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Compute the probability that a randomly chosen positive divisor of $10^{99}$ is an integer multiple of $10^{88}$

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What have you tried? Or, in the spirit of the 'question', maybe I should say: Show us what you tried. – TMM Mar 24 '12 at 19:16
Hint: a divisor is of the form $2^m5^n$ with $0\leq m,n \leq 99$. It is divisible bt $10^{88}$ only when $m,n\geq 88$ – Thomas Andrews Mar 24 '12 at 19:19

Choosing a random positive divisor of $10^{99}$ is equivalent to choosing two random integers $0 \leq a, b \leq 99$ (the corresponding divisor is $2^a\cdot 5^b$). Multiples of $10^{88}$ are of the form $2^x 5^y$ where $x,y \geq 88$. So we need to find the probability that $a,b \geq 88$. There are $100$ possible values for each of $a$ and $b$, and $12$ of these integers are at least 88 and at most 99. Therefore the desired probability is $(12/100)^2 = 9/625$.

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We have that $10^{99} = 2^{99}\cdot 5^{99}$, where both $2$ and $5$ are prime, so random divisor is equivalent to random choice of two numbers, say $\alpha$ and $\beta$, each from $\{0,1,\ldots 99\}$, i.e. $d = 2^\alpha\cdot 5^\beta$. For this random divisor to be multiple of $10^{88}$ we need that $\alpha, \beta \in \{88,89,\ldots, 99\}$. With this constraint both $\alpha$ and $\beta$ can be picked in $99+1-88 = 12$ ways and there are $100^2$ different divisiors, so the probability in question equals $$\frac{12^2}{100^2} = 0.0144$$

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If $n$ is a natural number and $d$ a divisor of $n$ then there is a bijection between the divisors of $n/d$ and the divisors of $n$ that are multiples of $d$ (the bijection is given by multiplication/division by $d$). So there are $\tau(n/d)$ divisors of $n$ that are multiples of $d$ where $\tau$ denotes "number of divisors". The ratio of divisors of $n$ that are multiples of $d$ is therefore $\frac{\tau(n)}{\tau(n/d)}$. In the special case where $n = 10^{99}$ and $d = 10^{88}$ this gives $$\frac{\tau(10^{11})}{\tau(10^{99})} = \frac{\tau(2^{11})\tau(5^{11})}{\tau(2^{99})\tau(5^{99})} = \frac{12^2}{100^2}.$$

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