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

First hint

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

Second Hint

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

Final Step

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

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Categories

## 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

First hint

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

Second Hint

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$\theta$

Final Step

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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## Roots of Equation | PRMO 2017 | Question 19

Try this beautiful problem from the Pre-RMO, 2017 based on roots of equation.

## Roots of equation – PRMO 2017

Suppose 1,2,3 are roots of the equation $x^{4}+ax^{2}+bx=c$. Find the value of c.

• is 107
• is 36
• is 840
• cannot be determined from the given information

### Key Concepts

Roots

Equations

Algebra

PRMO, 2017, Question 19

Higher Algebra by Hall and Knight

## Try with Hints

First hint

1,2,3 are the roots of $x^{4}+ax^{2}+bx-c=0$

Second Hint

since sum of roots=0 fourth root=-6 by using Vieta’s formula

Final Step

c=36.

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## 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

First hint

$f_1(11)=4$

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

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

Second Hint

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

Final Step

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

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

Categories

## Natural Numbers Problem | PRMO 2019 | Question 30

Try this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

## Natural numbers Problem – PRMO 2019

Let E denote the set of all natural number n such that $3< n<100$ and the set {1,2,3,…,n} can be partitioned in to 3 subsets with equal sums. Find the number of elements of E.

• is 107
• is 64
• is 840
• cannot be determined from the given information

### Key Concepts

Divisibility

Equations

Integer

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

## Try with Hints

First hint

{1,2,…,n}

This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3

$\frac{n(n+1)}{2}$ is divisible by 3.

or, n of form 3k, 3k+2

or, n=6k,6k+2,6k+3, 6k+5

Second Hint

case I n=6k, we group numbers in bundles of 6 for each bundle 1,2,3,4,5,6(16,25,34)

case II n=6k+2 then we club last bundle of 8 numbers rest can be partitioned and those eight numbers can be done 1,2,3,4,5,6,7,8 (1236,48)

case III n=6k+3 we club last nine number and rest can be partitioned 1,2,3,4,5,6,7,8,9 (12345,69,78)

Final Step

case IV 6k+5 we take last five numbers, rest can be aprtitioned 1,2,3,4,,5(14,25,5)

Hence we select any number of form 6k(16), 6k+2(16), 6K+3(16), 6K+5(16)

or, total=64 numbers.

Categories

## Diameter of a circle | PRMO 2019 | Question 25

Try this beautiful problem from the Pre-RMO, 2019 based on Diameter of a circle.

## Diameter of a circle – PRMO 2019

A village has a circular wall around it, and the wall has four gates pointing north, southeast and west. A tree stands outside the village, 16 m north of the north gate, and it can be just seen appearing on the horizon from a point 48 m east of the south gate. Find the diameter in meters of the wall that surrounds the village.

• is 107
• is 48
• is 840
• cannot be determined from the given information

### Key Concepts

Pythagoras Theorem

Equations

Integer

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

or,$AB=\sqrt{AO^{2}-OB^{2}}=\sqrt{(16+r)^{2}-r^{2}}$

$=\sqrt{256+32r}$

or, $AD^{2}+DC^{2}=CA^{2}$

Second Hint

or, $48^{2}+(2r+16)^{2}$

$=(48+\sqrt{256+32r})^{2}$

or, $r^{2}+8r=24\sqrt{256+32r}$

or, $r(r+8)=24(4\sqrt{2})(\sqrt{r+8})$

Final Step

or, r=24

or, 2r=diameter=48.

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

First hint

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

Second Hint

when they see again then

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

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

Final Step

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

## Equations and Roots | TOMATO B.Stat Objective 123

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Equations and Roots.

## Equation and Roots ( B.Stat Objective Question )

Consider the equation of the form $x^{2}+bx+c=0$. The number of such equations that have real roots and have coefficients b and c in the set {1,2,3,4,5,6}, (b may be equal to c), is

• 1113
• 18
• 53361
• 5082

### Key Concepts

Equation

Integers

Roots

B.Stat Objective Problem 123

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

We know that if a quadratic equations have real roots then it’s discriminant is >=0 so here $b^{2}-4c \geq 0$. Now we will go casewise . First we will choose a particular Value for b then check what are the values of c that satisfies the above inequality.

Second Hint

for b=2, c=1

for b=3, c=1,2

for b=4, c=1,2,3,4

for b=5, c=1,2,3,4,5

for b=6, c=1,2,3,4,5,6

Final Step

we get required number =18.