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Algebra Combinatorics Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 667.

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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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.

.

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Arithmetic Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 997.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


 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

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

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

.

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Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 276.

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


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

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\(\theta\)

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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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 169.

AIME I, 1988, Question 2

Functional Equation by Venkatchala

Try with Hints


\(f_1(11)=4\)

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

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

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

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

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

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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


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

when they see again then

\(\frac{-x}{y}=\frac{-100}{t}\)

or, \(y=\frac{xt}{100}\)

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.

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AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Complex roots and equations | AIME I, 1994 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Complex roots and equations.

Complex roots and equations – AIME I, 1994


\(x^{10}+(13x-1)^{10}=0\) has 10 complex roots \(r_1\), \(\overline{r_1}\), \(r_2\),\(\overline{r_2}\).\(r_3\),\(\overline{r_3}\),\(r_4\),\(\overline{r_4}\),\(r_5\),\(\overline{r_5}\) where complex conjugates are taken, find the values of \(\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}\)

  • is 107
  • is 850
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Complex Roots

Equation

Check the Answer


Answer: is 850.

AIME I, 1994, Question 13

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


here equation gives \({13-\frac{1}{x}}^{10}=(-1)\)

\(\Rightarrow \omega^{10}=(-1)\) for \(\omega=13-\frac{1}{x}\)

where \(\omega=e^{i(2n\pi+\pi)(\frac{1}{10})}\) for n integer

\(\Rightarrow \frac{1}{x}=13- {\omega}\)

\(\Rightarrow \frac{1}{(x)(\overline{x})}=(13-\omega)(13-\overline{\omega})\)

=\(170-13(\omega+\overline{\omega})\)

adding over all terms \(\frac{1}{(r_1)(\overline{r_1})}+\frac{1}{(r_2)(\overline{r_2})}+\frac{1}{(r_3)(\overline{r_3})}+\frac{1}{(r_4)(\overline{r_4})}+\frac{1}{(r_5)(\overline{r_5})}\)

=5(170)

=850.

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Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Equations and Complex numbers | AIME I, 2019 Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2019 based on Equations and Complex numbers.

Equations and Complex numbers – AIME 2019


For distinct complex numbers \(z_1,z_2,……,z_{673}\) the polynomial \((x-z_1)^{3}(x-z_2)^{3}…..(x-z_{673})^{3}\) can be expressed as \(x^{2019}+20x^{2018}+19x^{2017}+g(x)\), where g(x) is a polynomial with complex coefficients and with degree at most 2016. The value of \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\) can be expressed in the form \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n

  • is 107
  • is 352
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Complex Numbers

Integers

Check the Answer


Answer: is 352.

AIME, 2019, Question 10

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


here \(|\displaystyle\sum_{1 \leq j\leq k \leq 673}(z_j)(z_k)|\)=s=\((z_1z_2+z_1z_3+….z_1z_{673})+(z_2z_3+z_2z_4+…+z_2z_{673})\)

\(+…..+(z_{672}z_{673})\) here

P=\((x-z_1)(x-z_1)(x-z_1)(x-z_2)(x-z_2)(x-z_2)…(x-z_{673})(x-z_{673})(x-z_{673})\)

with Vieta’s formula,\(z_1+z_1+z_1+z_2+z_2+z_2+…..+z_{673}+z_{673}+z_{673}\)=-20 then \(z_1+z_2+…..+z_{673}=\frac{-20}{3}\) the first equation and \({z_1}^{2}+{z_1}^{2}+{z_1}^{2}+{z_1z_2}+{z_1z_2}+{z_1z_2}+…..\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+\(9({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})\)=\(3({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})\)+9s which is second equation

here \((z_1+z_2+…..+z_{673})^{2}=\frac{400}{9}\) from second equation then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2({z_1z_2}+{z_1z_3}+….+{z_{672}z_{673}})=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})+2s=\frac{400}{9}\) then \(({z_1}^{2}+{z_2}^{2}+…..+{z_{673}}^{2})=\frac{400}{9}\)-2s then with second equation and with vieta s formula \(3(\frac{400}{9}-2s)+9s\)=19 then s=\(\frac{-343}{9}\) then |s|=\(\frac{343}{9}\) where 343 and 9 are relatively prime then 343+9=352.

.

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AIME I Algebra Arithmetic Functions Math Olympiad USA Math Olympiad

Equations and Integers | AIME I, 2008 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Equations and integers.

Equations and integers – AIME I, 2008


There exists unique positive integers x and y that satisfy the equation \(x^{2}+84x+2008=y^{2}\)

  • is 107
  • is 80
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Integers

Check the Answer


Answer: is 80.

AIME I, 2008, Question 4

Elementary Number Theory by David Burton

Try with Hints


\(y^{2}=x^{2}+84x+2008=(x+42)^{2}+244\) then 244=\(y^{2}-(x+42)^{2}=(y-x-42)(y+x+42)\)

here 244 is even and 244=\(2^{2}(61)\)=\( 2 \times 122\) for \(x,y \gt 0\)

(y-x-42)=2 and (y+x+42)=122 then y+x=80 and x=18 y=62.

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AIME I Algebra Math Olympiad USA Math Olympiad

Logarithm and Equations | AIME I, 2012 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2012 based on logarithm and Equations.

Logarithm and Equations – AIME I, 2012


Let x,y,z be positive real numbers \(2log_{x}(2y)\)=\(2log_{2x}(4z)=log_{2x^4}(8yz)\neq0\) the value of (x)(\(y^{5}\))(z) may be expressed in the form \(\frac{1}{2^\frac{p}{q}}\) where p and q are relatively prime positive integers, find p+q.

  • is 107
  • is 49
  • is 840
  • cannot be determined from the given information

Key Concepts


Equations

Algebra

Logarithm

Check the Answer


Answer: is 49.

AIME I, 2012, Question 9

Higher Algebra by Hall and Knight

Try with Hints


Let \(2log_{x}(2y)\)=\(2log_{2x}(4z)=log_{2x^4}(8yz) =2\) then from first and last term x=2y from second and last term 2x=4z and from third and last term \(4x^{8}=8yz\)

taking these together \(4x^{8}\)=(4z)(2y)=x(2x) then x=\(2^\frac{-1}{6}\) then y=z=\(2^\frac{-7}{6}\)

(x)(\(y^{5}\))(z) =\(2^\frac{-43}{6}\) then p+q =43+6=49.

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