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## 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

## 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.