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## 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

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

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

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

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sds

Categories

## 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

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

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

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

Categories

## 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

$a_{n+1}a_{n-1}$=$a_{n}^{2}+2007$ 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

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}$

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.

Categories

## 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

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

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

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

Categories

## Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.

## Problem on Series and Sequences (SMO Test)

For each positive integer $n \geq 1$ , we define the recursive relation given by $a_{n+1} = \frac {1}{1+a_{n}}$.

Suppose that $a_{1} = a_{2012}$.Find the sum of the squares of all

possible values of $a_{1}$.

• 2
• 3
• 6
• 12

### Key Concepts

Series and Sequence

Functional Equation

Recursive Relation

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you got stuck start from here :

At first we have to understand the sequence it is following

Given that : $a_{n+1} = \frac {1}{1+a_{n}}$

Let $a_{1} = a$

so , $a_{2} = \frac {1}{1 + a_{1}}$ = $\frac {1}{1 + a}$

Again, $a_{3} = \frac {1}{1+a_{2}} = \frac {1+a}{2+a}$

For , $a_{4} = \frac {1}{1+a_{3}} = \frac {2+a}{3+2a}$

And , $a_{5} = \frac {1}{1+a_{4}} = \frac {3+2a}{5+3a}$

and so on……..

Try to do the rest …………………………….

Looking at the previous hint ………………

In general we can say ……………..

$a_{n} = \frac {F_{n} + F_{n-1}a}{F_{n+1} +F_{n}a}$

Where $F_{1} = 0 , F_{ 2} = 1$ and $F_{n+1} = F_{n}$ for all value of $n\geq 1$

Try to do the rest …….

Here is the rest of the solution,

If $a_{2012} = \frac {F_{2012}+F_{2011}a}{F_{2013} + F{2012}a} = a$

Then $(a^2+a-1 )F_{2012} = 0$

Since $F_{2012}>0$ we have $a^2 +a -1 = 0$ ……………………….(1)

Assume x and y are the two roots of the $eq^n (1)$, then

$x^2 + y^2 = (x+y)^2 -2xy = (-1)^2 – 2(-1) = 3$ (Answer)

Categories

## Trigonometry Problem | AIME I, 2015 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Trigonometry.

## Trigonometry Problem – AIME 2015

With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n), where (m) and (n) are integers greater than 1. Find (m+n).

• is 107
• is 91
• is 840
• cannot be determined from the given information

### Key Concepts

Trigonometry

Sequence

Algebra

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

## Try with Hints

Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x – \frac{1}{x}}{2i} = \frac{x^2 – 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 – 1|.]

But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ – z)$, if we let our product be $M$ then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 – 1| |x^6 – 1| |x^{10} – 1| \dots |x^{354} – 1| |x^{358} – 1|]because $\sin$ is positive in the first and second quadrants.

Now, notice that $x^2, x^6, x^{10}, \dots, x^{358}$ are the roots of $z^{90} + 1 = 0.$ Hence, we can write $(z – x^2)(z – x^6)\dots (z – x^{358}) = z^{90} + 1$, and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \dots |1 – x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that $M = 2^{89}$ and that our answer is $2+89=91$.