This is a problem from the Indian Statistical Institute, ISI BStat 2006 Subjective Problem 3 based on Sophie Germain Identity. Try to solve it. Problem: Prove that $\mathbf{n^4 + 4^{n}}$ is composite for all values of $n$ greater than $1$. Discussion: Teacher: This problem uses an identity that has a fancy name: Sophie Germain identity. […]
Problem: In the figure below, $E$ is the midpoint of the arc $ABEC$ and the segment $ED$ is perpendicular to the chord $BC$ at $D$. If the length of the chord $AB$ is $\mathbf{\ell_1} $, and that of the segment $BD$ is $\mathbf{\ell_2} $, determine the length of $DC$ in terms of $\mathbf{\ell_1, \ell_2} $. […]
Problem: Find all real solutions of the equation $sin^{5}x+cos^{3}x=1$ . Discussion: Teacher: Notice that $|\sin x| \leq 1 , |\cos x | \leq 1 $ . So if you raise $\sin x$ and $\cos x$ to higher powers you necessarily lower the value. Take for example the number $\frac{1}{2}$. If you raise that to the […]
This is problem number 5 from Indian Statistical Institute, ISI BStat 2005 based on Geometric inequality. Consider an acute angled triangle $PQR$ such that $C,I$ and $O$ are the circumcentre, incentre and orthocentre respectively. Suppose $\displaystyle{ \angle QCR, \angle QIR } $ and $ \displaystyle{ \angle QOR } $, measured in degrees, are $ \displaystyle{ […]
Gauss called it the 'fundamental theorem' and published 6 proofs of it. Since then quadratic reciprocity has been an obsession of the mathematical community. Over 200 proofs has been published. I encountered a very simple and elegant proof. Here is a pdf file with a simple 2-page proof. quadratic reciprocity
This is an objective problem 151 from TOMATO based on Consecutive composites, useful for Indian Statistical Institute Entrance Exam. Let $n = 51! + 1$. Then the number of primes among $n+1, n+2, ... , n+50$ is (A) $0$; (B) $1$; (C) $2$; (D) more than $2$; Discussion: $51!$ is divisible by $2, 3,... 51$. […]
If three prime numbers, all greater than $3$, are in A.P. , then their common difference (A) must be divisible by $2$ but not necessarily by $3$; (B) must be divisible by $3$ but not necessarily by $2$; (C) must be divisible by both $2$ and $3$; (D) need not be divisible by any of […]
Let N be a positive integer not equal to 1. Then note that none of the numbers 2, 3, ... , N is a divisor of (N! -1). From this we can conclude that: (A) (N! - 1) is a prime number; (B) at least one of the numbers N+1 , N+2 , ...., N! […]
The number $1000! = 1.2.3...1000$ ends exactly with (A) $249$ zeroes; (B) $250$ zeroes; (C) $240$ zeroes; (D) $200$ zeroes; Discussion: To find the number of zeroes at the end of n! we just need to figure out the number of 5's occurring in prime factorization of it. Why? Because there are much more 2's […]
This post consists of Problems and solutions from TIFR 2013 Paper. Try to solve them and then read their solutions. TIFR 2013 Paper PART A (Linear and Abstract Algebra) Problem 1 Problem 2 - Automorphism of the Additive Group of Rationals Problem 3 - Existence of Real Root Problem 4 - Existence of Complex Root […]