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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

GCD and Sequence – AIME I, 1985


The numbers in the sequence 101, 104,109,116,…..are of the form \(a_n=100+n^{2}\) where n=1,2,3,——-, for each n, let \(d_n\) be the greatest common divisor of \(a_n\) and \(a_{n+1}\), find the maximum value of \(d_n\) as n ranges through the positive integers.

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

Key Concepts


GCD

Sequence

Integers

Check the Answer


Answer: is 401.

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

Try with Hints


\(a_n=100+n^{2}\) \(a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1\) and \(a_{n+1}-a_{n}=2n +1\)

\(d_{n}|(2n+1)\) and \(d_{n}|(100 +n^{2})\) then \(d_{n}|[(100+n^{2})-100(2n+1)]\) then \(d_{n}|(n^{2}-200n)\)

here \(n^{2} -200n=0\) then n=200 then \(d_{n}\)=2n+1=2(200)+1=401.

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