Categories

## Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

## Positive Integer – AIME I, 1996

For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that $n \lt 1000$ and that $[log_{2}n]$ is a positive even integer.

• is 107
• is 340
• is 840
• cannot be determined from the given information

### Key Concepts

Inequality

Greatest integer

Integers

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

## Try with Hints

here Let $[log_{2}n]$=2k for k is an integer

$\Rightarrow 2k \leq log_{2}n \lt 2k+1$

$\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}$ and $n \lt 1000$

$\Rightarrow 4 \leq n \lt 8$

$16 \leq n \lt 32$

$64 \leq n \lt 128$

$256 \leq n \lt 512$

$\Rightarrow 4+16+64+256$=340.

Categories

## Trapezoid Problem | AIME I, 1992 | Question 9

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

## Trapezoid – AIME I, 1992

Trapezoid ABCD has sides AB=92, BC=50,CD=19,AD=70 with AB parallel to CD. A circle with centre P on AB is drawn tangent to BC and AD. Given that AP=$\frac{m}{n}$, where m and n are relatively prime positive integers, find m+n.

• is 107
• is 164
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Trapezoid

Angle Bisectors

AIME I, 1992, Question 9

Coordinate Geometry by Loney

## Try with Hints

Let AP=y or, PB=92-y

extending AD and BC to meet at Y

and YP bisects angle AYB

Let F be point on CD where it meets

Taking angle bisector theorem,

let YB=z(92-y), YA=zy for some z

YD=zy-70, YC=z(92-y)-50

$\frac{yz-79}{z(92-y)-50}=\frac{YD}{YC}=\frac{FD}{FC}=\frac{AP}{PB}=\frac{y}{42-y}$

solving we get 120y=(70)(92)

or, AP=y=$\frac{161}{3}$

or, 161+3=164.

Categories

## Problem on Fibonacci sequence | AIME I, 1988 | Question 13

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

## Fibonacci sequence Problem – AIME I, 1988

Find a if a and b are integers such that $x^{2}-x-1$ is a factor of $ax^{17}+bx^{16}+1$.

• is 107
• is 987
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Sets

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

## Try with Hints

Let F(x)=$ax^{17}+bx^{16}+1$

Let P(x) be polynomial such that

$P(x)(x^{2}-x-1)=F(x)$

constant term of P(x) =(-1)

now $(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$ where $c_{i}$=coefficient

comparing the coefficients of x we get the terms

since F(x) has no x term, then $c_{15}$=1

getting $c_{14}$

$(x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)$

=terms +$0x^{2}$ +terms

or, $c_{14}=-2$

proceeding in the same way $c_{13}=3$, $c_{12}=-5$, $c_{11}=8$ gives a pattern of Fibonacci sequence

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=$c_1=F_{16}$ where $F_{16}$ is 16th Fibonacci number

or, a=987.

Categories

## 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

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.

Categories

## 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

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, $ax^{5}+by^{5}=20$.

Categories

## Digits and Integers | AIME I, 1990 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Digits and Integers.

## Digits and Integers – AIME I, 1990

Let T={$9^{k}$: k is an integer, $0 \leq k \leq 4000$} given that $9^{4000}$ has 3817 digits and that its first (leftmost) digit is 9, how many elements of T have 9 as their leftmost digit?

• is 107
• is 184
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Sets

AIME I, 1990, Question 13

Elementary Number Theory by David Burton

## Try with Hints

here $9^{4000}$ has 3816 digits more than 9,

or, 4000-3816=184

or, 184 numbers have 9 as their leftmost digits.

Categories

## Complex numbers and Sets | AIME I, 1990 | Question 10

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

## Complex Numbers and Sets – AIME I, 1990

The sets A={z:$z^{18}=1$} and B={w:$w^{48}=1$} are both sets of complex roots with unity, the set C={zw: $z \in A and w \in B$} is also a set of complex roots of unity. How many distinct elements are in C?.

• is 107
• is 144
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Complex Numbers

Sets

AIME I, 1990, Question 10

Complex Numbers from A to Z by Titu Andreescue

## Try with Hints

18th and 48th roots of 1 found by de Moivre’s Theorem

=$cis(\frac{2k_1\pi}{18})$ and $cis(\frac{2k_2\pi}{48})$

where $k_1$, $K_2$ are integers from 0 to 17 and 0 to 47 and $cis \theta = cos \theta +i sin \theta$

zw= $cis(\frac{k_1\pi}{9}+\frac{k_2\pi}{24})=cis(\frac{8k_1\pi+3k_2\pi}{72})$

and since the trigonometric functions are periodic every period ${2\pi}$

or, at (72)(2)=144 distinct elements in C.

Categories

## Consecutive positive Integers | AIME I, 1990| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

## Consecutive positive integer – AIME I, 1990

Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.

• is 107
• is 23
• is 634
• cannot be determined from the given information

### Key Concepts

Integers

Inequality

Algebra

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

## Try with Hints

The product of (n-3) consecutive integers=$\frac{(n-3+a)!}{a!}$ for a is an integer

$n!=\frac{(n-3+a)!}{a!}$ for $a \geq 3$ $(n-3+a)! \geq n!$

or, $n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}$

for a=4, n+1=4! or, n=23 which is greatest here

n=23.

Categories

## Convex polyhedron Problem | AIME I, 1988 | Question 10

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

## Convex polyhedron Problem – AIME I, 1988

A convex polyhedron has for its faces 12 squares, 8 regular hexagons, and 6 regular octagons. At each vertex of the polyhedron one square, one hexagon, and one octagon meet. How many segments joining vertices of the polyhedron lie in the interior of the polyhedron rather than along an edge or a face?

• is 107
• is 840
• is 634
• cannot be determined from the given information

### Key Concepts

Integers

Edges

Algebra

AIME I, 1988, Question 10

Geometry Revisited by Coxeter

## Try with Hints

${48 \choose 2}$=1128

Every vertex lies on exactly one vertex of a square/hexagon/octagon

V=(12)(4)=(8)(6)=(6)(8)=48

each vertex is formed by the trisection of three edges and every edge is counted twice, once at each of its endpoints, the number of edges E=$\frac{3V}{2}$=72

each of the segment on face of polyhedron is diagonal of that face, so each square gives $\frac{n(n-3)}{2}=2$ diagonals, each hexagon=9,each octagon=20. The number of diagonals is $(2)(12)+(9)(8)+(20)(6)$=216

or, number of space diagonals =1128-72-216=840.

Categories

## Fair coin Problem | AIME I, 1990 | Question 9

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

## Fair Coin Problem – AIME I, 1990

A fair coin is to be tossed 10 times. Let i|j, in lowest terms, be the probability that heads never occur on consecutive tosses, find i+j.

• is 107
• is 73
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Combinatorics

Algebra

AIME I, 1990, Question 9

Elementary Algebra by Hall and Knight

## Try with Hints

5 tails flipped, any less,

by Pigeonhole principle there will be heads that appear on consecutive tosses

(H)T(H)T(H)T(H)T(H)T(H) 5 tails occur there are 6 slots for the heads to be placed but only 5H remaining, ${6 \choose 5}$ possible combination of 6 heads there are

$\sum_{i=6}^{11}{i \choose 11-i}$=${6 \choose 5} +{7 \choose 4}+{8 \choose 3}+{9 \choose 2} +{10 \choose 1} +{11 \choose 0}$=144

there are $2^{10}$ possible flips of 10 coins

or, probability=$\frac{144}{1024}=\frac{9}{64}$ or, 9+64=73.