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If you pick a number at random from $1 \ldots 10^k$ where $k$ is large, call this number $Z$. $M(Z)$ is defined as 0 if $Z$ is not divisible by two repeated primes. Find the probability that $M(Z) = 0$. I'm having a hard time getting started on this question, I know you use a particular identity but I'm not sure why you use it.

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    I am having trouble understanding what "two repeated primes" means. Is the number $17^2$ supposed to be counted? What about $17^2 \cdot 19^2$? Finally, how about $17^2 \cdot 19$ and $17 \cdot 19$?2011-09-21

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The mention of "Möbius" makes me suspect that you want the probability that $\mu(Z) = 0$, i.e. that $Z$ is divisible by the square of a prime. For each nonempty set $S$ of primes whose product $Q_S \le 10^{k/2}$, let $A_S$ be the event that $Z$ is divisible by $Q_S^2$. Then $P(A_S) = \lfloor 10^k/Q_S^2 \rfloor / 10^k$. By inclusion-exclusion, $P(\mu(Z)=0) = \sum_S (-1)^{|S|-1} P(A_S)$.

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    I corrected "mobius" in the title.2011-09-21