Categories
Algebra Functional Equations Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 5.

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.

.

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations Math Olympiad Math Olympiad Videos USA Math Olympiad

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

Check the Answer


Answer: is 25.

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 \(y_{1}+y_{2}=25\).

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 23.

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.

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations Math Olympiad PRMO

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

Check the Answer


Answer: is 25.

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 \(17=-(\alpha + \beta) \Rightarrow (\alpha,\beta)=(-1,-16),(-2,-15),\)

\((-8,-9)\)

min b=(-1)(-16)=16 from all products of roots possible

Final Step

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

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations Math Olympiad PRMO

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

Check the Answer


Answer: is 36.

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.

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations Math Olympiad PRMO

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

Check the Answer


Answer: is 47.

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.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 13.

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.

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


Answer:2

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

Number of roots - graph

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

Final Step

or, number of roots is 2.

Subscribe to Cheenta at Youtube


Categories
Algebra Arithmetic Functional Equations Functions I.S.I. and C.M.I. Entrance ISI Entrance Videos

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

Check the Answer


Answer:2

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.

Subscribe to Cheenta at Youtube


Categories
AIME I Algebra Arithmetic Functional Equations Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 163.

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.

Subscribe to Cheenta at Youtube