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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.

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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