Categories

# Understand the problem

Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. If $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$, then prove that the triangle is isosceles.

##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 6 from 2016
Trigonometry

7 out of 10

##### Suggested Book
Plane Trigonometry by S.L. Loney

Do you really need a hint? Try it first!

Let $a,b,c$ be the sides of a triangle and $A,B,C$ be the angles opposite to those sides respectively. Given $\sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$

$\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B$ $\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})$ $\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$

$\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B$ $\Rightarrow b\sin A\cos B=a\cos A \sin B$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$

$\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A}$ $\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}$  [since $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$] $\Rightarrow \frac{\cos B}{\cos A}=1$ $\cos B=\cos A$ $A=B$.

$\Rightarrow \Delta ABC$ is isosceles  (Proved).

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Similar Problems

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## What is Parity in Mathematics ? 🧐

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Categories

## Understand the problem

Consider a circle of radius 6 as given in the diagram below. Let $B,C,D$ and $E$ be points on the circle such that $BD$ and $CE$, when extended, intersect at $A$. If $AD$ and $AE$ have length 5 and 4 respectively, and $DBC$ is a right angle, then show that the length  of $BC$ is $\frac{12+\sqrt{15}}{5}$.

##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 2 from 2017
Plane Geometry

##### Difficulty Level
6 out of 10

Do you really need a hint? Try it first!

Given, $AD=5, AE=4$ and $\angle DBC=90^\circ$.

As $D,B,C$ are points on the circle having radius 6.

Therefore $DC$ is the diameter of the circle

$\Rightarrow DC=6×2=12$.

Now  DC is diameter of the circle $\Rightarrow \angle DEC=90^\circ$.

Therefore $\angle DEA$ is also right angle.

The length of $DE=\sqrt{5^2-4^2}=3$.

And the length of $EC=\sqrt{12^2-3^2}=3\sqrt{15}$.

Therefore $AC=AE+EC=4+3\sqrt{15}$.

From $∆AED$ and $∆ABC$ we have,

$\frac{AD}{DE}=\frac{AC}{BC} \Rightarrow BC=\frac{AC\cdot DE}{AD}=\frac{(4+3\sqrt{15})\cdot 3}{5}$.

Therefore the length of $BC$ is $\frac{12+9\sqrt{15}}{5}$.   (Ans.)

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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## What is Parity in Mathematics ? 🧐

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Categories

## Understand the problem

Let $g : \mathbb{N} \to \mathbb{N}$ with $g(n)$ being the product of digits of $n$.

(a) Prove that $g(n)\le n$ for all $n \in \mathbb{N}$ .

(b) Find all $n \in \mathbb{N}$ , for which $n^2-12n+36=g(n)$.

##### Source of the problem
I.S.I. (Indian Statistical Institute, B.Stat, B.Math) Entrance. Subjective Problem 5 from 2017
Inequality

9 out of 10

##### Suggested Book

Challenge and Thrill of Pre-College Mathematics by V.Krishnamuthy , C.R.Pranesachar, ect.

Do you really need a hint? Try it first!

Let $n$ be a $k$ digit(s) number number , then $n$ can be written as

$n=a_0+10a_1+10^2a_2+\cdots+10^{k-1}a_{k-1}$

Where ,$a_0,a_1,…,a_{k-1}$ are digits of $n$.

$a_0,a_1,…,a_{k-1} \in [1,9]$ as the range of the function $g$ is $\mathbb{N}$

$\Rightarrow a_0,a_1,…,a_{k-1}\neq 0$ .

Now $g(n)=a_0a_1a_2\cdots a_{k-1}\le \underbrace{10\cdot10\cdot10\cdots 10}_{(k-1) times} \cdot a_{k-1}$   [Since $a_0,a_1,…,a_{k-1} \le 10$]

Equality holds when $k=1$ .

$\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}$

$\Rightarrow g(n)\le 10^{k-1}\cdot a_{k-1}+\cdots+10^2a_2+10a_1+a_0$    [Since $a_0,a_1,…,a_{k-1} >0$]

$\Rightarrow g(n)\le n$        (Proved) .

$n^2-12n+36=g(n)$

$\Rightarrow n^2-12n+36 \le n$     [Since $g(n) \le n$ ]

$\Rightarrow n^2-13n+36 \le 0$

$\Rightarrow (n-9)(n-4) \le 0$

$\Rightarrow 4\le n\le 9$

$\Rightarrow n={4,5,6,7,8,9}$ (Ans.)  .

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Triangle Problem | PRMO-2018 | Problem No-24

Try this beautiful Problem on Trigonometry from PRMO -2018.You may use sequential hints to solve the problem.

## What is Parity in Mathematics ? 🧐

Parity in Mathematics is a term which we use to express if a given integer is even or odd. It is basically depend on the remainder when we divide a number by 2. Parity can be divided into two categories – 1. Even Parity 2. Odd Parity Even Parity : If we...

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Try this Integer Problem from Number theory from PRMO 2018, Question 16 You may use sequential hints to solve the problem.

## Chessboard Problem | PRMO-2018 | Problem No-26

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Categories

## Understand the problem

Let $p_1,p_2,p_3$  be primes with $p_2\neq p_3$, such that $4+p_1p_2$ and $4+p_1p_3$ are perfect squares. Find all possible values of $p_1,p_2,p_3$.

Do you really need a hint? Try it first!

Let $4+p_1p_2=m^2$ and $4+p_1p_3=n^2$, where $m,n \in \mathbb{N}$.

$\Rightarrow p_1p_2=(m-2)(m+2)$ and $p_1p_3=(n+2)(n-2)$.

Since $p_1,p_2,p_3$ are primes with $p_2\neq p_3$ $\Rightarrow m\neq n$.

Case –I: $p_2<p_3 \Rightarrow m<n$.

Clearly, $p_1=m+2=n-2, \Rightarrow n=m+4$

$p_2=m-2$    and

$p_3=n+2=m+6$.

Therefore, $(m+2),(m-2),(m+6)$ are all prime numbers.

We see that $m=5$ satisfy the above condition. And then,$p_1=7,p_2=3,p_3=11$ .

Case-II: $p_2>p_3 \Rightarrow m>n$.

$\Rightarrow p_1=m-2=n+2 \Rightarrow m=n+4$

$p_2=m+2=n+6$   and

$p_3=n-2$.

Now $(n+2),(n+6), (n-2)$ all are primes. Again , $n=5$ satisfy this condition. Hence $p_1=7,p_2=11,p_3=3$.

Thus all possible values of $p_1,p_2,p_3$ are ( 7,3,11)  and (7,11,3).

Now need to conclude that there does not exist any more triple of prime numbers satisfying the given condition.

Consider these numbers:

$p_1=m+2,p_2=m-2,p_3=m+6$ , now the gaps between $p_1,p_2,p_3$ are given by:

$p_1-p_2=4, p_3-p_2=8$  and  $p_3-p_1=4$.

We see that for $m>9$ these three gaps cannot be 4,8 and 4 simultaneously . That is at least one of these three gaps is greater than 4 for $m>9$ . And between 1 to 9 only $m=5$ satisfy the given condition. Hence there does not exist any more triples.

## Connected Program at Cheenta

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2017. Subjective Problem no. 6.

Number Theory

8.5 out of 10

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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Categories

# Understand the problem

For $n\ge3$, determine all real solutions of the system of $n$ equations :

$x_1+x_2+\cdots+x_{n-1}=\frac{1}{x_n}$

………………………….

$x_1+x_2+\cdots+x_{i-1}+x_{i+1}+\cdots+x_{n-1}+x_n=\frac{1}{x_i}$

…………………………..

$x_2+x_3+\cdots+x_{n-1}+x_n=\frac{1}{x_1}$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Math.(Hons.) Entrance Examination 2008. Subjective Problem no. 9.

##### Topic

System of $n$ Equations

9 out of 10

##### Suggested Book

Mathematical Olympiad Challenges by Titu Andreescu & Razvan Gelca.

Do you really need a hint? Try it first!

Let $s = x_1+x_2+\cdots+x_i+\cdots +x_{n-1}+x_n$.

Then the system of $n$ equations are equivalent to $x_i^2-sx_i+1=0$ for $i=1,2,….,n$.

It follows that $x_1,x_2,…..,x_{n-1},x_n$ are solutions to the quadratic equation: $u^2-su+1=0$.                                                   $\cdots\cdots\cdots\cdots (i)$

Now we have two possible cases.

Case – I :  Two roots of the equation (i) are equal, then , all $x_i$ are equal.

i.e. , $x_1=x_2=\cdots=x_{n-1}=x_n=u$.

$\Rightarrow s=nu$.

Putting this value of $s$ in equation (i) we get, $(n-1)u^2=1$.

$\Rightarrow u=\frac{1}{\pm \sqrt{n-1}}$.

Case – II : Two roots of equation (i) are not equal. Then let these two roots are $u_1$ and $u_2$. Also let among {$x_1,x_2,……,x_{n-1},x_n$}   $k$  of them equal to $u_1$ and $(n-k)$ of them equal to $u_2$, where $0<k<n$.

In this case, we have , (a) $u_1+u_2=s$.    [Sum of roots of equation (i)]

(b) $u_1\cdot u_2=1$.    [ Product of roots of equation (i)].

Now,.   $s=x_1+x_2+\cdots+x_{n-1}+x_n=k\cdot u_1+(n-k)\cdot u_2$

$\Rightarrow s=(u_1+u_2)+(k-1)\cdot u_1+(n-k-1)\cdot u_2$

$\Rightarrow (k-1)\cdot u_1+(n-k-1)\cdot u_2=0$   [using (a)]

$\Rightarrow (k-1)\cdot u_1^2=(k+1-n)\cdot u_1\cdot u_2=k+1-n\le 0$ [using (b) and since ,$k<n\Rightarrow k+1\le n$].

$\Rightarrow u_1^2\le 0$

But $u_1\neq 0$ as the coefficient of $u^0$ is 1($\neq 0$) in equation (i).

$\Rightarrow u_1^2<0$

$\Rightarrow u_1$ is not real. So we have no real solution to the system of $n$ equations in this case.

Thus the total no. of real solutions to the system of $n$ equations is two and these two solutions are:

$x_1=x_2=\cdots=x_{n-1}=x_n=\frac{1}{\sqrt{n-1}}$      and

$x_1=x_2=\cdots=x_{n-1}=x_n=-\frac{1}{\sqrt{n-1}}$.(Ans.)

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

# Similar Problem

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Categories

## Understand the problem

For all natural numbers$n$, let

$A_n=\sqrt{2-\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}}$           ($n$ many radicals)

(a) Show that for $n\ge 2, A_n=2\sin \frac{π}{2^{n+1}}$.

(b) Hence, or otherwise, evaluate the limit

$\displaystyle\lim _{n\to \infty} 2^n A_n$ .

##### Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 6.
##### Topic
Trigonometric Substitution

##### Difficulty Level
7.5 out of 10

Do you really need a hint? Try it first!

For $n\ge 2$,

$\sqrt{2+\sqrt{2+\cdots +\sqrt{2}}}$       ($n$ many radicals)

=$\sqrt{2+\sqrt{2+\cdots +\sqrt{2+0}}}$

=$\sqrt{2+\sqrt{2+\cdots +\sqrt{2+2\cos \frac{π}{2}}}}$.

=$\sqrt{2+\sqrt{2+\cdots +\sqrt{2(1+\cos \frac{π}{2})}}}$

=$\sqrt{2+\sqrt{2+\cdots +\sqrt{4\cos^2 \frac{π}{2^2}}}}$

=$\sqrt{2+\sqrt{2+\cdots +2\cos \frac{π}{2^2}}}$. ($n-1$ many radicals)

……..    ………..     ……….      …   …….  …

…….     ………     …………    .….     ………

=$2\cos \frac{π}{2^n}$        $[n\ge 2]$.

Now $A_n=\sqrt{2-2\cos \frac{π}{2^n}}$

$\Rightarrow A_n= \sqrt{2(1-\cos \frac{π}{2^n})}$

$\Rightarrow A_n= \sqrt{4\sin^2 \frac{π}{2^{n+1}}}$

$\Rightarrow A_n= 2\sin \frac{π}{2^{n+1}}$.

Now $\displaystyle\lim_{n\to \infty} 2^nA_n=\displaystyle\lim_{n\to \infty} 2^{n+1}\cdot \sin \frac{π}{2^{n+1}}$.

Since , $n \to \infty$

$\Rightarrow 2^{n+1}\to \infty$

$\Rightarrow \frac{π}{2^{n+1}}\to 0$

Let $\frac{π}{2^{n+1}}=z \Rightarrow z \to 0$, when $n \to \infty$.

Therefore, $\displaystyle\lim_{n\to \infty} 2^n A_n$=$\displaystyle\lim_{z \to 0}\frac{\sin z}{z}\cdot π =1\times π=π$. (Ans.)

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

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## AM-GM Inequality Problem | ISI Entrance

Here is a video solution for ISI Entrance Number Theory Problems based on AM-GM Inequality Problem. Watch and Learn! Here goes the question… a, b, c, d are positive real numbers. Prove that: (1+a)(1+b)(1+c)(1+d) <= 16. We will recommend you to try the problem...

## Sum of 8 fourth powers | ISI Entrance Problem

Here is a video solution for ISI Entrance Number Theory Problems based on Sum of 8 fourth powers. Watch and Learn! Can you show that the sum of 8 fourth powers of integers never adds up to 1993? How can you solve this fourth-degree diophantine equation? Let’s...

## ISI MStat Entrance 2020 Problems and Solutions

Problems and Solutions of ISI MStat Entrance 2020 of Indian Statistical Institute.

## ISI Entrance 2020 Problems and Solutions – B.Stat & B.Math

Problems and Solutions of ISI BStat and BMath Entrance 2020 of Indian Statistical Institute.

## Testing of Hypothesis | ISI MStat 2016 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination,2016 making us realize the beautiful connection between exponential and geometric distribution and a smooth application of Central Limit Theorem.

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

Categories

## Two Similar Triangles, ISI Entrance Subjective 2018

##### Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 2.

5.5 out of 10

##### Suggested Book

‘Challenge and Thrill of Pre-College Mathematics’ by V,Krishnamurthy, C.R.Pranesachar, ect.

Do you really need a hint? Try it first!

$PQ$ and $RS$ are two chords of the circle $C$ , intersecting at the point $O$. See figure: click here.

Given $PO=3$ cm

$SO=4$ cm

$[\triangle POR]= 7 cm^2$.

From the triangles $POS$ and $QOS$ we have,

$\angle POR=\angle SOQ$ [Opposite angles]

$\angle SRP=\angle SQO$ [Angle on the same semi-circle $STP$]

$\angle QSO= \angle OPR$ [Angle on the same semi-circle $ST’P$]

Therefore the $\triangle POR$ and $\triangle SOQ$ are similar triangles .

$\frac{[\triangle POR]}{OP^2}=\frac{[\triangle SOQ]}{SO^2}.$

$\Rightarrow [\triangle SOQ]=\frac{SO^2}{PO^2}\cdot [\triangle POR]$=$\frac{4^2}{3^2}\cdot 7=12\frac{4}{9}$.(Ans.)

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Carpet Strategy in Geometry | Watch and Learn

Here is a video solution for a Problem based on Carpet Strategy in Geometry. This problem is helpful for Math Olympiad, ISI & CMI Entrance, and other math contests. Watch and Learn! Here goes the question… Suppose ABCD is a square and X is a point on BC such that...

## Bijection Principle Problem | ISI Entrance TOMATO Obj 22

Here is a video solution for a Problem based on Bijection Principle. This is an Objective question 22 from TOMATO for ISI Entrance. Watch and Learn! Here goes the question… Given that: x+y+z=10, where x, y and z are natural numbers. How many such solutions are...

## What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

Here is a video solution for a Problem based on finding the area of a quadrilateral. This question is from American Mathematics Competition, AMC 12, 2018. Watch and Learn! Here goes the question… Connect the centroids of the four triangles in a square. Can you find...

## Solving Weird Equations using Inequality | TOMATO Problem 78

Here is a video solution for ISI Entrance Number Theory Problems based on solving weird equations using Inequality. Watch and Learn! Here goes the question… Solve: 2 \cos ^{2}\left(x^{3}+x\right)=2^{x}+2^{-x} We will recommend you to try the problem yourself. Done?...

## AM-GM Inequality Problem | ISI Entrance

Here is a video solution for ISI Entrance Number Theory Problems based on AM-GM Inequality Problem. Watch and Learn! Here goes the question… a, b, c, d are positive real numbers. Prove that: (1+a)(1+b)(1+c)(1+d) <= 16. We will recommend you to try the problem...

## Sum of 8 fourth powers | ISI Entrance Problem

Here is a video solution for ISI Entrance Number Theory Problems based on Sum of 8 fourth powers. Watch and Learn! Can you show that the sum of 8 fourth powers of integers never adds up to 1993? How can you solve this fourth-degree diophantine equation? Let’s...

## ISI MStat Entrance 2020 Problems and Solutions

Problems and Solutions of ISI MStat Entrance 2020 of Indian Statistical Institute.

## ISI Entrance 2020 Problems and Solutions – B.Stat & B.Math

Problems and Solutions of ISI BStat and BMath Entrance 2020 of Indian Statistical Institute.

## Testing of Hypothesis | ISI MStat 2016 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination,2016 making us realize the beautiful connection between exponential and geometric distribution and a smooth application of Central Limit Theorem.

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

Categories

## Understand the problem

Let $a,b,c \in \mathbb{N}$ be such that
$a^2+b^2=c^2$ and $c-b=1$.
Prove that
(i) $a$ is odd,
(ii) $b$ is divisible by 4,
(iii) $a^b+b^a$ is divisible by $c$.

##### Source of the problem
I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2018. Subjective Problem no. 7.
Number Theory

8 out of 10

##### Suggested Book

Elementary Number Theory‘ by David M. Burton
‘Challenge and Thrill of Pre-College Mathematics’ by V,Krishnamurthy, C.R.Pranesachar, ect.

Do you really need a hint? Try it first!

$a^2+b^2=c^2=(b+1)^2=b^2+2b+1$

$\Rightarrow a^2=2b+1$ Which implies $a^2$ is odd integer.

$\Rightarrow a$ is also an odd integer =$2k+1$ (say).

$(2k+1)^2+b^2=c^2$

$\Rightarrow 4k^2+4k+1+b^2=b^2+2b+1$

$\Rightarrow b=2k(k+1)$.

Now $k(k+1)$ is always even =$2l$ (say).

Therefore, $b=4l$, i.e. $b$ is divisible by 4.

$a^b+b^a=a^{4l}+b^{2k+1}=(a^2)^{2l}+(c-1)^{2k+1}$

=$(2b+1)^{2l}+(c-1)^{2k+1}$

=$(2c-1)^{2l}+(c-1)^{2k+1}.$

Now $2l$ is even , therefore , $(2c-1)^{2l}$ is of the form : $2cp+1$ where $p \in \mathbb{N}$.

And $2k+1$ is odd , therefore $(c-1)^{2k+1}$ is of the form : $cq-1$, where $q \in \mathbb{N}$.

Therefore $a^b+b^a=(2cp+1)+(cq-1)=c\cdot (2p+q)$.

$\Rightarrow c|a^b+b^a$.

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Carpet Strategy in Geometry | Watch and Learn

Here is a video solution for a Problem based on Carpet Strategy in Geometry. This problem is helpful for Math Olympiad, ISI & CMI Entrance, and other math contests. Watch and Learn! Here goes the question… Suppose ABCD is a square and X is a point on BC such that...

## Bijection Principle Problem | ISI Entrance TOMATO Obj 22

Here is a video solution for a Problem based on Bijection Principle. This is an Objective question 22 from TOMATO for ISI Entrance. Watch and Learn! Here goes the question… Given that: x+y+z=10, where x, y and z are natural numbers. How many such solutions are...

## What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

Here is a video solution for a Problem based on finding the area of a quadrilateral. This question is from American Mathematics Competition, AMC 12, 2018. Watch and Learn! Here goes the question… Connect the centroids of the four triangles in a square. Can you find...

## Solving Weird Equations using Inequality | TOMATO Problem 78

Here is a video solution for ISI Entrance Number Theory Problems based on solving weird equations using Inequality. Watch and Learn! Here goes the question… Solve: 2 \cos ^{2}\left(x^{3}+x\right)=2^{x}+2^{-x} We will recommend you to try the problem yourself. Done?...

## AM-GM Inequality Problem | ISI Entrance

Here is a video solution for ISI Entrance Number Theory Problems based on AM-GM Inequality Problem. Watch and Learn! Here goes the question… a, b, c, d are positive real numbers. Prove that: (1+a)(1+b)(1+c)(1+d) <= 16. We will recommend you to try the problem...

## Sum of 8 fourth powers | ISI Entrance Problem

Here is a video solution for ISI Entrance Number Theory Problems based on Sum of 8 fourth powers. Watch and Learn! Can you show that the sum of 8 fourth powers of integers never adds up to 1993? How can you solve this fourth-degree diophantine equation? Let’s...

## ISI MStat Entrance 2020 Problems and Solutions

Problems and Solutions of ISI MStat Entrance 2020 of Indian Statistical Institute.

## ISI Entrance 2020 Problems and Solutions – B.Stat & B.Math

Problems and Solutions of ISI BStat and BMath Entrance 2020 of Indian Statistical Institute.

## Testing of Hypothesis | ISI MStat 2016 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination,2016 making us realize the beautiful connection between exponential and geometric distribution and a smooth application of Central Limit Theorem.

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !

Categories

## Understand the problem

Prove that the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r>1$ ,are of the form $n=2^l$ for some $l\ge 0$.

##### Source of the problem

I.S.I. (Indian Statistical Institute) B.Stat/B.Math Entrance Examination 2019. Subjective Problem no. 1.

Number Theory

8.5 out of 10

##### Suggested Book

Do you really need a hint? Try it first!

Claim: Any positive integer $n$ can be written as $n=2^k\cdot m$ , where $k\ge0$ and $m$ is an odd positive integer.

To prove this claim use the fact :  $n=2^k\cdot p_1^{k_1}\cdot p_2^{k_2}\cdots p_i^{k_i}$, where $k$ and all $k_i$ are non-negetive integers and all $p_i$ are odd primes.

The sum of (any) $r$ consecutive positive integers is given by,

$(q+1)+(q+2)+(q+3)+\cdots+(q+r)$

=   $q\cdot r+(1+2+3+\cdots+r)$

=   $q\cdot r+\frac{r(r+1)}{2}$.

Equating this sum to $n$ we get,

$2^k\cdot m = q\cdot r+\frac{r(r+1)}{2}$

Or,     $2^{k+1}\cdot m = 2q\cdot r+r(r+1)$

Or,     $2^l\cdot m = r(2q+r+1)$ , where $l=k+1\ge1$.

In particular if we take $n=2^l$ then $m$ is equal to 1.

Since both $r$ and $(2q+r+1)$ are greater than 1, so they can’t be equal to $m$ in this case. Again one of $r, (2q+r+1)$ is odd integer which implies the product $r(2q+r+1)$ can’t be equal to $2^l$.

$\Rightarrow 2^l \neq r(2q+r+1)$

$\Rightarrow 2^l$ can’t be expressed as the sum of $r$ consecutive positive integers with $r>1$ and $l\ge 1$.

Now, $n=2^0=1$ also can’t be written in the same manner when $l=0$. Therefore the positive integers $n$ that cannot be written as a sum of $r$ consecutive positive integers, with $r>1$ , are of the form $n=2^l$ for some $l\ge 0$.

To show that any number other than of the form $2^l$ is sum of consecutive integers:
Observe that for $n=2^l.m$, where $m$ is an odd number greater than 1, there are two cases:
1. $m < 2^l$
Select $r$ and $q$ such that $r = m$ and $2q+r+1 = 2^l$.

2. $m > 2^l$
Select $r$ and $q$ such that $2q+r+1 = m$ and $r = 2^l$.

(QED).

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

The entrances to these programs are far more challenging than usual engineering entrances. Cheenta offers an intense, problem-driven program for these two entrances.

## Carpet Strategy in Geometry | Watch and Learn

Here is a video solution for a Problem based on Carpet Strategy in Geometry. This problem is helpful for Math Olympiad, ISI & CMI Entrance, and other math contests. Watch and Learn! Here goes the question… Suppose ABCD is a square and X is a point on BC such that...

## Bijection Principle Problem | ISI Entrance TOMATO Obj 22

Here is a video solution for a Problem based on Bijection Principle. This is an Objective question 22 from TOMATO for ISI Entrance. Watch and Learn! Here goes the question… Given that: x+y+z=10, where x, y and z are natural numbers. How many such solutions are...

## What is the Area of Quadrilateral? | AMC 12 2018 | Problem 13

Here is a video solution for a Problem based on finding the area of a quadrilateral. This question is from American Mathematics Competition, AMC 12, 2018. Watch and Learn! Here goes the question… Connect the centroids of the four triangles in a square. Can you find...

## Solving Weird Equations using Inequality | TOMATO Problem 78

Here is a video solution for ISI Entrance Number Theory Problems based on solving weird equations using Inequality. Watch and Learn! Here goes the question… Solve: 2 \cos ^{2}\left(x^{3}+x\right)=2^{x}+2^{-x} We will recommend you to try the problem yourself. Done?...

## AM-GM Inequality Problem | ISI Entrance

Here is a video solution for ISI Entrance Number Theory Problems based on AM-GM Inequality Problem. Watch and Learn! Here goes the question… a, b, c, d are positive real numbers. Prove that: (1+a)(1+b)(1+c)(1+d) <= 16. We will recommend you to try the problem...

## Sum of 8 fourth powers | ISI Entrance Problem

Here is a video solution for ISI Entrance Number Theory Problems based on Sum of 8 fourth powers. Watch and Learn! Can you show that the sum of 8 fourth powers of integers never adds up to 1993? How can you solve this fourth-degree diophantine equation? Let’s...

## ISI MStat Entrance 2020 Problems and Solutions

Problems and Solutions of ISI MStat Entrance 2020 of Indian Statistical Institute.

## ISI Entrance 2020 Problems and Solutions – B.Stat & B.Math

Problems and Solutions of ISI BStat and BMath Entrance 2020 of Indian Statistical Institute.

## Testing of Hypothesis | ISI MStat 2016 PSB Problem 9

This is a problem from the ISI MStat Entrance Examination,2016 making us realize the beautiful connection between exponential and geometric distribution and a smooth application of Central Limit Theorem.

## ISI MStat PSB 2006 Problem 8 | Bernoullian Beauty

This is a very simple and regular sample problem from ISI MStat PSB 2009 Problem 8. It It is based on testing the nature of the mean of Exponential distribution. Give it a Try it !