You are trying to crack someone's password. You don't have any password cracking software, nor do you have time to get any. From sources unknown, you know that:

A. The password is alphanumeric (only letters and numbers)

B. It's length is between 6 and 10

C. 25% of all the characters are numbers

D. The password is case insensitive

E. All the digits are at the end of the password (after the letters)

F. The first letter of the password is J

G. The sum of the digits are equal to the numerical value of the first letter

H. There is one E, one U and one Z in the password inclusively (they appear in that order)

I. The first digit is equal to the numerical value of the second letter

J. The product of all the digits is equal to the last letter

K. The letter that was unmentioned (not declared) is d upside down and mirrored.

L. The numerical difference between the third letter and the Z is more than 6

No computers please (except for posting your answer). Search, there is an answer to this!

  • $\begingroup$ @JLee Checked it just now, yes that is right. $\endgroup$ – bitbyte Jun 14 '15 at 15:50
  • $\begingroup$ Should condition I be "The first digit is equal to the numerical value of the second character/letter"? $\endgroup$ – Julian Rosen Jun 14 '15 at 15:50
  • $\begingroup$ @Julian Ronsen The second character is a letter, from condition E, All the digits are at the end of the password (after the letters). $\endgroup$ – bitbyte Jun 14 '15 at 15:54
  • 2
    $\begingroup$ Since the second character is not a number, maybe this should say "the numerical value of the second letter" instead of "the numerical value of the second number" $\endgroup$ – Julian Rosen Jun 14 '15 at 15:59
  • 1
    $\begingroup$ @BitCoder rule K may be ambiguous, when you flip something upside down you can either rotate it 180 degrees or flip over its horizontal axis. Another possible interpretation is that "upside down and mirrored" both refer to the same transformation (e.g. q). Is this solvable as is or should it be clarified? $\endgroup$ – Quark Jun 14 '15 at 16:15

The password is:

JEQUZY55. 25% of the password are numbers, and the only number between 6 and 10 that can be broken up this way is 8. Therefore there are 6 letters and 2 numbers. The 6 letters include J,E,U,Z, and Q so far. One of these must be the first of these two digits which make the first digit 5. The sum of these digits is 10 making both digits 5, making the product 25. This makes the last letter Y. The letters JEUZY are in order, and Q can be anywhere between the E (Rule I) and the Y. For Rule L, the third letter can't be U, meaning it is Q.

| improve this answer | |
  • $\begingroup$ That is correct! Good job :) $\endgroup$ – bitbyte Jun 14 '15 at 16:30
  • 1
    $\begingroup$ To the edit, rule D says the password is case insensitive but I guess it doesn't really matter either way. $\endgroup$ – Quark May 21 '16 at 20:27
  • $\begingroup$ @JorVa I hit the rollback as the post explicitly stated that the password is case INsensitive. Perhaps a comment would serve better here -- just mention that's where the logic comes in for that particular rule (K). $\endgroup$ – feelinferrety May 24 '16 at 0:08

The password is:



25 percent are numbers ... Length between 6 and 10. This means the length is 8: 6 characters and 2 numbers. Numerical value of the 2 digits is 10(j), and their product is equal to the last letter. This can be y, x, u, p, i. None of these have been mentioned, which means the first five letters are jequz or jqeuz (q must occupy the 2nd or 3rd position, from L). Jqeuz does not satisfy I, so it must be jequz, which forces the first number to be 5 and thus the last letter/second number to be Y/5, giving us jequzy55

| improve this answer | |
  • $\begingroup$ I don't see any Q $\endgroup$ – bitbyte Jun 14 '15 at 16:19
  • 1
    $\begingroup$ @BitCoder Give him a little time; he posted this before you explicitly said that the unmentioned letter was a Q. Interpreting that hint differently, you could have gotten a P instead. $\endgroup$ – Dennis Meng Jun 14 '15 at 16:23
  • $\begingroup$ When this answer was posted, it was spot on. Doesn't seem fair not to accept it. $\endgroup$ – r.e.s. Jun 14 '15 at 16:35
  • $\begingroup$ @r.e.s. I know, but when this answer was first posted (puzzling.stackexchange.com/posts/16390/revisions) it had the wrong response. Then Quark (puzzling.stackexchange.com/users/7722/quark) posted the correct answer, and shortly this guy caught on and edited his answer. $\endgroup$ – bitbyte Jun 14 '15 at 19:58
  • $\begingroup$ @BitCoder r.e.s. posted that before I edited. I believe they are pointing out that this answer was correct when first posted if we take 'd upside down and mirrored' as p rather than q. $\endgroup$ – frodoskywalker Jun 14 '15 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.