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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

Sequence and fraction – AIME I, 2000


A sequence of numbers \(x_1,x_2,….,x_{100}\) has the property that, for every integer k between 1 and 100, inclusive, the number \(x_k\) is k less than the sum of the other 99 numbers, given that \(x_{50}=\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 173
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

Try with Hints


Let S be the sum of the sequence \(x_k\)

given that \(x_k=S-x_k-k\) for any k

taking k=1,2,….,100 and adding

\(100S-2(x_1+x_2+….+x_{100})=1+2+….+100\)

\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)

\(\Rightarrow S=\frac{2525}{49}\)

for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=75+98

=173.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

Sequence and greatest integer – AIME I, 2000


Let S be the sum of all numbers of the form \(\frac{a}{b}\),where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed \(\frac{S}{10}\).

  • is 107
  • is 248
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 248.

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints


We have 1000=(2)(2)(2)(5)(5)(5) and \(\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3\)

sum of all numbers of form \(\frac{a}{b}\) such that a and b are relatively prime positive divisors of 1000

=\((2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})\)

\(\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times\) \(\frac{(5^{-3})(5^{7}-1)}{5-1}\)

=2480 + \(\frac{437}{1000}\)

\(\Rightarrow [\frac{s}{10}]\)=248.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Reflection Problem | AIME I, 1988 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988, Question 14, based on Reflection.

Reflection Problem – AIME I, 1988


Let C be the graph of xy=1 and denote by C’ the reflection of C in the line y=2x. let the equation of C’ be written in the form \(12x^{2}+bxy +cy^{2}+d=0\), find the product bc.

  • is 107
  • is 84
  • is 840
  • cannot be determined from the given information

Key Concepts


Geometry

Equation

Algebra

Check the Answer


Answer: is 84.

AIME I, 1988, Question 14

Coordinate Geometry by Loney

Try with Hints


Let P(x,y) on C such that P'(x’,y’) on C’ where both points lie on the line perpendicular to y=2x

slope of PP’=\(\frac{-1}{2}\), then \(\frac{y’-y}{x’-x}\)=\(\frac{-1}{2}\)

or, x’+2y’=x+2y

also midpoint of PP’, \((\frac{x+x’}{2},\frac{y+y’}{2})\) lies on y=2x

or, \(\frac{y+y’}{2}=x+x’\)

or, 2x’-y’=y-2x

solving these two equations, x=\(\frac{-3x’+4y’}{5}\) and \(y=\frac{4x’+3y’}{5}\)

putting these points into the equation C \(\frac{(-3x’+4y’)(4x’+3y’)}{25}\)=1

which when expanded becomes

\(12x’^{2}-7x’y’-12y’^{2}+25=0\)

or, bc=(-7)(-12)=84.

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AIME I Algebra Arithmetic Complex Numbers Math Olympiad USA Math Olympiad

Problem on Complex plane | AIME I, 1988| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Complex Plane.

Problem on Complex Plane – AIME I, 1988


Let w_1,w_2,….,w_n be complex numbers. A line L in the complex plane is called a mean line for the points w_1,w_2,….w_n if L contains points (complex numbers) z_1,z_2, …..z_n such that \(\sum_{k=1}^{n}(z_{k}-w_{k})=0\) for the numbers \(w_1=32+170i, w_2=-7+64i, w_3=-9+200i, w_4=1+27i\) and \(w_5=-14+43i\), there is a unique mean line with y-intercept 3. Find the slope of this mean line.

  • is 107
  • is 163
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 163.

AIME I, 1988, Question 11

Elementary Algebra by Hall and Knight

Try with Hints


\(\sum_{k=1}^{5}w_k=3+504i\)

and \(\sum_{k-1}^{5}z_k=3+504i\)

taking the numbers in the form a+bi

\(\sum_{k=1}^{5}a_k=3\) and \(\sum_{k=1}^{5}b_k=504\)

or, y=mx+3 where \(b_k=ma_k+3\) adding all 5 equations given for each k

or, 504=3m+15

or, m=163.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Problem on Real Numbers | AIME I, 1990| Question 15

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

Problem on Real Numbers – AIME I, 1990


Find \(ax^{5}+by^{5}\) if real numbers a,b,x,y satisfy the equations

ax+by=3

\(ax^{2}+by^{2}=7\)

\(ax^{3}+by^{3}=16\)

\(ax^{4}+by^{4}=42\)

  • is 107
  • is 20
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Equations

Algebra

Check the Answer


Answer: is 20.

AIME I, 1990, Question 15

Elementary Algebra by Hall and Knight

Try with Hints


Let S=x+y, P=xy

\((ax^{n}+by^{n})(x+y)\)

\(=(ax^{n+1}+by^{n+1})+(xy)(ax^{n-1}+by^{n-1})\)

or,\( (ax^{2}+by^{2})(x+y)=(ax^{3}+by^{3})+(xy)(ax+by)\) which is first equation

or,\( (ax^{3}+by^{3})(x+y)=(ax^{4}+by^{4})+(xy)(ax^{2}+by^{2})\) which is second equation

or, 7S=16+3P

16S=42+7P

or, S=-14, P=-38

or, \((ax^{4}+by^{4})(x+y)=(ax^{5}+by^{5})+(xy)(ax^{3}+by^{3})\)

or, \(42S=(ax^{5}+by^{5})+P(16)\)

or, \(42(-14)=(ax^{5}+by^{5})+(-38)(16)\)

or, \(ax^{5}+by^{5}=20\).

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AMC 10 Math Olympiad USA Math Olympiad

Algebraic Equation | AMC-10A, 2001 | Problem 10

Try this beautiful problem from Algebra based on Algebraic Equation.

Algebraic Equation – AMC-10A, 2001- Problem 10


If $x$, $y$, and $z$ are positive with $xy = 24$, $xz = 48$, and $yz = 72$, then $x + y + z$ is

  • \(5\)
  • \(20\)
  • \(22\)
  • \(25\)
  • \(36\)

Key Concepts


algebra

Equation

sum

Check the Answer


Answer: \(22\)

AMC-10A (2001) Problem 10

Pre College Mathematics

Try with Hints


The given equations are $xy=24$ and $xz=48$.we have to find out \(x+y+z\)

Now using two relations we can write \(\frac{xy}{xz}=\frac{24}{48}\)\(\Rightarrow 2y=z\)

Can you now finish the problem ……….

Now if we substitute the value $z = 2y$ into the equation $yz = 72$ then we will get \(2y^2=72\) \( \Rightarrow y=6\) and $2y=z=12$. We replace $y$ into the first equation to obtain $x=4$.

Therefore The sum of three numbers are \((x+y+z)\)=\(4+6+12\)=\(22\)

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AMC 10 Math Olympiad USA Math Olympiad

Sum of whole numbers | AMC-10A, 2012 | Problem 8

Try this beautiful problem from Algebra based on Sum of whole numbers.

Sum of whole numbers – AMC-10A, 2012- Problem 8


The sums of three whole numbers taken in pairs are 12, 17, and 19. What is the middle number?

  • \(8\)
  • \(9\)
  • \(7\)
  • \(6\)
  • \(5\)

Key Concepts


Algebra

Equation

Sum of digits

Check the Answer


Answer: \(7\)

AMC-10A (2012) Problem 8

Pre College Mathematics

Try with Hints


Let us assume three numbers are \(x\),\(y\) and \(z\)

Now according to the problem

\(x+y=12\)…………………(1)

\(y+z=17\)………………..(2)

\(z+x=19\)………………….(3).

Can you find out the values of \(x\),\(y\) and \(z\)…………….?

can you finish the problem……..

Adding three equations we get \(2(x=y+z)=48\)

\(\Rightarrow x+y+z=24\)………….(3)

Now subtract (3) from (1),we will get \(z=12\) and similarly if we subtract (2) and (3) from (1) we will get \(x=7\) and \(y=5\)

can you finish the problem……..

Therefore,the numbers are \(12\), \(7\), and \(5\) and middle number is \(7\)

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