alpha_{1},\alpha_{2}, \alpha_{3}, \ldots, \alpha_{n})=\sum \alpha_{1}^{i_{1}}\alpha_{2}^{i_{2}}\ldots \alpha_{n}^{i_{n}} the summation being taken over all ordered partitions of r, where the parts are also allowed to be zero. How many terms are there in h_{r}?

Test yourself ! Improve your mettle in math !
Nalin Pithwa.

Pre RMO or PRMO problem set in elementary combinatorics

1) How many maps are there from an n-set to an m-set? How many of these are onto? How many are one-one? Under what conditions?

2) Consider the letters of the word DELHI. Let us form new words, whether or not meaningful, using these letters. The *length* of a word is the number of letters in it, e.g., the length of “Delhi” is 5; the length of “Hill” is 4. Answer the following questions when (a) repetition of letters is not allowed and (b) repetition of the letters is allowed:
(i) How many words can be formed of length 1,2,3,4,…?
(ii) How many words in (i) will consist of all the letters?
(iii) How many words of the words in (i) will consist of 1,2,3,4, …specified letters?
(iv) How many of the words in (i) will consist of only 1,2,3,4,…letters?
(v) How many of the words in (i) will be in the alphabetical order of the letters?

3) Repeat Problem 2 with the word MISSISSIPPI.

4) Suppose there are 5 distinct boxes and we want to sort out 1,2,3,…,n objects into these boxes.
4i) In how many ways can this be done?
4ii) In how many of these situations would no box be empty?
4iii) In how many of the above would only 1,2,3,4 … specified boxes be occupied?
4iv) In how many would only 1,2,3,4…boxes be occupied?
4iv) If the objects are indistinguishable from one another, how would the answers to (i) to (iv) change, it at all?

If there is an added restriction that each box can hold only one object and no more, what will be the answers to (i) to (v)?

5) Repeat Problem 4 with 9 boxes.

6) Repeat Problem 4 with 5 non-distinct (=indistinguishable identical) boxes.

7) Repeat Problem 6 with 9 boxes.

8) How many 5-letter words of binary digits are there?

9) Ten teams participate in a tournament. The first team is awarded a gold medal, the second a silver medal, and the third a bronze medal. In how many ways can the medals be distributed?

10) The RBI prints currency notes in denominations of One Rupee, Two Rupees, Five Rupees, Ten Rupees, Twenty Rupees, Fifty Rupees, and One hundred rupees. In how many ways can it display 10 currency notes, not necessarily of different denominations? How many of these will have all denominations?

11) In how many ways can an employer distribute INR 100/- as Holiday Bonus to his 5 employees? No fraction of a rupee is allowed. Also, do not worry about question of equity and fairness!

12) The results of 20 chess games (win, lose, or draw) have to be predicted. How many different forecasts can contain exactly 15 correct results?

13) How many distinct results can we obtain from one throw of four dice? five dice? Can you generalize this?

14) In how many ways can 8 rooks be placed on a standard chess board so that no rook can attack another? How many if the rooks are labelled? How would the answer be modified if we remove the restriction that “no rook can attack another”?

15) Show that there are 7 partitions of the integer 5, and 33 partitions of the integer 9. How many of these have 4 parts ? How many have the largest part equal to 4? Experiment with other partitions and other numbers.

Nalin Pithwa.

Elementary problems in Ramsey number theory for RMO

Question 1:

Show that in any group of 6 people there will always be a subgroup of 3 people who are pairwise acquainted or a subgroup of 3 people who are pairwise strangers.

Solution 1:

Let \{ A, B, C, D, E, F\} be a group of 6 people. Suppose that the people known to A are seated in room Y and the people NOT known to A are seated in room Z; A is not in either room. Then, there are necessarily at least 3 people in either room Y or in room Z; (a) Suppose B, C, D to be in room Y. Either these 3 people are mutual strangers (and so the given theorem is true), or, at least two of them (say, B and C) know each other. In the latter case, A, B and C form a group of 3 mutual acquaintances — and again, the theorem is true. (b) In (a), replace room Y by Z and interchange the notion of ‘”acquaintances” and “strangers”‘.

Question 2:

Show that in any group of 10 people there is always (a) a subgroup of 3 mutual strangers or a subgroup of 4 mutual acquaintances, and (b) a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 2:

(a) Let A be one of the ten people; the remaining 9 people can be assigned to two rooms: those who are known to A are in room Y and those who are not known to A are in room Z. Either room Y has at least 6 people or room Z has at least 6 people. For, (i) suppose room Y has at least 6 people. Then, by previous problem number 1, there is either a subgroup of 3 mutual acquaintances or a subgroup of 3 mutual strangers (thus, the theorem is true) in this room. In the former case, A and these 3 people constitute 4 mutual acquaintances (ii) Suppose room Z has at least 4 people. Either these 4 people know one another or at least 2 of them, say B and C, do not know each other. In the former case, we have a subgroup of 4 mutual acquaintances. In the latter case A, B and C constitute 3 mutual strangers.

(b) In the previous scenario, let people who are strangers become acquaintances, and let people who are acquaintances pretend they are strangers. The situation is symmetric.

Question 3:

Show that in any subgroup of 20 people there will always be either a subgroup of 4 mutual acquaintances or a subgroup of 4 mutual strangers.

Solution 3:

Suppose A is one of these 20 people. People known to A are in room Y and people not known to A are room Z. Either room Y has at least 10 people or room Z has at least 10 people. (i) If Y has at least 10 people, then by part B of problem number 2 here, there is either a subgroup of 3 mutual acquaintances or a subgroup of 4 mutual strangers — as asserted — in this room. In the former case, A and these mutual acquaintances will form a subgroup of 4 mutual acquaintances. (ii) Switch ‘”acquaintances” and “strangers”‘ in (i).

Question 4:

Let p and q be 2 positive integers. A positive integer r is said to have the (p,q) – Ramsey property, if in any group of r people either there is a subgroup of p people known to one another or there is a subgroup of q people not known to one another. {By Ramsey’s theorem, all sufficiently large integers r have the (p,q)-Ramsey property.} The smallest r with the (p,q)-Ramsey property is called the Ramsey number R(p,q). Show that (a) R(p,q) = R (q,p). (b) R(p,1)=1, and (c) R(p,2)=p.

Solution 4:

(a) By parts (b) of the previous three questions, we have proved part a of the proof here.

(b) This is obvious.

(c) In any group of p people, if all of them are not known to one another, there will be at least 2 people who do not know each other.

Question 5:

Prove that R(3,3)=6.

Solution 5:

Question 1 and its proof in this blog article imply that R(3,3) \leq 6.

To prove that R(3,3)>5, it is sufficient to consider a seating arrangement of 5 people about a round table in which each person knows only the 2 people on either side. In such a situation, there is no set of 3 mutual acquaintances and no set of 3 people not known to one another.

Question 6:

Show that if m and n are integers both greater than 2, then

R(m,n) \leq R(m-1,n) + R(m,n-1).

(this recursive inequality gives a non-sharp upper bound for R(m,n)).

Solution 6:

Let p \equiv R(m-1,n), q=R(m,n-1) and r \equiv p + q. Consider a group \{ 1,2, 3, \ldots, r\} of r people. Let L be the set of people known to person 1 and M be the set of people NOT known to person 1. The two sets together have r-1 people, so either L has at least p people or M has at least q people. (a) If L has p \equiv R(m-1,n) people, then, by definition, it contains a subset of (m-1) people known to one another or it contains a subset of n people unknown to one another. In the former case, the (m-1) people and person 1 constitute m people known to one another.

Thus, in their case, a group of R(m-1,n) + R(m,n-1) people necessarily includes m mutual acquaintances or n mutual strangers. That is, R(m,n) \leq R(m-1,n) + R(m,n-1).

(b) By the usual symmetry argument, the same conclusion follows when M contains q people.

Question 7:

(Remark: A pretty property of Ramsey numbers related to combinatorics).

Show that if m and n are integers greater than 1, then R(m,n) \leq { {m+n-2} \choose {m-1}} — a non-recursive upper bound.

Solution 7:

When m=2, or n=2, (i) holds with equality (see problem 4 in this blog article). The proof is by induction on k=m+n. As we have just seen, the result is true when k=4. Assume the result true for k-1. Then,

R(m-1,n) \leq {{m+n-3} \choose {m-2}}  and R(m,n-1) \leq {{m+n-3} \choose {m-1}}

Now, Pascal’s identity gives:

{{m+n-3} \choose {m-2}} + {{m+n-3} \choose {m-1}} = {{m+n-2}} \choose {m-1} so that R(m-1,n) + R(m,n-1) \leq {{m+n-2}} \choose {m-1}

But, from the previous question and its solution, we get R(m,n) \leq R(m-1,n) + R(m, n-1)

PS: As Richard Feynman, used to say, you will have to “piddle” with smallish problems as particular cases of these questions in order to get a grip over theory or formal language of this introduction.

PS: Additionally, you can refer to any basic Combinatorics text like Brualdi, or Alan Tucker or even Schaum Series outline ( V K Balakrishnan).

How to find the number of proper divisors of an integer and other cute related questions

Question 1:

Find the number of proper divisors of 441000. (A proper divisor of a positive integer n is any divisor other than 1 and n):

Solution 1:

Any integer can be uniquely expressed as the product of powers of prime numbers (Fundamental theorem of arithmetic); thus, 441000 = (2^{3})(3^{2})(5^{3})(7^{2}). Any divisor, proper or improper, of the given number must be of the form (2^{a})(3^{b})(5^{c})(7^{d}) where 0 \leq a \leq 3, 0 \leq b \leq 2, 0 \leq c \leq 3, and 0 \leq d \leq 2. In this paradigm, the exponent a can be chosen in 4 ways, b in 3 ways, c in 4 ways, d in 3 ways. So, by the product rule, the total number of proper divisors will be (4)(3)(4)(3)-2=142.

Question 2:

Count the proper divisors of an integer N whose prime factorization is: N=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} p_{3}^{\alpha_{3}}\ldots p_{k}^{\alpha_{k}}

Solution 2:

By using the same reasoning as in previous question, the number of proper divisors of N is (\alpha_{1}+1)(\alpha_{2}+1)(\alpha_{3}+1)\ldots (\alpha_{k}+1)-2, where we deduct 2 because choosing all the factors means selecting the given number itself, and choosing none of the factors means selecting the trivial divisor 1.

Question 3:

Find the number of ways of factoring 441000 into 2 factors, m and n, such that m>1, n>1, and the GCD of m and n is 1.

Solution 3:

Consider the set A = {2^{3}, 3^{2}, 5^{3}, 7^{2}} associated with the prime factorization of 441000. It is clear that each element of A must appear in the prime factorization of m or in the prime factorization of n, but not in both. Moreover, the 2 prime factorizations must be composed exclusively of elements of A. It follows that the number of relatively prime pairs m, n is equal to the number of ways of partitioning A into 2 unordered nonempty, subsets (unordered as mn and nm mean the same factorization; recall the fundamental theorem of arithmetic).

The possible unordered partitions are the following:

A = \{ 2^{3}\} + \{ 3^{2}, 5^{3}, 7^{2}\} = \{3^{2}\}+\{ 2^{3}, 5^{3}, 7^{2}\} = \{ 5^{3}\} + \{ 2^{3}, 3^{2}, 7^{2}\} = \{ 7^{2}\}+\{ 2^{3}, 3^{2}, 5^{3}\},

and A = \{ 2^{3}, 3^{2}\} + \{ 5^{3}, 7^{2}\}=\{ 2^{3}, 5^{3}\} + \{3^{2}, 7^{2} \} = \{ 2^{3}, 7^{2}\} + \{ 3^{2}, 5^{3}\}

Hence, the required answer is 4+3=2^{4-1}+1=7.

Question 4:

Generalize the above problem by showing that any integer has 2^{k-1}-1 factorizations into relatively prime pairs m, n (m>1, n>1).

Solution 4:

Proof by mathematical induction on k:

For k=1, the result holds trivially.

For k=2, we must prove that a set of k distinct elements, Z = \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}, a_{k}\} has 2^{k-1}-1 sets. Now, one partition of Z is

Z = \{ a_{k}\} \bigcup \{ a_{1}, a_{2}, a_{3}, \ldots, a_{k-1}\} \equiv \{ a_{k}\} \bigcup W

All the remaining partitions may be obtained by first partitioning W into two parts — which, by the induction hypothesis, can be done in 2^{k-2}-1 ways — and then, including a_{k} in one part or other — which can be done in 2 ways. By the product rule, the number of partitions of Z is therefore

1 + (2^{k-2})(2)=2^{k-1}-1. QED.

Remarks: Question 1 can be done by simply enumerating or breaking it into cases. But, the last generalized problem is a bit difficult without the refined concepts of set theory, as illustrated; and of course, the judicious use of mathematical induction is required in the generalized case. 


Nalin Pithwa.

Practice questions based on combinatorics for RMO Training and IITJEE Mathematics

Question 1:

Prove that if n is an even integer, then

\frac{1}{(1!)(n-1)!} + \frac{1}{3! (n-3)!} + \frac{1}{5! (n-5)!} + \ldots + \frac{1}{(n-1)! 1!} = \frac{2^{n-1}}{n!}

Question 2:

If {n \choose 0}, {n \choose 1}, {n \choose 2}, ….{n \choose n} are the coefficients in the expansion of (1+x)^{n}, when n is a positive integer, prove that

(a) {n \choose 0} - {n \choose 1}  + {n \choose 2} - {n \choose 3} + \ldots + (-1)^{r}{n \choose r} = (-1)^{r}\frac{(n-1)!}{r! (n-r-1)!}

(b) {n \choose 0} - 2{n \choose 1} + 3{n \choose 2} - 4{n \choose 3} + \ldots + (-1)^{n}(n+1){n \choose n}=0

(c) {n \choose 0}^{2} - {n \choose 1}^{2} + {n \choose 2}^{2} - {n \choose 3}^{2} + \ldots + (-1)^{n}{n \choose n}^{2}=0, or (-1)^{\frac{n}{2}}{n \choose {n/2}}, according as n is odd or even.

Question 3:

If s_{n} denotes the sum of the first n natural numbers, prove that

(a) (1-x)^{-3}=s_{1}+s_{2}x+s_{3}x^{2}+\ldots + s_{n}x^{n-1}+\ldots

(b) 2(s_{1}s_{2m} + s_{2}s_{2n-1} + \ldots + s_{n}s_{n+1}) = \frac{2n+4}{5! (2n-1)!}

Question 4:

If q_{n}=\frac{}{}, prove that

(a) q_{2n+1}+q_{1}q_{2n}+ q_{2}q_{2n-1} + \ldots + q_{n-1}q_{n+2} + q_{n}q_{n+1}= \frac{1}{2}

(b) 2(q_{2n}-q_{1}q_{2m-1}+q_{2}q_{2m-2}+\ldots + (-1)^{m}q_{m-1}q_{m+1}) = q_{n} + (-1)^{n-1}{q_{n}}^{2}.

Question 5:

Find the sum of the products, two at a time, of the coefficients in the expansion of (1+x)^{n}, where n is a positive integer.

Question 6:

If (7+4\sqrt{3})^{n} = p + \beta, where n and p are positive integers, and \beta, a proper fraction, show that (1-\beta)(p+\beta)=1.

Question 7:

If {n \choose 0}, {n \choose 1}, {n \choose 2}, …,, {n \choose n} are the coefficients in the expansion of (1+x)^{n}, where n is a positive integer, show that

{n \choose 1} - \frac{{n \choose 2}}{2} + \frac{{n \choose 3}}{3} - \ldots + \frac{(-1)^{n-1}{n \choose n}}{n} = 1 + \frac{1}{2} + \frac {1}{3} + \frac{1}{4} + \ldots + \frac{1}{n}.

That’s all for today, folks!

Nalin Pithwa.









Combinatorics for RMO : some basics and examples: homogeneous products of r dimensions


Find the number of homogeneous products of r dimensions that can be formed out of the n letters a, b, c ….and their powers.


By division, or by the binomial theorem, we have:

\frac{1}{1-ax} = 1 + ax + a^{2}x^{2} + a^{3}x^{3} + \ldots

\frac{1}{1-bx} = 1+ bx + b^{2}x^{2} + a^{3}x^{3} + \ldots

\frac{1}{1-cx} = 1 + cx + c^{2}x^{2} + c^{3}x^{3} + \ldots

Hence, by multiplication,

\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \times \ldots

= (1+ax + a^{2}x^{2}+a^{3}x^{3}+ \ldots)(1+bx + b^{2}x^{2} + b^{3}x^{3}+ \ldots)(1+cx + c^{2}x^{2} + c^{3}x^{3}+ \ldots)\ldots

= 1 + x(a + b + c + \ldots) +x^{2}(a^{2}+ab+ac+b^{2}+bc + c^{2} + \ldots) + \ldots

= 1 + S_{1}x + S_{2}x^{2} + S_{3}x^{3} + \ldots suppose;

where S_{1}, S_{2}, S_{3}, \ldots are the sums of the homogeneous products of one, two, three, … dimensions that can be formed of a, b, c, …and their powers.

To obtain the number of these products, put a, b, c, …each equal to 1; each term in S_{1}, S_{2}, S_{3}, …now becomes 1, and the values of S_{1}, S_{2}, S_{3}, …so obtained give the number of the homogeneous products of one, two, three, ….dimensions.


\frac{1}{1-ax} \times \frac{1}{1-bx} \times \frac{1}{1-cx} \ldots

becomes \frac{1}{(1-x)^{n}}, or (1-x)^{-n}

Hence, S_{r} = the coefficient of x^{r} in the expansion of (1-x)^{-n}

= \frac{n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}= \frac{(n+r-1)!}{r!(n-1)!}


Find the number of terms in the expansion of any multinomial when the index is a positive integer.


In the expansion of (a_{1}+ a_{2} + a_{3} + \ldots + a_{r})^{n}

every term is of n dimensions; therefore, the number of terms is the same as the number of homogeneous products of n dimensions that can be formed out of the r quantities a_{1}, a_{2}, a_{3}, …a_{r}, and their powers; and therefore by the preceding question and solution, this is equal to

\frac{(r+n-1)!}{n! (r-1)!}

A theorem in combinatorics:

From the previous discussion in this blog article, we can deduce a theorem relating to the number of combinations of n things.

Consider n letters a, b, c, d, ….; then, if we were to write down all the homogeneous products of r dimensions, which can be formed of these letters and their powers, every such product would represent one of the combinations, r at a time, of the n letters, when any one of the letters might occur once, twice, thrice, …up to r times.

Therefore, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of homogeneous products of r dimensions which can be formed out of n letters, and therefore equal to \frac{(n+r-1)!}{r!(n-1)!}, or {{n+r-1} \choose r}.

That is, the number of combinations of n things r at a time when repetitions are allowed is equal to the number of combinations of n+r-1 things r at a time when repetitions are NOT allowed.

We conclude this article with a few miscellaneous examples:

Example 1:

Find the coefficient of x^{r} in the expansion of \frac{(1-2x)^{2}}{(1+x)^{3}}

Solution 1:

The expression = (1-4x+4x^{2})(1+p_{1}x+p_{2}x^{2}+ \ldots + p_{r}x^{r}+ \ldots), suppose.

The coefficients of x^{r} will be obtained by multiplying p_{r}, p_{r-1}, p_{r-2} by 1, -4, and 4 respectively, and adding the results; hence,

the required coefficient is p_{r} - 4p_{r-1}+4p_{r-2}

But, with a little work, we can show that p_{r} = (-1)^{r}\frac{(r+1)(r+2)}{2}.

Hence, the required coefficient is

= (-1)^{r}\frac{(r+1)(r+2)}{2} - 4(-1)^{r-1}\frac{r(r+1)}{2} + 4 (-1)^{r-2}\frac{r(r-1)}{2}

= \frac{(-1)^{r}}{2}\times ((r+1)(r+2) + 4r(r+1) + 4r(r-1))

= \frac{(-1)^{r}}{2}(9r^{2}+3r+2)

Example 2:

Find the value of the series

2 + \frac{5}{(2!).3} + \frac{5.7}{3^{2}.(3!)} + \frac{5.7.9}{3^{3}.(4!)} + \ldots

Solution 2:

The expression is equal to

2 + \frac{3.5}{2!}\times \frac{1}{3^{2}} + \frac{3.5.7}{3!}\times \frac{1}{3^{3}} + \frac{}{4!}\times \frac{1}{3^{4}} + \ldots

= 2 + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times \frac{2^{2}}{3^{2}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times \frac{2^{3}}{3^{3}} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times \frac{2^{4}}{3^{4}} + \ldots

= 1 + \frac{\frac{3}{2}}{1} \times \frac{2}{3} + \frac{\frac{3}{2}.\frac{5}{2}}{2!} \times (\frac{2}{3})^{2} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}}{3!} \times (\frac{2}{3})^{3} + \frac{\frac{3}{2}.\frac{5}{2}.\frac{7}{2}.\frac{9}{2}}{4!} \times (\frac{2}{3})^{4} + \ldots

= (1-\frac{2}{3})^{\frac{-3}{2}} = (\frac{1}{3})^{-\frac{3}{2}} = 3^{\frac{3}{2}} = 3 \sqrt{3}.

Example 3:

If n is any positive integer, show that the integral part of (3+\sqrt{7})^{n} is an odd number.

Solution 3:

Suppose I to denote the integral and f the fractional part of (3+\sqrt{7})^{n}.

Then, I + f = 3^{n} + {n \choose 1}3^{n-1}\sqrt{7} + {n \choose 2}3^{n-2}.7 + {n \choose 3}3^{n-3}.(\sqrt{7})^{3}+ \ldots…call this relation 1.

Now, 3 - \sqrt{7} is positive and less than 1, therefore (3-\sqrt{7})^{n} is a proper fraction; denote it by f^{'};

Hence, f^{'} = 3^{n} - {n \choose 1}.3^{n-1}.\sqrt{7} + {n \choose 2}.3^{n-2}.7 - {n \choose 3}.3^{n-3}.(\sqrt{7})^{3}+ \ldots…call this as relation 2.

Add together relations 1 and 2; the irrational terms disappear, and we have

I + f + f^{'} = 2(3^{n} + {n \choose 2}.3^{n-2}.7+ \ldots ) = an even integer

But, since f and f^{'} are proper fractions their sum must be 1;

Hence, I is an odd integer.

Hope you had fun,

Nalin Pithwa.

Inequalities and mathematical induction: RMO sample problems-solutions


1. Prove the inequality —- 2^{n}(n!)^{2} \leq (2n)! for all natural numbers greater than or equal to 1.

Proof 1:

First consider the following: 2.6. 10.14 \ldots (4n-2)=\frac{(2n)!}{n!}. Let us prove this claim first and then use it to prove what is asked: Towards, that end, consider

RHS = \frac{(2n)(2n-1)(2n-2)(2n-3)(2n-4)\ldots 4.2.1}{\ldots (n-1)n}=LHS, cancelling off the common factors in numerator and denominator of RHS. (note this can also be proved by mathematical induction! 🙂 )

In the given inequality:

we need to prove 2^{n}(n!)^{2} \leq (2n)!

consider \ldots (4n-2)=\frac{(2n)!}{n!} where

LHS = (2.1)(2.3) (2.5) (2.7) \ldots 2(n-1), this is an AP with first term 2 and nth term (4n-2) and common difference 4; there are n factors “2”; hence, we 2^{n} \ldots (n-1)=\frac{(2n)!}{n!} so we get

2^{n} (n!)\ldots (n-1) = (2n)!; multiplying and dividing RHS of this by\ldots n, we get the desired inequality. Remember the inequality is less than or equal to.

Problem 2:

Establish the Bernoulli inequality: If (1+a) > 0, then (1+a)^{n} \geq 1+na.

Solution 2:

Apply the binomial theorem, which in turn, is proved by mathematical induction ! 🙂

Problem 3:

For all natural numbers greater than or equal to 1, prove the following by mathematical induction:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{n^{2}} \leq 2-\frac{1}{n}

Proof 3:

Let the given proposition be P(n).

Step 1: Check if P(1) is true. Towards that end:

LHS=\frac{1}{1^{2}}=1 and $latex RHS=2-\frac{1}{1}=2-1=1$ and hence, P(1) is true.

Step 2: Let P(n) be true for some n=k, k \in N. That is, the following is true:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} \leq 2 -\frac{1}{k}

Add \frac{1}{(k+1)^{2}} to both sides of above inequality, we get the following:

\frac{1}{1^{2}} + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \ldots + \frac{1}{k^{2}} + \frac{1}{(k+1)^{2}} \leq 2-\frac{1}{k}+\frac{1}{(k+1)^{2}}

Now, the RHS in above is 2-\frac{1}{k} +\frac{1}{(k+1)^{2}}=2-\frac{k^{2}+k+1}{k(k+1)^{2}}. We want this to be less than or equal to 2-\frac{1}{k+1}. Now, k \in N, k>1, so what we have to prove is the following:

-\frac{k^{2}+k+1}{k(k+1)^{2}} \leq -\frac{1}{k+1}, that is, we want to prove that

(k+1)(k^{2}+k+1) \geq k(k^{2}+2k+1), that is, we want k^{3}+k^{2}+k+k^{2}+k+1 \geq k^{3}+2k^{2}+k, that is, we want k+1 \geq 0, which is obviously true. QED.


Nalin Pithwa.

RMO Training: more help from Nordic mathematical contest


32 competitors participate in a tournament. No two of them are equal and in a one against one match the better always wins. (No tie please). Show that the gold, silver and bronze medal can be found in 39 matches.


We begin by determining the gold medallist using classical elimination, where we organize 16 pairs and matches, then 8 matches of the winners, 4 matches of the winners in the second round, then 2-semifinal matches and finally one match making 31 matches altogether.

Now, the second best player must have at some point lost to the best player, and as there were 5 rounds in the elimination, there are 5 candidates for the silver medal. Let C_{i} be the candidate who  lost to the gold medalist in round i. Now, let C_{1} and C_{2} play, the winner play against C_{3}, and so forth. After 4 matches, we know the silver medalist; assume this was C_{k}.

Now, the third best player must have lost against the gold medalist or against C_{k} or both. (If the player had lost to someone else, there would be at least three better players.) Now, C_{k} won k-1 times in the elimination rounds, the 5-k players C_{k+1}\ldots C_{5} and if k is greater than one, one player C_{j} with j<k. So there are either (k-1)+(5-k)=4 or (k-1)+(5-k)+1=5 candidates for the third place. At most 4 matches are again needed to determine the bronze winners.

Cheers to Norway mathematicians!

Nalin Pithwa.

Reference: Nordic mathematical contests, 1987-2009.

Amazon India link:


RMO 2017 Warm-up: Two counting conundrums

Problem 1:

There are n points in a circle, all joined with line segments. Assume that no three (or more) segments intersect in the same point. How many regions inside the circle are formed in this way?

Problem 2:

Do there exist 10,000 10-digit numbers divisible by 7, all of which can be obtained from one another by a re-ordering of their digits?

Solutions will be put up in a couple of days.

Nalin Pithwa.