Puzzling Asked by Omega Krypton on January 10, 2021
It’s me, KryptonOmega, back with a password puzzle. Enjoy!
Add "unique" in condition 2 for clarification.
Length of answer is five. (Why would you expect more when there are five underscores only???)
The answer is
Because
Explanation
Correct answer by MacGyver88 on January 10, 2021
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 $gammageq10$.
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
Answered by Rand al'Thor on January 10, 2021
Answer:
Checking
Partial explanation (to be continued):
Answered by trolley813 on January 10, 2021
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.
Answered by Omega Krypton on January 10, 2021
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