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 1

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

Solution 2

Essentially, the problem asks us to compute $\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b}$ which is pretty easy: $\sum_{a=-3}^3 \sum_{b=-3}^3 \frac{2^a}{5^b} = \sum_{a=-3}^3 2^a \sum_{b=-3}^3 \frac{1}{5^b} = \sum_{a=-3}^3 2^a 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg) \sum_{a=-3}^3 2^a = 5^{3}\bigg( \frac{1-5^{-7}}{1-\frac{1}{5}} \bigg)2^{-3} \bigg( \frac{1-2^7}{1-2} \bigg) = 2480 + \frac{437}{1000}$ so our answer is $\left\lfloor \frac{2480 + \frac{437}{1000}}{10} \right\rfloor = \boxed{248}$ .

Solution 3

The sum is equivalent to $\sum_{i | 10^6}^{} \frac{i}{1000}$ Therefore, it's the sum of the factors of $10^6$ divided by $1000$ . The sum is $\frac{127 \times 19531}{1000}$ by the sum of factors formula. The answer is therefore $\boxed{248}$ after some computation. - whatRthose

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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2000 AIME I Problems/Problem 11

Contents

Problem

Solution 1

Solution 2

Solution 3

See also