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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 987.

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.

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Categories
AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 184.

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.

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AIME I Complex Numbers Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 144.

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.

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