Categories

## 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

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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Categories

## 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

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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Categories

## 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

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.

Categories

## Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

## Finding smallest positive Integer – AIME I, 1996

Find the smallest positive integer solution to $tan19x=\frac{cos96+sin96}{cos96-sin96}$.

• is 107
• is 159
• is 840
• cannot be determined from the given information

Functions

Trigonometry

Integers

## Check the Answer

Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

## Try with Hints

$\frac{cos96+sin96}{cos96-sin96}$

=$\frac{sin(90+96)+sin96}{sin(90+96)-sin96}$

=$\frac{sin186+sin96}{sin186-sin96}$

=$\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}$

=$\frac{2sin141cos45}{2cos141sin45}$

=tan141

here $tan(180+\theta)$=$tan\theta$

$\Rightarrow 19x=141+180n$ for some integer n is first equation

multiplying equation with 19 gives

$x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)$ [since 2679 divided by 180 gives remainder 159]

$\Rightarrow x=159$.

Categories

## 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

Rearranging equations we get $-logxlogy+logx+logy-1=3-log2000$ and $-logylogz+logy+logz-1=-log2$ and $-logxlogz+logx+logz-1=-1$

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

from both equations (q-1)=+-(log2) gives (logy)+-(log2)=1 gives $y_{1}=20$,$y_{2}=5$ then $y_{1}+y_{2}=25$.

Categories

## 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.

Categories

## 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

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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Categories

## Tetrahedron Problem | AIME I, 1992 | Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.

## Tetrahedron Problem – AIME I, 1992

Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.

• is 107
• is 320
• is 840
• cannot be determined from the given information

Area

Volume

Tetrahedron

## Check the Answer

Answer: is 320.

AIME I, 1992, Question 6

Coordinate Geometry by Loney

## Try with Hints

Area BCD=80=$\frac{1}{2} \times {10} \times {16}$,

where the perpendicular from D to BC has length 16.

The perpendicular from D to ABC is 16sin30=8

[ since sin30=$\frac{perpendicular}{hypotenuse}$ then height = perpendicular=hypotenuse $\times$ sin30 ]

or, Volume=$\frac{1}{3} \times 8 \times 120$=320.

Categories

## Triangle and integers | AIME I, 1995 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Triangle and integers.

## Triangle and integers – AIME I, 1995

Triangle ABC is isosceles, with AB=AC and altitude AM=11, suppose that there is a point D on AM with AD=10 and $\angle BDC$=3$\angle BAC$. then the perimeter of $\Delta ABC$ may be written in the form $a+\sqrt{b}$ where a and b are integers, find a+b.

• is 107
• is 616
• is 840
• cannot be determined from the given information

Integers

Triangle

Trigonometry

## Check the Answer

Answer: is 616.

AIME I, 1995, Question 9

Plane Trigonometry by Loney

## Try with Hints

Let x= $\angle CAM$

$\Rightarrow \angle CDM =3x$

$\Rightarrow \frac{tan3x}{tanx}=\frac{\frac{CM}{1}}{\frac{CM}{11}}$=11 [by trigonometry ratio property in right angled triangle]

$\Rightarrow \frac{3tanx-tan^{3}x}{1-3tan^{2}x}=11tanx$

solving we get, tanx=$\frac{1}{2}$

$\Rightarrow CM=\frac{11}{2}$

$\Rightarrow 2(AC+CM)$ where $AC=\frac{11\sqrt {5}}{2}$ by Pythagoras formula

=$\sqrt{605}+11$ then a+b=605+11=616.

Categories

## 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

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.