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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and greatest integer | AIME I, 2000 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and the greatest integer.

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and greatest integer.

Sequence and greatest integer – AIME I, 2000


Let S be the sum of all numbers of the form \(\frac{a}{b}\),where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed \(\frac{S}{10}\).

  • is 107
  • is 248
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


But try the problem first…

Answer: is 248.

Source
Suggested Reading

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints


First hint

We have 1000=(2)(2)(2)(5)(5)(5) and \(\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3\)

Second Hint

sum of all numbers of form \(\frac{a}{b}\) such that a and b are relatively prime positive divisors of 1000

=\((2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})\)

Final Step

\(\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times\) \(\frac{(5^{-3})(5^{7}-1)}{5-1}\)

=2480 + \(\frac{437}{1000}\)

\(\Rightarrow [\frac{s}{10}]\)=248.

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