# Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

## Sequence and permutations - AIME II, 2015

Call a permutation $$a_1,a_2,....,a_n$$ of the integers 1,2,...,n quasi increasing if $$a_k \leq a_{k+1} +2$$ for each $$1 \leq k \leq n-1$$, find the number of quasi increasing permutations of the integers 1,2,....,7.

• is 107
• is 486
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Permutations

Integers

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

## Try with Hints

First hint

While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Second Hint

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 $$\times$$ number of permutations of (n-1) elements

or, number of permutations for n elements=$$3^{2}$$ number of permutations of (n-2) elements

......

or, number of permutations for n elements=$$3^{n-2}$$ number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 $$\times$$ $$3^{n-2}$$

forming recurrence relation as the number of permutations =2 $$\times$$ $$3^{n-2}$$

for n=3 all six permutations taken and go up 18, 54, 162, 486

Final Step

for n=7, here $$2 \times 3^{5} =486.$$

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sds

# ISI MStat 2018 PSA Problem 11 | Sequence & it's subsequence

This is a problem from ISI MStat 2018 PSA Problem 11 based on Sequence and subsequence.

## Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11

Let $${a_{n}}_{n \geq 1}$$ be a sequence such that $$a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots$$
Suppose the subsequence $${a_{2 n}}_{n \geq 1}$$ is bounded. Then

• (A) $$\{a_{2 n}\}_{n \geq 1}$$ is always convergent but $$\{a_{2 n+1}\}_{n \geq 1}$$ need not be convergent.
• (B) both $$\{a_{2 n}\}_{n \geq 1}$$ and $$\{a_{2 n+1}\}_{n \geq 1}$$ are always convergent and have the same limit.
• (C) $$\{a_{3 n}\}_{n \geq 1}$$ is not necessarily convergent.
• (D) both $$\{a_{2 n}\}_{n \geq 1}$$ and $$\{a_{2 n+1}\}_{n \geq 1}$$ are always convergent but may have different limits.

### Key Concepts

Sequence

Subsequence

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Given that $$a_{2n}$$ is bounded . Again we have $$a_{1} \leq a_{2} \leq \cdots \leq a_{2n} \leq a_{2n+1} \leq a_{2n+2} \leq \cdots$$ which shows that if $$a_{2n}$$ is bounded then $$a_{2n+1}$$ is also bounded .

Again both $$a_{2n}$$ and $$a_{2n+1}$$ both are monotonic sequence . Hence both converges .

Now we have to see whether they converges to same limit or not ?

As both $$a_{2n}$$ and $$a_{2n+1}$$ are bounded hence $$a_{n}$$ is bounded and it's already given that it is monotonic . Hence $$a_{n}$$ converges . So, it's subsequences must converges to same limit . Hence option (B) is correct .

# ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

This is a problem from ISI MStat 2018 PSA Problem 12 based on Sequence of positive numbers

## Sequence of positive numbers - ISI MStat Year 2018 PSA Question 12

Let $$a_n$$ ,$$n \ge 1$$ be a sequence of positive numbers such that $$a_{n+1} \leq a_{n}$$ for all n, and $$\lim {n \rightarrow \infty} a{n}=a .$$ Let $$p_{n}(x)$$ be the polynomial $$p_{n}(x)=x^{2}+a_{n} x+1$$ and suppose $$p_{n}(x)$$ has no real roots for every n . Let $$\alpha$$ and $$\beta$$ be the roots of the polynomial $$p(x)=x^{2}+a x+1 .$$ What can you say about $$(\alpha, \beta)$$?

• (A) $$\alpha=\beta, \alpha$$ and $$\beta$$ are not real
• (B) $$\alpha=\beta, \alpha$$ and $$\beta$$ are real.
• (C) $$\alpha \neq \beta, \alpha$$ and $$\beta$$ are real.
• (D) $$\alpha \neq \beta, \alpha$$ and $$\beta$$ are not real

### Key Concepts

Sequence

Discriminant

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

## Try with Hints

Write the discriminant. Use the properties of the sequence $$a_n$$ .

Note that as  has no real root so discriminant is  so  and 's are positive and decreasing so  . So , what can we say about a ?

Therefore we can say that $$0 \le a < 2$$ hence discriminant of P  , $$a^2-4$$ must be strictly negative so option D.

# Telescopic Continuity | ISI MStat 2015 PSB Problem 1

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

## Problem- Telescopic Continuity

Let $$f: R \rightarrow R$$ be a function which is continuous at 0 and $$f(0)=1$$
Also assume that $$f$$ satisfies the following relation for all $$x$$ :
$$f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x$$ Find $$f(3)$$.

## Solution

$$f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3$$

$$f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}$$

$$f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}$$

$$\cdots$$

$$f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}$$

Add them all up. That's the telescopic elegance.

$$f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]$$

Observe that $$a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1$$ since, $$f(x)$$ is continuous at $$x=0$$.

Hence take limit $$n \to \infty$$ on $$[*]$$, and we get $$f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15$$.

## Food for Thought

• Find the general function from the given condition.
• $$f(x)-f(\frac{x}{2})=g(x)$$ and g(x) is continuous, then prove that $$g(0) =0$$.
• What if $$g(x)$$ is not continuous?

# Combination of Sequence | B.Stat Objective | TOMATO 79

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Combination of Sequence.

## Logic and combination of sequence (B.Stat Objective)

The two sequence of numbers {1,4,16,64,.....} and {3,12,48,192,.....} are mixed as follows {1,3,4,12,16,48,64,192,....}. One of the numbers in the mixed series is 1048576. Then the number immediately preceeding it is

• 262144
• 786432
• 814572
• 786516

### Key Concepts

Logic

Sequence

Integers

B.Stat Objective Question 79

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

The first series is of form $$4^{r}$$ for $$r \geq 0$$ $$r \in$$ set of natural numbers the second series is of form $$3 \times 4^{r}$$ for $$r \geq 0$$ $$r \in$$ set of natural numbers and the third series is of $$4^{r}$$,$$3 \times 4^{r}$$ in alternate element form for $$r \geq 0$$ $$r \in$$ set of natural numbers

Second Hint

given that 1048576=$$4^{r}$$=$$4^{10}$$

Final Step

then preceeding term $$3 \times 4^{9}$$=(3)(262144)=786432.

# Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

## Geometric Sequence Problem - AIME 2009

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

• is 500
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Series

Real Analysis

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

## Try with Hints

First hint

3-digit sequence a, ar, $$ar^{2}$$. The largest geometric number must have a<=9.

Second Hint

ar $$ar^{2}$$ less than 9 r fraction less than 1 For a=9 is $$\frac{2}{3}$$ then number 964.

Final Step

a>=1 ar and $$ar^{2}$$ greater than 1 r is 2 and number is 124. Then difference 964-124=840.

# Composite number Problem | B.Stat Objective | TOMATO 75

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Sequence and composite number.

## Composite number Problem (B.Stat Objective)

Consider the sequence $$a_1$$=101, $$a_2$$=10101,$$a_3$$=1010101 and so on. Then $$a_k$$ is a composite number ( that is not a prime number)

• if and only if $$k \geq 2$$ and $$11|(10^{k+1}+1)$$
• if and only if $$k \geq 2$$ and k-2 is divisible by 3
• if and only if $$k \geq 2$$ and $$11|(10^{k+1}-1)$$
• if and only if $$k \geq 2$$

### Key Concepts

Logic

Sequence

Composite number

Answer: if and only if $$k \geq 2$$ and k-2 is divisible by 3

B.Stat Objective Question 75

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

for $$a_k$$ $$k \geq 2$$ may be prime also then not considering this here

Second Hint

for $$a_{8}$$ $$10^{9}-1$$ and $$10^{9}+1$$ not divisible by 11

Final Step

8-2 is divisible by 3 and $$a_{8}$$ is composite number then $$a_{k}$$ is composite if and only if $$k \geq 2$$ and k-2 is divisible by 3.

# Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

## Sequence and Integers - AIME I, 2007

A sequence is defined over non negetive integral indexes in the following way $$a_0=a_1=3$$, $$a_{n+1}a_{n-1}=a_n^{2}+2007$$, find the greatest integer that does not exceed $$\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}$$

• is 107
• is 224
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Inequalities

Integers

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

## Try with Hints

First hint

$$a_{n+1}a_{n-1}$$=$$a_{n}^{2}+2007$$ then $$a_{n-1}^{2} +2007 =a_{n}a_{n-2}$$ adding these $$\frac{a_{n-1}+a_{n+1}}{a_{n}}$$=$$\frac{a_{n}+a_{n-2}}{a_{n-1}}$$, let $$b_{j}$$=$$\frac{a_{j}}{a_{j-1}}$$ then $$b_{n+1} + \frac{1}{b_{n}}$$=$$b_{n}+\frac{1}{b_{n-1}}$$ then $$b_{2007} + \frac{1}{b_{2006}}$$=$$b_{3}+\frac{1}{b_{2}}$$=225

Second Hint

here $$\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}$$=$$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$ then $$b_{2007}$$=$$\frac{a_{2007}}{a_{2006}}$$=$$\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}$$$$\gt$$$$\frac{a_{2006}}{a_{2005}}$$=$$b_{2006}$$

Final Step

then $$b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}$$=225 which is small less such that all $$b_{j}$$ s are greater than 1 then $$\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}$$=$$b_{2007}+\frac{1}{b_{2007}}$$=224.

# GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

## GCD and Sequence - AIME I, 1985

The numbers in the sequence 101, 104,109,116,.....are of the form $$a_n=100+n^{2}$$ where n=1,2,3,-------, for each n, let $$d_n$$ be the greatest common divisor of $$a_n$$ and $$a_{n+1}$$, find the maximum value of $$d_n$$ as n ranges through the positive integers.

• is 107
• is 401
• is 840
• cannot be determined from the given information

### Key Concepts

GCD

Sequence

Integers

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

## Try with Hints

First hint

$$a_n=100+n^{2}$$ $$a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1$$ and $$a_{n+1}-a_{n}=2n +1$$

Second Hint

$$d_{n}|(2n+1)$$ and $$d_{n}|(100 +n^{2})$$ then $$d_{n}|[(100+n^{2})-100(2n+1)]$$ then $$d_{n}|(n^{2}-200n)$$

Final Step

here $$n^{2} -200n=0$$ then n=200 then $$d_{n}$$=2n+1=2(200)+1=401.

# Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

## Problem on Series | SMO Test

Given that $$x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$$ where each $$a_r$$ is an integer , $$r = 0,1,2,...,n$$

Find the value of n such that $$a_0 +a_1 +a_2+a_3 +...........+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}$$?

• 2
• 5
• 6
• 0

### Key Concepts

Series Problem

Algebra

Challenges and Thrills - Pre - college Mathematics

## Try with Hints

If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

$$x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n$$

The right hand side of this equation will be

$$a_0 +a_1 + a_2 +a_3+..................+ a_n$$

and the left hand side of the given equation be like

$$1 + 2^2 + 2^ 3 + ......+2^n$$

$$1 + 2^2 + 2^ 3 + ......+2^n$$ = $$a_0 +a_1 + a_2 +a_3+..................+ a_n$$

Try the rest of the sum..............

In the next hint we continue from the previous hint:

so the expression , $$1 + 2^2 + 2^ 3 + ......+2^n$$ = $$2^{n+1} - 3$$

Again , $$a_1 = 1+2+3+..........+n = \frac {n(n+1)}{2}$$

so $$a_n = 1$$

Now we have almost reach the final step.I am not showing it now.Try first.........

Now coming back to the last step :

$$60 - \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} - 3$$