Categories

## Algebra and Combination | AIME I, 2000 Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebra and Combination.

## Algebra and combination – AIME 2000

In expansion $(ax+b)^{2000}$ where a and b are relatively prime positive integers the coefficient of $x^{2}$ and $x^{3}$ are equal, find a+b

• is 107
• is 667
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Combination

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

here coefficient of $x^{2}$= coefficient of $x^{3}$ in the same expression

then ${2000 \choose 1998}a^{2}b^{1998}$=${2000 \choose 1997}a^{3}b^{1997}$

then $b=\frac{1998}{3}$a=666a where a and b are relatively prime that is a=1,b=666 then a+b=666+1=667.

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## Algebraic Equation | AIME I, 2000 Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation – AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, $x+\frac{1}{z}=5$ and $y+\frac{1}{x}=29$ then $z+\frac{1}{y}$=$\frac{m}{n}$ where m and n are relatively prime, find m+n

• is 107
• is 5
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Integers

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here $x+\frac{1}{z}=5$ then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and $y=(29-\frac{1}{x}$) together gives 5-x=x$(29-\frac{1}{x}$) then x=$\frac{1}{5}$

then y=29-5=24 and z=$\frac{1}{5-x}$=$\frac{5}{24}$

$z+\frac{1}{y}$=$\frac{1}{4}$ then 1+4=5.

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## Arithmetic and geometric mean | AIME I, 2000 Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

## Arithmetic and geometric mean with Algebra – AIME 2000

Find the number of ordered pairs (x,y) of integers is it true that $0 \lt y \lt 10^{6}$ and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

• is 107
• is 997
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Ordered pair

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

given that $\frac{x+y}{2}=2+({xy})^\frac{1}{2}$ then solving we have $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and-2

given that $y \gt x$ then $y^\frac{1}{2}$-$x^\frac{1}{2}$=+2 and here maximum integer value of $y^\frac{1}{2}$=$10^{3}-1$=999 whose corresponding $x^\frac{1}{2}$=997 and decreases upto $y^\frac{1}{2}$=3 whose corresponding $x^\frac{1}{2}$=1

then number of pairs ($x^\frac{1}{2}$,$y^\frac{1}{2}$)=number of pairs of (x,y)=997.

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## Amplitude and Complex numbers | AIME I, 1996 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1996 based on Amplitude and Complex numbers.

## Amplitude and Complex numbers – AIME 1996

Let P be the product of the roots of $z^{6}+z^{4}+z^{2}+1=0$ that have a positive imaginary part and suppose that P=r(costheta+isintheta) where $0 \lt r$ and $0 \leq \theta \lt 360$ find $\theta$

• is 107
• is 276
• is 840
• cannot be determined from the given information

### Key Concepts

Equations

Complex Numbers

Integers

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here$z^{6}+z^{4}+z^{2}+1$=$z^{6}-z+z^{4}+z^{2}+z+1$=$z(z^{5}-1)+\frac{(z^{5}-1)}{(z-1)}$=$\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$ then $\frac{(z^{5}-1)(z^{2}-z+1)}{(z-1)}$=0

gives $z^{5}=1 for z\neq 1$ gives $z=cis 72,144,216,288$ and $z^{2}-z+1=0 for z \neq 1$ gives z=$\frac{1+-(-3)^\frac{1}{2}}{2}$=$cis60,300$ where cis$\theta$=cos$\theta$+isin$\theta$

taking $0 \lt theta \lt 180$ for positive imaginary roots gives cis72,60,144 and then P=cis(72+60+144)=cis276 that is theta=276.

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Categories

## Function Problem | AIME I, 1988 | Question 2

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

## Function Problem – AIME I, 1988

For any positive integer k, let $f_1(k)$ denote the square of the sum of the digits of k. For $n \geq 2$, let $f_n(k)=f_1(f_{n-1}(k))$, find $f_{1988}(11)$.

• is 107
• is 169
• is 634
• cannot be determined from the given information

### Key Concepts

Functions

Equations

Algebra

AIME I, 1988, Question 2

Functional Equation by Venkatchala

## Try with Hints

$f_1(11)=4$

or, $f_2(11)=f_1(4)=16$

or, $f_3(11)=f_1(16)=49$

or, $f_4(11)=f_1(49)=169$

or, $f_5(11)=f_1(169)=256$

or, $f_6(11)=f_1(256)=169$

or, $f_7(11)=f_1(169)=256$

This goes on between two numbers with this pattern, here 1988 is even,

or, $f_1988(11)=f_4(11)=169$.

Categories

## Equation of X and Y | AIME I, 1993 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1993 based on Equation of X and Y.

## Equation of X and Y – AIME I, 1993

Jenny and Kenny are walking in the same direction, Kenny at 3 feet per second and Jenny at 1 foot per second, on parallel paths that are 200 feet apart. A tall circular building 100 feet in diameter is centred mid way between the paths . At the instant when the building first blocks the line of sight between Jenny and Kenny, they are 200 feet apart. Let t be amount of time, in seconds, Before Jenny and Kenny, can see each other again. If t is written as a fraction in lowest terms, find the sum of numerator and denominator.

• is 107
• is 163
• is 840
• cannot be determined from the given information

### Key Concepts

Variables

Equations

Algebra

AIME I, 1993, Question 13

Elementary Algebra by Hall and Knight

## Try with Hints

Let circle be of radius 50

Let start points be (-50,100),(-50,-100) then at time t, end points (-50+t,100),(-50+3t,-100)

or, equation and equation of circle is

y=$\frac{-100}{t}x+200 -\frac{5000}{t}$ is first equation

$50^2=x^2+y^2$ is second equation

when they see again then

$\frac{-x}{y}=\frac{-100}{t}$

or, $y=\frac{xt}{100}$

solving in second equation gives $x=\frac{5000}{\sqrt{100^2+t^2}}$

or, $y=\frac{xt}{100}$

solving in first equation for t gives $t=\frac{160}{3}$

or, 160+3=163.

Categories

## Complex roots and equations | AIME I, 1994 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.

## Complex roots and equations – AIME I, 1994

$x^{10}+(13x-1)^{10}=0$ has 10 complex roots $r_1$, $\overline{r_1}$, $r_2$,$\overline{r_2}$.$r_3$,$\overline{r_3}$,$r_4$,$\overline{r_4}$,$r_5$,$\overline{r_5}$ where complex conjugates are taken, find the values of $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

• is 107
• is 850
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Complex Roots

Equation

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

here equation gives ${13-\frac{1}{x}}^{10}=(-1)$

$\Rightarrow \omega^{10}=(-1)$ for $\omega=13-\frac{1}{x}$

where $\omega=e^{i(2n\pi+\pi)(\frac{1}{10})}$ for n integer

$\Rightarrow \frac{1}{x}=13- {\omega}$

$\Rightarrow \frac{1}{(x)(\overline{x})}=(13-\omega)(13-\overline{\omega})$

=$170-13(\omega+\overline{\omega})$

adding over all terms $\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}$

=5(170)

=850.

Categories

## 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.

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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.

Categories

## Logarithm and Equations | AIME I, 2012 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on logarithm and Equations.

## Logarithm and Equations – AIME I, 2012

Let x,y,z be positive real numbers $2log_{x}(2y)$=$2log_{2x}(4z)=log_{2x^4}(8yz)\neq0$ the value of (x)($y^{5}$)(z) may be expressed in the form $\frac{1}{2^\frac{p}{q}}$ where p and q are relatively prime positive integers, find p+q.

• is 107
• is 49
• is 840
• cannot be determined from the given information

### Key Concepts

Equations

Algebra

Logarithm

AIME I, 2012, Question 9

Higher Algebra by Hall and Knight

## Try with Hints

Let $2log_{x}(2y)$=$2log_{2x}(4z)=log_{2x^4}(8yz) =2$ then from first and last term x=2y from second and last term 2x=4z and from third and last term $4x^{8}=8yz$

taking these together $4x^{8}$=(4z)(2y)=x(2x) then x=$2^\frac{-1}{6}$ then y=z=$2^\frac{-7}{6}$

(x)($y^{5}$)(z) =$2^\frac{-43}{6}$ then p+q =43+6=49.