# 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.

.

# 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 $$y_{1}+y_{2}=25$$.

# 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.

# 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.

# 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.

# 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.

# 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.

# 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.

# 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.

# 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.