Categories

## 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

## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

## Sequence and fraction – AIME I, 2000

A sequence of numbers $x_1,x_2,….,x_{100}$ has the property that, for every integer k between 1 and 100, inclusive, the number $x_k$ is k less than the sum of the other 99 numbers, given that $x_{50}=\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 173
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

## Try with Hints

Let S be the sum of the sequence $x_k$

given that $x_k=S-x_k-k$ for any k

$100S-2(x_1+x_2+….+x_{100})=1+2+….+100$

$\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050$

$\Rightarrow S=\frac{2525}{49}$

for $k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}$

$\Rightarrow x_{50}=\frac{75}{98}$

$\Rightarrow m+n$=75+98

=173.

Categories

## Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

## Sequence and greatest integer – AIME I, 2000

Let S be the sum of all numbers of the form $\frac{a}{b}$,where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed $\frac{S}{10}$.

• is 107
• is 248
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

## Try with Hints

We have 1000=(2)(2)(2)(5)(5)(5) and $\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3$

sum of all numbers of form $\frac{a}{b}$ such that a and b are relatively prime positive divisors of 1000

=$(2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})$

$\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times$ $\frac{(5^{-3})(5^{7}-1)}{5-1}$

=2480 + $\frac{437}{1000}$

$\Rightarrow [\frac{s}{10}]$=248.

Categories

## Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

## Series and sum – AIME I, 1999

given that $\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}$ where angles are measured in degrees, m and n are relatively prime positive integer that satisfy $\frac{m}{n} \lt 90$, find m+n.

• is 107
• is 177
• is 840
• cannot be determined from the given information

### Key Concepts

Angles

Triangles

Side Length

AIME I, 2009, Question 5

Plane Trigonometry by Loney

## Try with Hints

s=$\displaystyle\sum_{k=1}^{35}sin5k$

s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$

$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.

Categories

## Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

## Fibonacci sequence Problem – AIME I, 1988

Find a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.

• is 107
• is 987
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Sets

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

## Try with Hints

Let F(x)=$ax^{17}+bx^{16}+1$

Let P(x) be polynomial such that

$P(x)(x^{2}-x-1)=F(x)$

constant term of P(x) =(-1)

now $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$ where $c_{i}$=coefficient

comparing the coefficients of x we get the terms

since F(x) has no x term, then $c_{15}$=1

getting $c_{14}$

$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$

=terms +$0x^{2}$ +terms

or, $c_{14}=-2$

proceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number

or, a=987.

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

## Number and Series | Number Theory | AIME I, 2015

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.

## Number Theory and Series – AIME 2015

The expressions A = $(1 \times 2)+(3 \times 4)+….+(35 \times 36)+37$ and B = $1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.

• is 107
• is 648
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2015, Question 1

Elementary Number Theory by David Burton

## Try with Hints

B-A=$-36+(2 \times 3)+….+(2 \times 36)$

=$-36+4 \times (1+2+3+….+18)$

=$-36+(4 \times \frac{18 \times 19}{2})$=648.