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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


But try the problem first…

Answer: is 667.

Source
Suggested Reading

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


First hint

 here coefficient of \(x^{2}\)= coefficient of \(x^{3}\) in the same expression

Second Hint

then \({2000 \choose 1998}a^{2}b^{1998}\)=\({2000 \choose 1997}a^{3}b^{1997}\)

Final Step

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


But try the problem first…

Answer: is 5.

Source
Suggested Reading

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

Try with Hints


First hint

 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=\(\frac{1}{5}\)

Second Hint

then y=29-5=24 and z=\(\frac{1}{5-x}\)=\(\frac{5}{24}\)

Final Step

\(z+\frac{1}{y}\)=\(\frac{1}{4}\) 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


But try the problem first…

Answer: is 997.

Source
Suggested Reading

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


First hint

 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

Second Hint

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

Final Step

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


But try the problem first…

Answer: is 276.

Source
Suggested Reading

AIME, 1996, Question 11

Complex Numbers from A to Z by Titu Andreescue

Try with Hints


First hint

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

Second Hint

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\(\theta\)

Final Step

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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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


But try the problem first…

Answer: is 36.

Source
Suggested Reading

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.

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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


But try the problem first…

Answer: is 169.

Source
Suggested Reading

AIME I, 1988, Question 2

Functional Equation by Venkatchala

Try with Hints


First hint

\(f_1(11)=4\)

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

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

Second Hint

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

Final Step

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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Algebra Arithmetic Math Olympiad PRMO

Natural Numbers Problem | PRMO 2019 | Question 30

Try this beautiful problem from the Pre-RMO, 2019 based on Natural Numbers.

Natural numbers Problem – PRMO 2019


Let E denote the set of all natural number n such that \(3< n<100\) and the set {1,2,3,…,n} can be partitioned in to 3 subsets with equal sums. Find the number of elements of E.

  • is 107
  • is 64
  • is 840
  • cannot be determined from the given information

Key Concepts


Divisibility

Equations

Integer

Check the Answer


But try the problem first…

Answer: is 64.

Source
Suggested Reading

PRMO, 2019, Question 30

Elementary Number Theory by David Burton

Try with Hints


First hint

{1,2,…,n}

This set can be partitioned into 3 subsets with equal sums so total sum is divisible by 3

\(\frac{n(n+1)}{2}\) is divisible by 3.

or, n of form 3k, 3k+2

or, n=6k,6k+2,6k+3, 6k+5

Second Hint

case I n=6k, we group numbers in bundles of 6 for each bundle 1,2,3,4,5,6(16,25,34)

case II n=6k+2 then we club last bundle of 8 numbers rest can be partitioned and those eight numbers can be done 1,2,3,4,5,6,7,8 (1236,48)

case III n=6k+3 we club last nine number and rest can be partitioned 1,2,3,4,5,6,7,8,9 (12345,69,78)

Final Step

case IV 6k+5 we take last five numbers, rest can be aprtitioned 1,2,3,4,,5(14,25,5)

Hence we select any number of form 6k(16), 6k+2(16), 6K+3(16), 6K+5(16)

or, total=64 numbers.

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Algebra Arithmetic Geometry Math Olympiad PRMO

Diameter of a circle | PRMO 2019 | Question 25

Try this beautiful problem from the Pre-RMO, 2019 based on Diameter of a circle.

Diameter of a circle – PRMO 2019


A village has a circular wall around it, and the wall has four gates pointing north, southeast and west. A tree stands outside the village, 16 m north of the north gate, and it can be just seen appearing on the horizon from a point 48 m east of the south gate. Find the diameter in meters of the wall that surrounds the village.

  • is 107
  • is 48
  • is 840
  • cannot be determined from the given information

Key Concepts


Pythagoras Theorem

Equations

Integer

Check the Answer


But try the problem first…

Answer: is 48.

Source
Suggested Reading

PRMO, 2019, Question 25

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

Let radius =r

or,\(AB=\sqrt{AO^{2}-OB^{2}}=\sqrt{(16+r)^{2}-r^{2}}\)

\(=\sqrt{256+32r}\)

or, \(AD^{2}+DC^{2}=CA^{2}\)

Diameter of a circle - figure

Second Hint

or, \(48^{2}+(2r+16)^{2}\)

\(=(48+\sqrt{256+32r})^{2}\)

or, \(r^{2}+8r=24\sqrt{256+32r}\)

or, \(r(r+8)=24(4\sqrt{2})(\sqrt{r+8})\)

Final Step

or, r=24

or, 2r=diameter=48.

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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


But try the problem first…

Answer: is 163.

Source
Suggested Reading

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.

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

Equations and Roots | TOMATO B.Stat Objective 123

Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Equations and Roots.

Equation and Roots ( B.Stat Objective Question )


Consider the equation of the form \(x^{2}+bx+c=0\). The number of such equations that have real roots and have coefficients b and c in the set {1,2,3,4,5,6}, (b may be equal to c), is

  • 1113
  • 18
  • 53361
  • 5082

Key Concepts


Equation

Integers

Roots

Check the Answer


But try the problem first…

Answer: 18.

Source
Suggested Reading

B.Stat Objective Problem 123

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

We know that if a quadratic equations have real roots then it’s discriminant is >=0 so here \( b^{2}-4c \geq 0\). Now we will go casewise . First we will choose a particular Value for b then check what are the values of c that satisfies the above inequality.

Second Hint

for b=2, c=1

for b=3, c=1,2

for b=4, c=1,2,3,4

for b=5, c=1,2,3,4,5

for b=6, c=1,2,3,4,5,6

Final Step

we get required number =18.

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