Categories

## 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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## Logarithms and Equations | AIME I, 2000 | Question 9

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

## Logarithms and Equations – AIME I 2000

$log_{10}(2000xy)-log_{10}xlog_{10}y=4$ and $log_{10}(2yz)-(log_{10}y)(log_{10}z)=1$ and $log_{10}(zx)-(log_{10}z)(log_{10}x)=0$ has two solutions $(x_{1},y_{1},z_{1}) and (x_{2},y_{2},z_{2})$ find $y_{1}+y_{2}$.

• is 905
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Logarithms

Theory of Equations

Number Theory

AIME I, 2000, Question 9

Polynomials by Barbeau

## Try with Hints

First hint

Rearranging equations we get $-logxlogy+logx+logy-1=3-log2000$ and $-logylogz+logy+logz-1=-log2$ and $-logxlogz+logx+logz-1=-1$

Second Hint

taking p, q, r as logx, logy and logz, $(p-1)(q-1)=log2$ and $(q-1)(r-1)=log2$ and $(p-1)(r-1)=1$ which is first system of equations and multiplying the first three equations of the first system gives $(p-1)^{2}(q-1)^{2}(r-1)^{2}=(log 2)^{2}$ gives $(p-1)(q-1)(r-1)=+-(log2)$ which is second equation

Final Step

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives $y_{1}=20$,$y_{2}=5$ then $y_{1}+y_{2}=25$.

Categories

## Roots of Equation and Vieta’s formula | AIME I, 1996 Problem 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

## Roots of Equation and Vieta’s formula – AIME I, 1996

Suppose that the roots of $x^{3}+3x^{2}+4x-11=0$ are a,b and c and that the roots of $x^{3}+rx^{2}+sx+t=0$ are a+b,b+c and c+a, find t.

• is 107
• is 23
• is 840
• cannot be determined from the given information

### Key Concepts

Functions

Roots of Equation

Vieta s formula

AIME I, 1996, Question 5

Polynomials by Barbeau

## Try with Hints

With Vieta s formula

$f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0$

$\Rightarrow a+b+c=-3$, $ab+bc+ca=4$ and $abc=11$

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

$\Rightarrow t=-(p-c)(p-a)(p-b)$

$\Rightarrow t=-f(p)=-f(-3)$

$t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]$

=23.

Categories

## Roots and coefficients of equations | PRMO 2017 | Question 4

Try this beautiful problem from the PRMO, 2017 based on Roots and coefficients of equations.

## Roots and coefficients of equations – PRMO 2017

Let a,b be integers such that all the roots of the equation $(x^{2}+ax+20)(x^{2}+17x+b)$=0 are negetive integers, find the smallest possible values of a+b.

• is 107
• is 25
• is 840
• cannot be determined from the given information

### Key Concepts

Polynomials

Roots

Coefficients

PRMO, 2017, Question 4

Polynomials by Barbeau

## Try with Hints

First hint

$(x^{2}+ax+20)(x^{2}+17x+b)$

where sum of roots $\lt$ 0 and product $\gt 0$ for each quadratic equation $x^{2}$+ax+20=0 and

$(x^{2}+17x+b)=0$

$a \gt 0$, $b \gt 0$

now using vieta’s formula on each quadratic equation $x^{2}$+ax+20=0 and $(x^{2}+17x+b)=0$, to get possible roots of $x^{2}$+ax+20=0 from product of roots equation $20=(1 \times 20), (2 \times 10), (4 \times 5)$

min a=4+5=9 from all sum of roots possible

Second Hint

again using vieta’s formula, to get possible roots of $(x^{2}$+17x+b)=0 from sum of roots equation $17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),$

$(-8,-9)$

Final Step

$(a+b)_{min}=a_{min}+b_{min}$=9+16=25.

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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.

Categories

## Solving Equation | PRMO 2017 | Question 23

Try this beautiful problem from the Pre-RMO, 2017, Question 23, based on Solving Equation.

## Solving Equation – PRMO 2017, Question 23

Suppose an integer r, a natural number n and a prime number p satisfy the equation $7x^{2}-44x+12=p^{n}$. Find the largest value of p.

• is 107
• is 47
• is 840
• cannot be determined from the given information

### Key Concepts

Equation

Algebra

Integers

PRMO, 2017, Question 23

Higher Algebra by Hall and Knight

## Try with Hints

First hint

$7x^{2}-44x+12=p^{n}$

or, $7x^{2}-42x-2x+12=p^{n}$

or, $(7x-2)(x-6)=p^{n}$

Second Hint

or, $7x-2=p^{\alpha}$, $x-6=p^{\beta}$

or, $(7x-2)-7(x-6)$=$p^{\alpha}-7p^{\beta}$

$40=p^{\alpha}-7p^{\beta}$

Final Step

If ${\alpha}, {\beta}$ are natural numbers, p is a divisor of 40

or, p=2 or 5

If p=2, 40=$2^{\alpha}-7(2^{\beta})$ or, $(2^{3})(5)$=$2^{\alpha}-7(2^{\beta})$

or, ${\beta}$=3 and $2^{\alpha}$=40+56

or, ${\alpha}$ not an integer hence not possible

If p=5 then 40=$5^{\alpha}-7(5^{\beta})$

or, $(2)^{3}(5)=5^{\alpha}-7(5^{\beta})$

or, ${\beta}=1$ and $5^{\alpha}$=40+35

or, ${\alpha}$ not an integer hence not possible

so ${\beta}$=0 or, $p^{\alpha}$=47

or, p=47 and ${\alpha}$=1.

Categories

## Positive solution | AIME I, 1990 | Question 4

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

## Positive solution – AIME I, 1990

Find the positive solution to

$\frac{1}{x^{2}-10x-29}+\frac{1}{x^{2}-10x-45}-\frac{2}{x^{2}-10x-69}=0$

• is 107
• is 13
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Divisibility

Algebra

AIME I, 1990, Question 4

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

here we put $x^{2}-10x-29=p$

$\frac{1}{p}+\frac{1}{p-16}-\frac{2}{p-40}=0$

Second Hint

or, (p-16)(p-40)+p(p-40)-2p(p-16)=0

or, -64p+(40)(16)=0

or, p=10

Final Step

or, 10=$x^{2}-10x-29$

or, (x-13)(x+3)=0

or, x=13 positive solution.

Categories

## Number of roots Problem | TOMATO B.Stat Objective 712

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of roots.

## Number of roots – B.Stat Objective Problem

The number of roots of the equation $xsinx=1$ in the interval $[0,{2\pi}]$ is

• 0
• 2
• 53361
• 5082

### Key Concepts

Equation

Roots

Algebra

B.Stat Objective Problem 712

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

$f(x)=xsinx-1=0$

which can be written as sinx=$\frac{1}{x}$ and f(x) has solution at those points where sinx and $\frac{1}{x}$ intersects

So let us draw the graph

here we see two graphs y=sin x and y=$\frac{1}{x}$

Second Hint

both graphs intersect at two points between $(0,2\pi]$

Final Step

or, number of roots is 2.

Categories

## Roots of Equation | TOMATO B.Stat Objective 711

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Roots of Equation.

## Roots of Equations (B.Stat Objective Question )

The number of roots of the equation $x^2+sin^2{x}-1$ in the closed interval $[0,\frac{\pi}{2}]$ is

• 0
• 2
• 53361
• 5082

### Key Concepts

Equation

Roots

Algebra

B.Stat Objective Problem 711

Challenges and Thrills of Pre-College Mathematics by University Press

## Try with Hints

First hint

$x^2+sin^2{x}-1=0$

$\Rightarrow x^{2}=cos^{2}x$

we draw two graphs $y=x^{2} and y=cos^{2}x$

where intersecting point gives solution now we look for intersecting points

Second Hint

we get two intersecting points

Final Step

so number of roots is 2.

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.