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Whitepunk [10]
3 years ago
11

Suppose that a password for a computer system must have at least 8, but no more than 12, characters, where each character in the

password is a lowercase English letter, an uppercase English letter, a digit, or one of the six special characters ∗, >, <, !, +, and =.
a) How many different passwords are available for this computer system?
b) How many of these passwords contain at least one occurrence of at least one of the six special characters?
c) Using your answer to part (a), determine how long it takes a hacker to try every possible password, assuming that it takes one nanosecond for a hacker to check each possible password.
Mathematics
1 answer:
uranmaximum [27]3 years ago
5 0

Part a)

There are 52 letters (26 lowercase and 26 uppercase), 10 digits, and 6 symbols. There are 52+10+6 = 68 different characters to choose from.

  • If there are 8 characters for this password, then we have 68^8 = 4.5716 * 10^14 different passwords possible.
  • If there are 9 characters, then we have 68^9 = 3.1087 * 10^16 different passwords
  • If there are 10 characters, then we have 68^10 = 2.1139 * 10^18 different passwords
  • If there are 11 characters, then we have 68^11 = 1.4375 * 10^20 different passwords
  • If there are 12 characters, then we have 68^12 = 9.7748 * 10^21 different passwords

Adding up those subtotals gives

68^8+68^9+68^10+68^11+68^12 = 9.9207 * 10^21

different passwords possible.

<h3>Answer: Approximately 9.9207 * 10^21 </h3>

======================================================

Part b)

Let's find the number of passwords where we don't have a special symbol

There are 52+10 = 62 different characters to pick from

  • If there are 8 characters for this password, then we have 62^8 = 2.1834 * 10^14 different passwords possible.
  • If there are 9 characters, then we have 62^9 = 1.3537 * 10^16 different passwords
  • If there are 10 characters, then we have 62^10 = 8.3930 * 10^17 different passwords
  • If there are 11 characters, then we have 62^11 = 5.2037 * 10^19 different passwords
  • If there are 12 characters, then we have 62^12 = 3.2263 * 10^21 different passwords

Adding those subtotals gives

62^8+62^9+62^10+62^11+62^12 = 3.2792 * 10^21

different passwords where we do not have a special character. Subtract this from the answer in part a) above

( 9.9207 * 10^21)  - (3.2792 * 10^21) = 6.6415 * 10^21

which represents the number of passwords where we have one or more character that is a special symbol. I'm using the idea that we either have a password with no symbols, or we have a password with at least one symbol. Adding up those two cases leads to the total number of passwords possible.

<h3>Answer: Approximately 6.6415 * 10^21</h3>

======================================================

Part c)

The answer from part a) was roughly 9.9207 * 10^21

It will take about 9.9207 * 10^21  nanoseconds to try every possible password from part a).

Divide 9.9207 * 10^21  over 1*10^9 to convert to seconds

(9.9207 * 10^21 )/(1*10^9) = 9,920,700,000,000

This number is 9.9 trillion roughly.

It will take about 9.9 trillion seconds to try every password, if you try a password per second.

------

To convert to hours, divide by 3600 and you should get

(9,920,700,000,000)/3600 = 2,755,750,000

So it will take about 2,755,750,000 hours to try all the passwords.

------

Divide by 24 to convert to days

(2,755,750,000)/24= 114,822,916.666667

which rounds to 114,822,917

So it will take roughly 114,822,917 days to try all the passwords.

------

Then divide that over 365 to convert to years

314,583.334246576

which rounds to 314,583

It will take roughly 314,583 years to try all the passwords

------------------------------

<h3>Answers:</h3>
  • 9.9 trillion seconds
  • 2,755,750,000 hours
  • 114,822,917 days
  • 314,583 years

All values are approximate, and are roughly equivalent to one another.

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