Question

6. Suppose that a password for a computer system must have at least 6, but no more than 9 characters, where each character in the password is a lowercase English letter, or an uppercase English letter, or a digit, or one of the five 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 five special characters?

(c) Using your answer to part (b), determine how long it takes a hacker to try ev ery possible password, assuming that it takes one microsecond for a hacker to check each possible password.

Answer 1

Answer:

How many of these passwords contain at least one occurrence of at least one of the five special characters?

Answer 2

a. There are 26 letters in the English alphabet, with two cases for each letter; 10 numerical characters in the range 0-9; and 5 special characters; thus a total of 26•2 + 10 + 5 = 67 characters.

Any character can be used more than once, so there are

67⁶ + 67⁷ + 67⁸ + 67⁹

or 27,618,753,243,839,080 total possible passwords.

b. If we require at least 1 special character, then there are 62 choices for each ordinary character we use and 5 for each special character.

Suppose we use a password of length 6. Then there are

62⁵•5¹ + 62⁴•5² + 62³•5³ + 62²•5⁴ + 62¹•5⁵ + 62⁰•5⁶

or

$\displaystyle \sum_{k=1}^6 62^{6-k}\cdot5^k$

We count the longer passwords in a similar fashion:

$\displaystyle \sum_{m=6}^9 \left(\sum_{k=1}^m 62^{m-k}\cdot5^k\right)$

or 1,206,930,268,794,100 total passwords

c. 1 second (s) is equal to 10⁶ microseconds (μs). If it takes 1 μs to try 1 password, then it would take

(1,206,930,268,794,100 passwords) • (1 μs / password) • (1 s / 10⁶ μs)

= 1,206,930,268.7941 s

= (1,206,930,268.7941 s) • (1 min / 60 s)

≈ 20,115,504.4799 min

≈ (20,115,504.4799 min) • (1 h / 60 min)

≈ 335,258.408 h

≈ (335,258.408 h) • (1 day / 24 h)

≈ 13,969.1003 days

≈ (13,969.1003 days) • (1 year / 365 days)

≈ 38.2715 years