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An intruder has a cluster of 64 machines, each of which can try 10^6 passwords per second. How long does it take him to try all legal passwords if the requirements for the password are as follows:

  • passwords can be 6, 7, or 8 characters long
  • each character is either a lower-case letter or a digit
  • at least one character has to be a digit

My approach is that, for password of length k, there are 36^k possibilities - 26^k forbidden possibilities (because can't be all letters): ∑(k=6 to 8)(36^k−26^k) = 2,684,483,063,360 so 2,684,483,063,360 passwords × (1 second / 64*10^6 passwords) × (1 minute/ 60 seconds) × (1 hour / 60 minutes) = 11.651 hours

Am I approaching this the right way? Is my solution correct?

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  • $\begingroup$ This problem has already been asked before. $\endgroup$ – Rohan Jan 26 '17 at 6:45
  • $\begingroup$ Why express the result in years if it is less than a day? $\endgroup$ – TMM Jan 26 '17 at 6:53
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The methodology is correct, and if the calculator says that the end number is what you say it is, then it's right too :P

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The number of possible passwords is $\sum_{k=0}^{2}10\cdot(10+26)^{5+k}=806,014,126,080=8.06\cdot10^{11}$.

The rate of password input is $64\cdot10^6$ passwords per second. Thus the time to try all of them is

$$\frac{8.06\cdot10^{11}\,\text{passwords}}{64\cdot10^6\,\text{passwords/second}}=12,593.971\,\text{seconds}\approx4.5\,\text{hours}$$

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