2000 AIME I Problems/Problem 11

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 $\left\lfloor \frac{S}{10} \right\rfloor = \boxed{248}$ .

NOTE: Hi, I'm just someone wondering: isn't the question saying that S is the sum of fractions where a and b are relatively prime divisors of 1000? So for example 5/8 would be okay but 10/25 wouldn't. If that's the case, then a and b would only be 5, 25, 125, 2, 4, and 8. So S would be:

(5+25+125)/8 + (5+25+125)/4 + (5+25+125)/2 + (2+4+8)/5 + (2+4+8)/25 + (2+4+8)/125 = 434/125 + 1085/8 then our solution would be the largest integer less than (3+135)/10, which would be 13.

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2000 AIME I Problems/Problem 11

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