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.









Some Basics for RMO Number Theory: Congruences

I. The congruence notation:

It often happens that for the purposes of a particular calculation, two numbers which differ by a multiple of some fixed number are equivalent, in the sense that they produce the same result. For example, the value of (-1)^{n} depends only on whether n is odd or even, so that two values of n which differ by a multiple of 2 give the same result. Or again, if we are concerned only with the last digit of a number, then for that purpose two numbers which differ by a multiple of 10 are effectively the same.

The congruence notation, introduced by Gauss, serves to express in a convenient form the fact that the two integers a and b differ by a multiple of a fixed natural number m. We say that a is congruent to b with respect to the modulus m, or in symbols,

a \equiv b {\pmod m}

The meaning of this, then, is simply that a-b is divisible by m. The notation facilitates calculations in which numbers differing by a multiple of m are effectively the same, by stressing the analogy between congruence and equality. Congruence, in fact, means “equality except for the addition of some multiple of m”.

A few examples of valid congruences are :

63 \equiv 0 {\pmod 3}; 7 \equiv -1 {\pmod 8}; 5^{2} \equiv -1 {\pmod {13}}

A congruence to the modulus 1 is always valid, whatever the two numbers may be, since every number is a multiple of 1. Two numbers are congruent with respect to the modulus 2 if they are of the same parity, that is, both even or both odd.

Two congruences can be added, subtracted or multiplied together, in just the same way as two equations, provided all the congruences have the same modulus. If

a \equiv \alpha {\pmod m} and b \equiv \beta {\pmod m}


a + b \equiv {\alpha + \beta} {\pmod m}

a - b \equiv {\alpha - \beta}{\pmod m}

ab \equiv {\alpha\beta} {\pmod m}

The first two of these statements are immediate: for example, (a+b) - (\alpha + \beta) is a multiple of m because a- \alpha and b - \beta are both multiples of m. The third is not quite so immediate because ab - \alpha \beta = (a-\alpha)b, and a - \alpha is a multiple of m. Next, ab \equiv \alpha \beta, for a similar reason. Hence, ab \equiv {\alpha \beta} {\pmod m}.

A congruence can always be multiplied throughout by any integer; if a \equiv {\alpha} {\pmod m} the10n ka \equiv {k\alpha} {\pmod m}. Indeed this is a special case of the third result above, where b and \beta are both k. But, it is not always legitimate to cancel a factor from a congruence. For example, 42 \equiv 12 {\pmod 10}, but it is not permissible to cancel the factor 6 from the numbers 42 and 12, since this would give the false result 7 \equiv 2 {\pmod 10}. The reason is obvious: the first congruence states that 42-12 is a multiple of 10, but this does not imply that \frac{1}{6}(42-12) is a multiple of 10. The cancellation of a factor from a congruence is legitimate if the factor is relatively prime to the modulus. For, let the given congruence be ax \equiv ay {\pmod m}, where is the factor to be cancelled, and we suppose that a is relatively prime to m. The congruence states that a(x-y) is divisible by m, and hence, (x-y) is divisible by m.

An illustration of the use of congruences is provided by the well-known rules for the divisibility of a number by 3 or 9 or 11. The usual representation of a number n by digits in the scale of 10 is really a representation of n in the form

n = a + 10b + 100c + \ldots

where a, b, c, … are the digits of the number, read from right to left, so that a is the number of units, b the number of tens, and so on. Since 10 \equiv 1 {\pmod 9}, we have also 10^{2} \equiv 1 {\pmod 9}, and 10^3 \equiv 1 {\pmod 9}, and so on. Hence, it follows from the above representation of n that

n \equiv {a+b+c+\ldots} {\pmod 9}

In other words, any number n differs from the sum of its digits by a multiple of 9, and in particular n is divisible by 9 if and only if the sum of its digits is divisible by 9. The same applies with 3 in place of 9 throughout.

The rule for 11 is based on the fact that 10 \equiv -1 {\pmod 11} so that 10^2 \equiv {+1} {\pmod 11}, and so on. Hence,

n \equiv {a-b+c- \ldots} {\pmod 11}

It follows that n is divisible by 11 if and only if a-b+c-\ldots is divisible by 11. For example, to test the divisibility of 958 by 11 we form 1-8+5-9, or -11. Since this is divisible by 11, so is 9581.

Ref: The Higher Arithmetic by H. Davenport, Eighth Edition, Cambridge University Press.


Nalin Pithwa.


Solutions to two algebra problems for RMO practice

Problem 1.

If a, b, c are non-negative real numbers such that (1+a)(1+b)(1+c)=8, then prove that the product abc cannot exceed 1.

Solution I:

Given that a \geq 0, b \geq 0, c \geq 0, so certainly abc>0, ab>0, bc>0, and ac>0.

Now, (1+a)(1+b) = 1 + a + b + ab and hence, (1+a)(1+b)(1+c) = (1+a+b+ab)(1+c)= 1+a+b+ab+c +ac + bc + abc=8, hence we get:

a+b+c+ab+bc+ca+abc=7Clearly, the presence of a+b+c and abc reminds us of the AM-GM inequality.

Here it is AM \geq GM.

So, \frac{a+b+c}{3} \geq (abc)^{1/3}.

Also, we can say: \frac{ab+bc+ca}{3} \geq (^{1/3}. Now, let x=(abc)^{1/3}.

So, 8 \geq 1+3x+3x^{2}+x^{3}

that is, 8 \geq (1+x)^{3}, or 2 \geq 1+x, that is, x \leq 1So, this is a beautiful application of arithmetic mean-geometric mean inequality twice. 🙂 🙂

Problem 2:

If a, b, c are three rational numbers, then prove that :\frac{1}{(a-b)^{2}} + \frac{1}{(b-c)^{2}} + \frac{1}{(c-a)^{2}} is always the square of a rational number.

Solution 2:

Let x=\frac{1}{a-b}, y=\frac{1}{b-c}, z=\frac{1}{c-a}. It can be very easily shown that \frac{1}{x}+ \frac{1}{y} + \frac{1}{z} =0, or xy+yz+zx=0. So, the given expression x^{2}+y^{2}+z^{2}=(x+y+z)^{2} is a perfect square !!! BINGO! 🙂 🙂 🙂

Nalin Pithwa.

Euler Series question and solution


Mengoli had posed the following series to be evaluated:

1+ \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + \ldots.

Some great mathematicians, including Liebnitz, John Bernoulli and D’Alembert, failed to compute this infinite series. Euler established himself as the best mathematician of Europe (in fact, one of the greatest mathematicians in history) by evaluating this series initially by a not-so-rigorous method. Later on, he gave alternative and more rigorous ways of getting the same result.

Prove that the series converges and gets an upper limit. Then, try to evaluate the series.


Due Nicolas Oresine:

Consider the following infinite series: \phi(s)=1 + \frac{1}{2^{s}} + \frac{1}{3^{s}} + \frac{1}{4^{s}} + ldots

We can re-write the preceding series as follows: \phi(s) = 1+ (\frac{1}{2^{s}}+\frac{1}{3^{s}}) + (\frac{1}{4^{s}} + \frac{1}{5^{s}} + \frac{1}{6^{s}} + \frac{1}{7^{s}}) + \ldots, which in turn is less than

1 + (\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots. Now, the RHS of this can be re-written as

1+(\frac{2}{2^{s}}) + (\frac{4}{4^{s}}) + \ldots=1 + \frac{1}{2^{(s-1)}}+ (\frac{1}{2^{(s-1)}})^{2} + \ldots, which is a geometric series and it is given by


Now, we can say that \phi(s) will converge if \frac{1}{2^{(s-1)}}<1 \Longrightarrow s >1.

In order to prove what is asked, we start with \phi(s)=1 + \frac{1}{2^{s}}+ \frac{1}{3^{s}}+ \frac{1}{4^{s}}+\ldots

And, then multiply both sides by \frac{1}{2^{s}} and then subtract the resulting equation from the preceding equation to get


where all the terms containing the reciprocals of the sth power of even numbers vanished.

Repeating this procedure with \frac{1}{3^{s}} gives

(1-\frac{1}{2^{s}})(1-\frac{1}{3^{s}})\phi(s)=1+\frac{1}{5^{s}}+ \ldots

where all terms containing the reciprocals of the sth power of multiples of 3 vanished.

By continuing this with all prime numbers, we get

\prod_{p}(1-\frac{1}{p^{s}})\phi(s)=1, where p represents all prime numbers. Thus, we get

\phi(s)=1 + \frac{}{} + \frac{}{} + \frac{}{} + \ldots =\frac{1}{\prod_{p}(1-\frac{1}{p^{s}})}

This is a remarkable result because the LHS is concerned with only positive integers, whereas the RHS is concerned with only primes. This result is known as the “Golden Key of Euler”.

Riemann created his famous \zeta- function by extending the variable s to the entire complex plane, except s=1 with 

\zeta(s)=1+ \frac{1}{2^{s}} + \frac{1}{3^{s}} + \ldots .

This function is now very famous as the Riemann zeta function.

How can we apply the Golden Key of Euler to Mengoli’s question that we started with?

Ans. In the Golden Key of Euler, substitute s=2.

Hence, we get the upper limit of the given series is 2.

Euler’s proof (1775):

The proof ran as follows:

It is a little roundabout way of arriving at the correct answer from a known result. Consider McLaurin’s series expansion of sin x:

\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} -\frac{x^{7}}{7!} + \frac{x^{9}}{9!} + \ldots

By dividing both sides by x and then substituting y=x^{2} on the right side, we get the following:

\frac{\sin{(x)}}{x} = 1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{7!} + \ldots

By taking a special value of x=n\pi (and, hence y=n^{2}\pi^{2}), we get the following:

\frac{\sin (n\pi)}{(n\pi)}=0=1-\frac{y}{3!} + \frac{y^{2}}{5!} - \frac{y^{3}}{5!}+ \ldots

Note  that preceding equation is not a polynomial, but an infinite series. But, Euler still treated it as a polynomial (that is why it was not accepted as a rigorous result) and observed that this “infinite” polynomial has roots equal to y_{n}=n^{2}x^{2}. Then, Euler had used the fact that the sum of the reciprocals of the roots is determined by the coefficient of the linear term (here, the y-term) when the constant is made unity. (check this as homework quiz, for a quadratic to be convinced). So, Euler had arrived at the following result:

1-\sum_{i=1}^{\infty}\frac{6}{y_{n}}=0 \Longrightarrow \sum_{i=1}^{\infty}\frac{1}{y_{n}}=\frac{1}{6}. With y_{n}=n^{2}(\pi)^{2}, we get the following:

\sum_{i=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}}=\frac{1}{6} or, \sum_{1}^{n^{2}}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}.

Another proof also attributed to Euler that uses the series expansion of sin (x) goes as follows below:

\sin {(x)} has roots given by 0, \pm \pi, \pm 2\pi, \pm 3\pi, …So does this polynomial that Euler reportedly constructed:


So, Euler considered the preceding equation to be equivalent to:

\sin{(x)}=x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots=0

Then, he had equated the coefficient of x^{3} in both to get the result:

\sum_{n=1}^{\infty}\frac{1}{n^{2}(\pi)^{2}} = \frac{1}{3!} = \frac{1}{6}.

Thus, \sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{(\pi)^{2}}{6}.

Later on, Euler had provided a few more alternate and rigorous proofs of this result.

Reference: Popular Problems and Puzzles in Mathematics by Asok Kumar Mallik, IISc Press, Foundation Books.

Amazon India link:

Hope you all enjoyed it — learning to think like Euler !! By the way, it did take a long time for even analysis to become so rigorous as it is now….You might like this observation a lot. 🙂 🙂 🙂

Nalin Pithwa.