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

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

## 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

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

## 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

Functions

Equations

Algebra

## Check the Answer

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, $f_1988(11)=f_4(11)=169$.

Categories

## Remainders and Functions | AIME I, 1994 | Question 7

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Remainders and Functions.

## Remainders and Functions – AIME I, 1994

The function f has the property that, for each real number x, $f(x)+f(x-1)=x^{2}$ if f(19)=94, find the remainder when f(94) is divided by 1000.

• is 107
• is 561
• is 840
• cannot be determined from the given information

Integers

Remainder

Functions

## Check the Answer

AIME I, 1994, Question 7

Elementary Number Theory by David Burton

## Try with Hints

f(94)=$94^{2}-f(93)=94^{2}-93^{2}+f(92)$

=$94^{2}-93^{2}+92^{2}-f(91)$

=$(94^{2}-93^{2})+(92^{2}-91^{2})$

$+….+(22^{2}-21^{2})+20^{2}-f(19)$

=94+93+…..+21+400-94

=4561

$\Rightarrow$ remainder =561.

Categories

## Function of Complex numbers | AIME I, 1999 | Question 9

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Function of Complex Numbers and Integers.

## Function of Complex Numbers – AIME I, 1999

Let f(z) =(a+bi)z where a,b are positive numbers. This function has the property that the image of each point in the complex plane is equidistant from that point and the origin given that |a+bi|=8 and that $b^{2}$=$\frac{m}{n}$ where m and n are relatively prime positive integers, find m+n.

• is 107
• is 259
• is 840
• cannot be determined from the given information

Functions

Integers

Complex Numbers

## Check the Answer

AIME I, 1999, Question 9

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

Let z=1+i f(1+i)=(a+bi)(1+i)=(a-b)+(a+b)i The image point must be equidistant from (1,1) and(0,0) then the image point lie on the line with slope -1 and which passes through $(\frac{1}{2},\frac{1}{2})$ that is x+y=1

putting x=(a-b) and y=(a+b) gives 2a=1 and $a=\frac{1}{2}$

and $(\frac{1}{2})^{2} +b^{2}=8^{2}$ then $b^{2}=\frac{255}{4}$ then 255+4=259.

Categories

## Function and symmetry | AIME I, 1984 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1984 based on Function and symmetry.

## Function and Symmetry – AIME I 1984

A function f is defined for all real numbers and satisfies f(2+x)=f(2-x) and f(7+x)=f(7-x) for all x. If x=0 is root for f(x)=0, find the least number of roots f(x) =0 must have in the interval $-1000 \leq x\leq 1000$.

• is 107
• is 401
• is 840
• cannot be determined from the given information

Functions

Symmetry

Number Theory

## Check the Answer

AIME I, 1984, Question 12

Elementary Number Theory by David Burton

## Try with Hints

by symmetry with both x=2 and x=7 where x=0 is a root, x=4 and x=14 are also roots

here 0(mod 10) or 4(mod10) are roots there are 201 roots as multiples of 10 and 200 roots as for 4(mod10)

Then least number of roots as 401.