Pre RMO Practice Sheet 2

Question 1:

A man walks a certain distance and rides back in 3.75 hours, he could ride both ways in 2.5 hours. How many hours would it take him to walk both ways?

Answer 1:

Let the man walk distance x km in time t hours. His walking speed is \frac{x}{t} kmph.

Let him ride distance x km in time T hours. His riding speed is \frac{x}{T} kmph.

First journey:

\frac{x}{t} + \frac{x}{T} = 3.75

Second journey:

\frac{2x}{T} = 2.5

Hence, \frac{x}{T} = 1.25

Using above in first equation: \frac{x}{t} + 1.25 = 3.75

Hence, \frac{x}{t} = 2.50 = his speed of walking. Hence, it would take him 5 hours to walk both ways.

Question 2:

Positive integers a and b are such that a+b=\frac{a}{b} + \frac{b}{a}. What is the value of a^{2}+b^{2}?

Answer 2:

Given that a and b are positive integers.

Given also \frac{a}{b} + \frac{b}{a} = a+b.

Hence, a^{2}(b-1)+b^{2}(a-1)=0

As a and b are both positive integers, so a-1 and b-1 both are non-negative.

So, both the terms are non-negative and hence, sum is zero if both are zero or a=1 and b=1.

Hence, a^{2}+b^{2}=2

Question 3:

The equations x^{2}-4x+k=0 and x^{2}+kx-4=0 where k is a real number, have exactly one common root. What is the value of k?

Answer 3:

x^{2}-4x+k=0…equation I

x^{2}+kx-4=0…equation II

Let k \in \Re and let \alpha is a common root.

Hence, \alpha satisfies both the equations. So, by plugging in the value of \alpha we get the following:

\alpha^{2} -4\alpha + k =0 and x^{2}+kx-4=0 and so using these two equations, we get the following:


Case 1: \alpha \neq  1 then k=-4. But hold on, we havent’t checked thoroughly if this is the real answer. We got to check now if both equations with these values of alpha and k have only one common root.

Equation I now goes as : x^{2}-4x-4=0 so this equation has irrational roots. On further examination, we see that if \alpha=1, then k can be any value. So, what are the conditions on k? We get that from equation I: plug in the value of alpha:

1-4+k=0 so k-3=0 and k=3.

So, we now recheck if both equations have only one common root when alpha is 1 and k is 3:

x^{2}-4x+3=0 so the roots of first equation are 3 and 1.

x^{2}+3x-4=0 so the roots of second equation are -4 and 1.

Clearly so k=3 is the final answer. 🙂


Nalin Pithwa

Pre RMO practice sheet

Question 1:

What is the smallest positive integer k such that k(3^{3}+4^{3}+5^{3})=a^{n} for some positive integers a and n with n>1?

Solution 1:

We have k \times 216= a^{n} so that k \times 6^{3} = a^{n} giving k=1.

Question 2:

Let S_{n} = \sum_{k=0}^{n}\frac{1}{\sqrt{k+1}+\sqrt{k}}.

What is the value of \sum_{n=1}^{99} \frac{1}{S_{n}+S_{n-1}}

Answer 2:


S_{n} = \sum_{k=0}^{n} \frac{1}{\sqrt{k+1}+\sqrt{k}} \times \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k+1}-\sqrt{k}}

S_{n} = \sum_{k=0}^{n} (\sqrt{k+1}-\sqrt{k}) = \sqrt{n+1}


S_{n-1} = \sum_{k=0}^{n-1}(\sqrt{k+1}-\sqrt{k}) = \sqrt{n}

Now, \sum_{n=1}^{99} \frac{1}{S_{n}+S_{n-1}} = \sum_{n=1}^{99}\frac{1}{\sqrt{n+1}-\sqrt{n}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}

which in turn is equal to \sum_{n=1}^{99} (\sqrt{n+1}-\sqrt{n}) = (\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+ (\sqrt{4}-\sqrt{3}+ \ldots + (\sqrt{100}-\sqrt{99})) = \sqrt{100} -\sqrt{1}=10-1=9

Question 3: Homework:

It is given that the equation x^{2}+ax+20=0 has integer roots. What is the sum of all possible values of a?


Nalin Pithwa

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


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 ?


Nalin Pithwa