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If you happen happen to have a message m ∈ Zn \ Z*n, RSA works but not secure. How likely is it going to happen? |n|=1024 bits |p| = 512 bits |q| = 512 bits.

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Since this is homework: How many elements are in $Z_n$? There are $n=p\cdot q$ right? How many elements are there in $Z_n^*$? There are $\varphi(n)=(p-1)(q-1)$ right? Consequently, the bad event is hitting one of the $n-\varphi(n)$ elements. Is it likely to hit one of those? –  DrLecter Mar 26 at 15:49
    
@user12708: DrLecter was giving you a hint rather than asking you a question. –  figlesquidge Mar 26 at 16:06
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Also, you have the wrong value for $\phi(p)$. $\;$ –  Ricky Demer Mar 26 at 16:08
    
Yes to all of the assumptions. So, since there n is 1024 bits, φ(n) = φ(p)φ(q), φ(p) = 2^512(1-1/2), φ(q) = 2^512(1-1/2), φ(n) = 2^512(1-1/2)*2^512(1-1/2) = (1/4)2^1024? ... I guess my main problem was, how do I hit the bad condition? –  user12708 Mar 26 at 16:09
    
You still have the wrong value for $\phi(p)$. $\phi(p)$ is defined to be the number of values between 1 and $p-1$ which is relatively prime to $p$. If $p$ is prime, how many of the values between 1 and $p-1$ are relatively prime to $p$? –  poncho Mar 26 at 16:14
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