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## 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.

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## 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.

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## Reflection Problem | AIME I, 1988 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988, Question 14, based on Reflection.

## Reflection Problem – AIME I, 1988

Let C be the graph of xy=1 and denote by C’ the reflection of C in the line y=2x. let the equation of C’ be written in the form $12x^{2}+bxy +cy^{2}+d=0$, find the product bc.

• is 107
• is 84
• is 840
• cannot be determined from the given information

### Key Concepts

Geometry

Equation

Algebra

AIME I, 1988, Question 14

Coordinate Geometry by Loney

## Try with Hints

Let P(x,y) on C such that P'(x’,y’) on C’ where both points lie on the line perpendicular to y=2x

slope of PP’=$\frac{-1}{2}$, then $\frac{y’-y}{x’-x}$=$\frac{-1}{2}$

or, x’+2y’=x+2y

also midpoint of PP’, $(\frac{x+x’}{2},\frac{y+y’}{2})$ lies on y=2x

or, $\frac{y+y’}{2}=x+x’$

or, 2x’-y’=y-2x

solving these two equations, x=$\frac{-3x’+4y’}{5}$ and $y=\frac{4x’+3y’}{5}$

putting these points into the equation C $\frac{(-3x’+4y’)(4x’+3y’)}{25}$=1

which when expanded becomes

$12x’^{2}-7x’y’-12y’^{2}+25=0$

or, bc=(-7)(-12)=84.

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## Problem on Complex plane | AIME I, 1988| Question 11

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

## Problem on Complex Plane – AIME I, 1988

Let w_1,w_2,….,w_n be complex numbers. A line L in the complex plane is called a mean line for the points w_1,w_2,….w_n if L contains points (complex numbers) z_1,z_2, …..z_n such that $\sum_{k=1}^{n}(z_{k}-w_{k})=0$ for the numbers $w_1=32+170i, w_2=-7+64i, w_3=-9+200i, w_4=1+27i$ and $w_5=-14+43i$, there is a unique mean line with y-intercept 3. Find the slope of this mean line.

• is 107
• is 163
• is 634
• cannot be determined from the given information

### Key Concepts

Integers

Equations

Algebra

AIME I, 1988, Question 11

Elementary Algebra by Hall and Knight

## Try with Hints

$\sum_{k=1}^{5}w_k=3+504i$

and $\sum_{k-1}^{5}z_k=3+504i$

taking the numbers in the form a+bi

$\sum_{k=1}^{5}a_k=3$ and $\sum_{k=1}^{5}b_k=504$

or, y=mx+3 where $b_k=ma_k+3$ adding all 5 equations given for each k

or, 504=3m+15

or, m=163.

Categories

## Problem on Real Numbers | AIME I, 1990| Question 15

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Real Numbers.

## Problem on Real Numbers – AIME I, 1990

Find $ax^{5}+by^{5}$ if real numbers a,b,x,y satisfy the equations

ax+by=3

$ax^{2}+by^{2}=7$

$ax^{3}+by^{3}=16$

$ax^{4}+by^{4}=42$

• is 107
• is 20
• is 634
• cannot be determined from the given information

### Key Concepts

Integers

Equations

Algebra

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

## Try with Hints

Let S=x+y, P=xy

$(ax^{n}+by^{n})(x+y)$

$=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})$

or,$(ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)$ which is first equation

or,$(ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})$ which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

or, $(ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})$

or, $42S=(ax^{5}+by^{5})+P(16)$

or, $42(-14)=(ax^{5}+by^{5})+(-38)(16)$

or, $ax^{5}+by^{5}=20$.

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## Algebraic Equation | AMC-10A, 2001 | Problem 10

Try this beautiful problem from Algebra based on Algebraic Equation.

## Algebraic Equation – AMC-10A, 2001- Problem 10

If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

• $5$
• $20$
• $22$
• $25$
• $36$

### Key Concepts

algebra

Equation

sum

Answer: $22$

AMC-10A (2001) Problem 10

Pre College Mathematics

## Try with Hints

The given equations are $xy=24$ and $xz=48$.we have to find out $x+y+z$

Now using two relations we can write $\frac{xy}{xz}=\frac{24}{48}$$\Rightarrow 2y=z$

Can you now finish the problem ……….

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get $2y^2=72$ $\Rightarrow y=6$ and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Therefore The sum of three numbers are $(x+y+z)$=$4+6+12$=$22$

Categories

## Sum of whole numbers | AMC-10A, 2012 | Problem 8

Try this beautiful problem from Algebra based on Sum of whole numbers.

## Sum of whole numbers – AMC-10A, 2012- Problem 8

The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

• $8$
• $9$
• $7$
• $6$
• $5$

### Key Concepts

Algebra

Equation

Sum of digits

Answer: $7$

AMC-10A (2012) Problem 8

Pre College Mathematics

## Try with Hints

Let us assume three numbers are $x$,$y$ and $z$

Now according to the problem

$x+y=12$…………………(1)

$y+z=17$………………..(2)

$z+x=19$………………….(3).

Can you find out the values of $x$,$y$ and $z$…………….?

can you finish the problem……..

Adding three equations we get $2(x=y+z)=48$

$\Rightarrow x+y+z=24$………….(3)

Now subtract (3) from (1),we will get $z=12$ and similarly if we subtract (2) and (3) from (1) we will get $x=7$ and $y=5$

can you finish the problem……..

Therefore,the numbers are $12$, $7$, and $5$ and middle number is $7$