AoPS Wiki talk:Problem of the Day/June 28, 2011

Problem

AoPSWiki:Problem of the Day/June 28, 2011

Solution

We see that $144 = 2^43^2$

Each factor of 144 can be written as $2^a3^b$ for $a\leq 4$ and $b \leq 2$

Thus, the sum of of the reciprocals of the factors of 144 can be written as: $(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16})(1 + \frac{1}{3} + \frac{1}{9})$ ,

since each term will be the product of some term in the first factor and some term in the second factor. By the formula for the sum of finite geometric series, this is seen to be: $\left( \dfrac{1 - \frac{1}{32}}{1 - \frac{1}{2}}\right)\left( \dfrac{1 - \frac{1}{27}}{1 - \frac{1}3} \right)$

This simplifies to: $\left(\frac{31}{16}\right)\left(\frac{13}{9}\right)$ ,

which simplifies to $\frac{403}{144}$

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AoPS Wiki talk:Problem of the Day/June 28, 2011

Problem

Solution