2023 SSMO Team Round Problems/Problem 13

Problem

Let $D(n)$ denote the product of all divisors of $n$ Let $P(i,j)$ denote the set of all integers that are both a multiple of $i$ and a factor of $j.$ Let $-F(a) = \sqrt{\left|\log_{10}\left(\frac{D(10^{a})}{\prod_{\omega\in P(10^2,10^{a+2})}\omega}\right)\right|}\text{ and }G(n) = \sqrt[n-1]{\prod_{i=2}^{n}10^{F(i)}}.$ Suppose $\sum_{k=2}^{\infty}G(k)$ is $\frac{a+b\sqrt{c}}{d}$ . Find the value of $a+b+c+d$ .

Solution

Let $\tau(n)$ denote the number of factors of an integer $n$ . Then $D(n) = n^{\tau(n)/2}$ , which can be proved by pairing up factors whose product is $n$ .

So, $D(10^a) = (10^a)^{\frac{(a+1)^2}{2}} = 10^{\frac{a(a+1)^2}{2}}.$

The set $P(10^2, 10^{a+2})$ consists of integers of the form $100m$ where $m \mid 10^a$ . Therefore, the product of elements of $P$ is $\prod_{m \mid 10^a} 100m = 100^{(a+1)^2/2} \cdot D(10^a).$

Now, the value of $F(a)$ is $F(a) = -\sqrt{\left|\log_{10} \left( \frac{D(10^a)}{100^{(a+1)^2/2} \cdot D(10^a)} \right) \right|} = -\sqrt{|\log_{10}(10^{-(a+1)^2})|} = -a - 1.$

Next, calculating $G(n)$ : \begin{align*} G(n) &= \sqrt[n-1]{\prod_{i=2}^n 10^{-i-1}} = \sqrt[n-1]{10^{-\sum_{i=2}^n (i+1)}} \\ &= \sqrt[n-1]{10^{-(n(n+1)/2 - 1)}} = 10^{\frac{-(n+4)}{2}}. \end{align*}

Finally, $\sum_{k=2}^\infty 10^{\frac{-(k+4)}{2}} = \sum_{k=6}^\infty 10^{-k/2} = \frac{10^{-3}}{1 - 10^{-1/2}} = \frac{1}{1000} \cdot \frac{1}{1 - \frac{1}{\sqrt{10}}} = \frac{10 + \sqrt{10}}{9000},$

giving a final answer of $10 + 1 + 10 + 9000 = \boxed{9021}.$ ~ ~SMO_Team

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2023 SSMO Team Round Problems/Problem 13

Problem

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