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A Simple Password Puzzle. Or Maybe Not

Puzzling Asked by Omega Krypton on January 10, 2021

Prologue:

It’s me, KryptonOmega, back with a password puzzle. Enjoy!

Puzzle:

enter image description here

Edits:

  1. Add "unique" in condition 2 for clarification.

  2. Length of answer is five. (Why would you expect more when there are five underscores only???)

4 Answers

The answer is

Because

Explanation

Correct answer by MacGyver88 on January 10, 2021

Setting the scene

We have five numbers $alpha,beta,gamma,delta,epsilon$ satisfying seven conditions. These numbers must all be non-negative integers (we assume). I tried assuming they are all single digits (at most $9$), but got a contradiction as shown in the first revision of this answer, so we know $gammageq10$.

Step-by-step deduction

  • Firstly, can any of them be zero?

  • The password is prime, so

  • Substituting condition 3 into condition 1, we find

  • Going back to condition 6, we notice

  • Now conditions 4 and 6 are used up, and condition 5 gives

Note that after fixing $beta$ and $epsilon$, the others are completely determined:

Let's now just try possibilities starting from the smallest:

  • If $beta=3$, then we must have

  • If $beta=4$, then we must have

  • If $beta=5$, then we must have

Final solution for the password

Answered by Rand al'Thor on January 10, 2021

Answer:

Checking

Partial explanation (to be continued):

Answered by trolley813 on January 10, 2021

OP's Answer

The answer is, as shown by @MacGvyer88,

but they did not provide logical deduction, so here it is.

@RandAlThor has proven that a five-digit number is impossible as the answer. Yet it is mentioned in the clarifications that the length is 5. Therefore it may be a

Well, intrinsically, you can say that

but I will try to deduct this on the approach that we only know that it is a five-letter word and that A1Z26 is used.

From (1) the max of LHS is 26+25+25 = 76 since gamma is unique largest. In other words beta <= floor(76/11) = 6.

Therefore beta is

And in second position it is very likely a

Notice condition 6 regarding duplicates.

From (2) gamma isn't a duplicate.

From (3) we know that alpha ≠ epsilon.

Based on all this we know that either beta or delta is one of the duplicates.

(5) shows that beta ≠ epsilon and delta ≠ epsilon, so epsilon is out.

We now have alpha, beta and delta being possible duplicates.

(3) shows that alpha ≠ beta

Therefore either alpha = delta or beta = delta.

Given beta = A or E most likely, we now have

And this should be enough, given (7) for you to make out


I am challenged to prove why beta is A or E. First of all I said "most likely". Secondly, here it is if you need proof. enter image description here enter image description here enter image description here enter image description here

Answered by Omega Krypton on January 10, 2021

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