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## Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

## Function of Complex Numbers – AIME I, 1999

Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that $b^{2}$=$\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.

• is 107
• is 259
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Integers

Complex Numbers

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through $(\frac{1}{2},\frac{1}{2})$ that is x+y=1

putting x=(a-b) and y=(a+b) gives 2a=1 and $a=\frac{1}{2}$

and $(\frac{1}{2})^{2} +b^{2}=8^{2}$ then $b^{2}=\frac{255}{4}$ then 255+4=259.

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## A Parallelogram and a Line | AIME I, 1999 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on A Parallelogram and a Line.

## A Parallelogram and a Line – AIME I, 1999

Consider the parallelogram with vertices (10,45),(10,114),(28,153) and (28,84). A line through the origin cuts this figure into two congruent polygons. The slope of the line is $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 118
• is 840
• cannot be determined from the given information

### Key Concepts

Parallelogram

Slope of line

Integers

AIME I, 1999, Question 2

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

By construction here we see that a line makes the parallelogram into two congruent polygons gives line passes through the centre of the parallelogram

Centre of the parallogram is midpoint of (10,45) and (28,153)=(19,99)

Slope of line =$\frac{99}{19}$ then m+n=118.

Categories

## Least Positive Integer Problem | AIME I, 2000 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Least Positive Integer.

## Least Positive Integer Problem – AIME I, 2000

Find the least positive integer n such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit 0.

• is 107
• is 8
• is 840
• cannot be determined from the given information

### Key Concepts

Product

Least positive integer

Integers

AIME I, 2000, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

$10^{n}$ has factor 2 and 5

for n=1 $2^{1}$=2 $5^{1}$=5

for n=2 $2^{2}$=4 $5^{2}=25$

for n=3 $2^{3}$=8 $5^{3}=125$

……..

for n=8 $2^{8}$=256 $5^{8}=390625$

here $5^{8}$ contains the zero then n=8.

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## Equations and Complex numbers | AIME I, 2019 Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

## Equations and Complex numbers – AIME 2019

For distinct complex numbers $z_1,z_2,……,z_{673}$ the polynomial $(x-z_1)^{3}(x-z_2)^{3}…..(x-z_{673})^{3}$ can be expressed as $x^{2019}+20x^{2018}+19x^{2017}+g(x)$, where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of $|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|$ can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n

• is 107
• is 352
• is 840
• cannot be determined from the given information

### Key Concepts

Equations

Complex Numbers

Integers

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here $|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|$=s=$(z_1z_2+z_1z_3+….z_1z_{673})+(z_2z_3+z_2z_4+…+z_2z_{673})$

$+…..+(z_{672}z_{673})$ here

P=$(x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)…(x-z_{673})(x-z_{673})(x-z_{673})$

with Vieta’s formula,$z_1+z_1+z_1+z_2+z_2+z_2+…..+z_{673}+z_{673}+z_{673}$=-20 then $z_1+z_2+…..+z_{673}=\frac{-20}{3}$ the first equation and ${z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+…..$=$3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})$+$9({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})$=$3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})$+9s which is second equation

here $(z_1+z_2+…..+z_{673})^{2}=\frac{400}{9}$ from second equation then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})=\frac{400}{9}$ then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2s=\frac{400}{9}$ then $({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})=\frac{400}{9}$-2s then with second equation and with vieta s formula $3(\frac{400}{9}-2s)+9s$=19 then s=$\frac{-343}{9}$ then |s|=$\frac{343}{9}$ where 343 and 9 are relatively prime then 343+9=352.

.

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

## Patterns and Integers | AIME I, 2001 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2001 based on Patterns and integers.

## Patterns and integers – AIME I, 2001

A mail carrier delivers mail to the nineteen houses on the east side of Elm street. The carrier notices that no two adjacent houses ever get mail on the same day, but that there are never more than two houses in a row that get no mail on the same day, find the number of different patterns of mail delivery are possible.

• is 107
• is 351
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Patterns

Integers

AIME I, 2001, Question 14

Elementary Number Theory by David Burton

## Try with Hints

0=house receive no mail 1=house receives mail and last two digit is not 11 then 00, 01 and 10 let $a_{n}$=number of n digit string ending 00,$b_{n}$=number of n digit string ending 01 $c_{n}$=number of n digit string ending 10

when nth digit ends in 00 then previous digit is 1 and the last two digits in the n-1 th substring is 10 then $a_{n}=c_{n-1}$ when nth digit string ends in 01 then previous digit 0 or 1 and the last two digits in the n-1 digit substring is 00 or 10 then $b_{n}=a_{n-1}+c_{n-1}$ when nth digit string ends in 10 then previous digit 0 and the last two digit of the n-1 digit substring is 01 then $c_{n}=b_{n-1}$

$a_{n} b_{n} c_{n}$

for n=2 , 1 1 1

for n=3 , 1 2 1

for n=4 , 1 2 2

for n=5 , 2 3 2

for n=6 , 2 4 3

for n=7 , 3 5 4

for n=8 , 4 7 5

for n=9 , 5 9 7

for n-10 , 7 12 9

for n=11 , 9 16 12

for n=12, 12 21 16

for n=13 , 16 28 21

for n=14 , 21 37 28

for n=15 , 28 49 37

for n=16 , 37 65 49

for n=17 , 49 86 65

for n=18 , 65 114 86

for n=19 , 86 151 114

then $a_{19}+b_{19}+c_{19}$=86+151+114=351.

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

## Logic and speed | AIME I, 2008 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Logic and Speed.

## Logic and Speed – AIME I, 2008

Ed and Sue bike at equal and constant rates and they swim at equal and constant rates. The same way they jog at equal and constant rates. ed covers 74 kms after biking for 2 hrs, jogging for 3 hrs and swimming for 4 hrs while sue covers 91 kms after jogging for 2 hrs swimming for 3 hrs and biking for 4 hrs. Their biking jogging and swimming rates are whole numbers of km/hr, find the sum of the squares of Ed’s biking jogging and swimming rates.

• is 107
• is 314
• is 840
• cannot be determined from the given information

### Key Concepts

Logic

Speed

Integers

AIME I, 2008, Question 3

Elementary Number Theory by David Burton

## Try with Hints

Let a,b, c be biking jogging and swimming rates then 2a+3b+4c=74 first eqn and 4a+2b+3c=91 second eqn subtracting second from first eqn gives 2a-b-c=17 third eqn

third eqn multiplied by 3 + first eqn gives 8a+c=125 gives $a \leq 15$ third eqn multiplied by 4 +first eqn gives 10a-b=142 gives $a \gt 14$

then a=15 and b=8, c=5 and $a^{2} +b^{2} + c^{2}$=225+64+25=314.

Categories

## Equations and Integers | AIME I, 2008 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and integers.

## Equations and integers – AIME I, 2008

There exists unique positive integers x and y that satisfy the equation $x^{2}+84x+2008=y^{2}$

• is 107
• is 80
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Integers

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

## Try with Hints

$y^{2}=x^{2}+84x+2008=(x+42)^{2}+244$ then 244=$y^{2}-(x+42)^{2}=(y-x-42)(y+x+42)$

here 244 is even and 244=$2^{2}(61)$=$2 \times 122$ for $x,y \gt 0$

(y-x-42)=2 and (y+x+42)=122 then y+x=80 and x=18 y=62.