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) ?

Answer 2:

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

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