AoPSWiki
Art of Problem Solving celebrates the many
accomplishments of its students and community members.

2000 AIME I Problems/Problem 11

From AoPSWiki

Problem

Let S be the sum of all numbers of the form a/b, where a and b are relatively prime positive divisors of 1000. What is the greatest integer that does not exceed S/10?

Solution

Since all divisors of 1000 = 2^35^3 can be written in the form of 2^{m}5^{n}, it follows that \frac{a}{b} can also be expressed in the form of 2^{x}5^{y}, where -3 \le x,y \le 3. Thus every number in the form of a/b will be expressed one time in the product

(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)

Using the formula for a geometric series, this reduces to S = \frac{2^{-3}(2^7 - 1)}{2-1} \cdot \frac{5^{-3}(5^{7} - 1)}{5-1} = \frac{127 \cdot 78124}{4000} = 2480 + \frac{437}{1000}, and \left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}.

See also

2000 AIME I (ProblemsResources)
Preceded by
Problem 10
Followed by
Problem 12
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us