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:

Given

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}

Similarly,

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?

Cheers,

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

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

PROMYS India: Mehta Fellowships, Online July 4-Aug 14 2021 Challenging Math Training Program

PROMYS India

Program in Mathematics for Young Scientists

Dear Dr. Pithwa,

We are seeking mathematically talented students in Standards IX-XII as applicants to the Mehta Fellowships to PROMYS, a challenging summer program in mathematics. The Mehta Fellowships cover all costs of participation. PROMYS 2021 will run online from July 4 – August 14, 2021.

PROMYS is the parent program of PROMYS India whose launch has sadly been delayed by Covid-19. PROMYS has been running at Boston University in the U.S. since 1989. It will run as close to the in-person program as possible in 2021, with intense mathematical activity and social events to build community.

The 2021 Mehta Fellowship application is HERE on the PROMYS site (www.promys.org). The application deadline is March 30, 2021. Students should allow themselves plenty of time to work on the intriguing application problems. 

Successful PROMYS applicants need to show unusual enthusiasm for thinking deeply about mathematics. They do not need to have taken any advanced courses or participated in competitions. Lectures, problem sets, and mathematical discussions at PROMYS are all in English.

PROMYS is an intensive six-week program which draws talented secondary school students from across the U.S. and around the world. Each year, PROMYS creates a collaborative and supportive mathematical community of 85 pre-university students, 25 undergraduate counselors, faculty, research mentors and guest mathematicians. The focus is on developing the mathematical habits of mind that support independence and creativity in facing unfamiliar mathematical challenges. 

Approximately 50% of PROMYS alumni go on to earn doctorates. About half of these are PhDs in Mathematics.

Mathematical focus: First-year students focus primarily on a series of very challenging problem sets, a daily lecture, and exploration labs in Number Theory. Students receive daily feedback from their counselors on their Number Theory problem sets. There are also dozens of additional seminars, mini-courses, and guest lectures on a wide range of mathematical topics. Advanced seminars and mentored research are offered, and the program is individualized to ensure each student is fully challenged mathematically.

The Mehta Fellowship flyer is here. I hope you will share this information with mathematically talented students and with their teachers and schools. 

Please let me know if you have any questions.

Yours sincerely,

Professor Ila Varma
Director of PROMYS India
 “PROMYS is all about thinking hard and having fun. Thinking and exploring really helped me grow intellectually.” Vruddhi Shah, 2020 Mehta Fellow

“The path of discovery that one follows as a student of PROMYS is unique and will change your perception of mathematics,” Siddharth Sridhar, 2017 and 2018 Mehta Fellow