# A series question : pre RMO, RMO, IITJEE

Question 1:

Let $x_{1}, x_{2}, \ldots, x_{2014}$ be real numbers different from 1, such that

$x_{1}+x_{2}+\ldots+x_{2014}=1$ and

$\frac{x_{1}}{1-x_{1}} + \frac{x_{2}}{1-x_{2}} + \ldots + \frac{x_{2014}}{1-x_{2014}} = 1$ also.

Then, what is the value of

$\frac{x_{1}^{2}}{1-x_{1}} + \frac{x_{2}^{2}}{1-x_{2}} + \frac{x_{3}^{2}}{1-x_{3}} + \ldots + \frac{x_{2014}^{2}}{1-x_{2014}}$ ?

Solution 1:

Note that

$\sum \frac{x_{i}^{2}}{1-x_{i}} = \sum \frac{x_{i} + (x_{i} - x_{i}^{2})}{1-x_{i}} = \sum (\frac{x_{i}}{1-x_{i}} - x_{i}) = \sum \frac{x_{i}}{1-x_{i}} -\sum x_{i} = 1 - 1 = 0$ which is required answer.

Note that the maximum index 2014 plays no significant role here.

Question 2:

Let f be a one-to-one function from the set of natural numbers to itself such that

$f(mn) = f(m)f(n)$ for all natural numbers m and n.

What is the least possible value of $f(999)$ ?

From elementary number theory, we know that given f is a multiplicative function and hence, the required function is such that if p and q are prime, then

$f(pq)=f(p)f(q)$

That is we need to decompose 999 into its unique prime factorization.

So, we have $999 = 3 \times 333 = 3^{2} \times 111 = 3^{3} \times 97$ where both 3 and 97 are prime.

We have $f(999) = f(3)^{3} \times f(97)$ and we want this to be least positive integer. Clearly, then f(3) cannot be greater than 97. Also, moreover, we need both f(3) and f(97) to be as least natural number as possible. So, $f(3)=2$ and $f(97)=3$ so that required answer is 24.

Question 3:

HW :

What is the number of ordered pairs (A,B) where A and B are subsets of $\{ 1,2,3,4,5\}$ such that neither $A \subset B$ nor $B \subset A$ ?

Cheers,

Nalin Pithwa

# Basic algebra facts of HCF and LCM

1. If x and y are two natural numbers, HCF(x,y) x LCM (x,y) = xy
2. If a, b, c are three natural numbers, then the product abc=HCF(a,b,c) x LCM(ab,bc,ca) and also abc=HCF(ab, bc,ca) x LCM(a,b,c).

# Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in $m+n$ ways. More generally, if $E_{i}$ ($i=1,2,\ldots,k$) are k events such that no two of them can occur at the same time, and if $E_{i}$ can occur in $n_{i}$ ways, then one of the k events can occur in $n_{1}+n_{2}+\ldots+n_{k}$ ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, \$if $E_{1}$ can occur in $n_{1}$, $E_{2}$ can occur in $n_{2}$ ways (no matter how $E_{1}$ occurs), $E_{3}$ can occur in $n_{3}$ ways (no matter how $E_{1}$ and $E_{2}$ occur), $\ldots$, $E_{k}$ can occur in $n_{k}$ ways (no matter how the previous $k-1$ events occur), then the k events can occur simultaneously in $n_{1}n_{2}n_{3}\ldots n_{k}$ ways.

)3. Definitions and some basic relations:

Suppose X is a collection of n distinct objects and r is a nonnegative integer less than or equal to n. An r-permutation of X is a selection of r out of the n objects but the selections are ordered.

An n-permutation of X is called a simply a permutation of X.

The number of r-permutations of a collection of n distinct objects is denoted by $P(n,r)$; this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of $(n-1)$ objects can be chosen to occupy the second position in $(n-1)$ ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in $n(n-1)$ ways,….and all r positions can be filled in

$P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}!$ ways.

In particular, $P(n,n) = n!$

Note: An unordered selection of r out of the n elements of X is called an r-combination of X. In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by $n \choose r$ (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary $(n-r)$-subset, whence the important relation ${n \choose r}$ = $n \choose {n-r}$.

To evaluate $n \choose r$, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule:

$P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)$

The number of terms on the right is the number of r-subsets of X. That is, $n \choose r$. Thus, $P(n,r)=P(r,r) \times {n \choose r}$=$r! \times {n \choose r}$.

The following is our summary:

1. $P(n,r) = \frac{n!}{(n-r)!}$
2. ${n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}$=$n \choose {n-r}$

4. The Pigeonhole Principle: Basic Version:

If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters).

5. The number of ways in which $m+n$ things can be divided into two groups containing m and n equal things respectively is given by : $\frac{(m+n)!}{m!n!}$

Note: If $m=n$, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is $\frac{(2m)!}{2!m!m!}$

6. The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: $\frac{(m+n+p)!}{m!n!p!}$

Note: If we put $m=n=p$, we obtain $\frac{(3m)!}{m!m!m !}$ but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in $\frac{(3m)!}{m!m!m!3!}$ ways.

7. The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: $\frac{n!}{p!q!r!}$

8. The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: $n^{r}$. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-))

9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : $2^{n}-1$

10. The total number of ways in which it is possible to make a selection by taking some or all out of $p+q+r+\ldots$ things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : $(p+1)(q+1)(r+1)\ldots-1$.

Regards,

Nalin Pithwa.

# III. Tutorial problems. Symmetric and alternating functions. RMO and IITJEE math

1. Simplify: $(b^{-1}+c^{1})(b+c-a)+(c^{-1}+a^{-1})(c+a-b)+(a^{-1}+b^{-1})(a+b=c)$
2. Simplify: $\frac{(x-b)(x-c)}{(a-b)(a-c)} + \frac{(x-c)(x-a)}{(b-c)(b-a)} + \frac{(x-a)(x-b)}{(c-a)(c-b)}$
3. Simplify: $\frac{b^{2}+c^{2}-a^{2}}{(a-b)(a-c)} + \frac{c^{2}+a^{2}-b^{2}}{(b-c)(b-a)} + \frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}$
4. Simplify: $\frac{b-c}{1+bc} + \frac{c-a}{1+ca} + \frac{a-b}{1+ab}$
5. Simplify: $\frac{a(b-c)}{1+bc} + \frac{b(c-a)}{1+ca} + \frac{c(a-b)}{1+ab}$
6. Factorize: $(b-c)^{2}(b+c-2a)+(c-a)^{2}(c+a-2b)+(a-b)^{2}(a+b-2c)$. Put $b-c=x, c-a=y, a-b=z$and $b+c-2a=y-z$
7. Factorize: $8(a+b+c)^{2}-(b+c)^{2}-(c+a)^{2}-(a+b)^{2}$. Put $b+c=x, c+a=y, a+b=z$.
8. Factorize: $(a+b+c)^{2}-(b+c-a)^{2}-(c+a-b)^{2}+(a+b-c)^{2}$
9. Factorize: $(1-a^{2})(1-b^{2})(1-c^{2})+(a-bc)(b-ac)(c-ab)$
10. Express the following substitutions as the product of transpositions: (i) $\left(\begin{array}{cccccc}123456\\654321\end{array}\right)$ (ii) $\left(\begin{array}{cccccc}123456\\246135\end{array}\right)$ (iii) $\left(\begin{array}{cccccc}123456\\641235\end{array}\right)$

Regards,

Nalin Pithwa.