A Simple Password Puzzle. Or Maybe Not

Question

Prologue:

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

Puzzle:

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???)

Answer 1

The answer is

S E V E N

Because

Using a letter's sequence number value:

1. (s + v + n) = (11 * e) or (19 + 14 + 22) = (11 * 5) = 55

2. v is the largest valued letter among the 5 letters

3. n + e = s or 14 + 5 = 19

4. e >= e

5. (e + e) < n or (5 + 5) < 14

6. e is the duplicate

7. seven is a prime number

Explanation

Well, the way I figured out this answer was noticing the 5 spaces for the answer and the fact that the answer was prime, but wouldn't be so easy as to actually be the word "PRIME".

So, my next thought was "SEVEN" because condition 6 said their were duplicates in the variables.

I then saw condition 2 was fulfilled with "v" and condition 3 worked out mathematically and wished I was better at writing up this type of answer because this took forever.

So, I ran the math for the remaining conditions and it actually was a valid answer. So, I decided to answer.

Answer 2

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 $\gamma\geq10$.

Step-by-step deduction

• Firstly, can any of them be zero?

  $\alpha\neq0$ since it's the first digit, so $\beta\neq0$ by condition 1. Also $\gamma\neq0$ by condition 2, and $\epsilon\neq0$ by condition 5. The only one which might be zero is $\delta$, since all we know about it is inequalities.

• The password is prime, so

  $\epsilon$ must be, or end in, one of $1,3,7,9$.

• Substituting condition 3 into condition 1, we find

  $\gamma+2\epsilon=10\beta$, so $\gamma$ is even and $\beta\geq2$, meaning $\epsilon\geq3$.

• Going back to condition 6, we notice

  $\gamma$ is not the same as any of the others (by condition 2), neither is $\alpha$ (it's bigger than all the others, by condition 3, nonzeroness, and condition 4), neither is $\epsilon$ (again bigger than all the others, by condition 5). So we have $\beta=\delta<\epsilon<\alpha<\gamma$.

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

  $\epsilon>2\beta$, therefore $\epsilon\geq7$ by the prime condition and $\beta\geq2$. Using $\gamma+2\epsilon=10\beta$, that means $\beta\geq3$.

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

$\delta=\beta,\alpha=\beta+\epsilon,\gamma=10\beta-2\epsilon$. So we need $10\beta-2\epsilon>\beta+\epsilon$, which means $\epsilon<3\beta$. Overall, $2\beta<\epsilon<3\beta$, and also $\epsilon$ is a prime ending.

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

• If $\beta=3$, then we must have

  $\beta=\delta=3$, $\epsilon=7$, $\alpha=10$, $\gamma=16$, giving the password $1031637$, but that's a multiple of 3.

• If $\beta=4$, then we must have

  $\beta=\delta=4$, $\epsilon=9\text{ or }11$, $\alpha=13\text{ or }15$, $\gamma=22\text{ or }18$, giving the password $1342249$ or $15418411$, but these are not prime.

• If $\beta=5$, then we must have

  $\beta=\delta=5$, $\epsilon=11\text{ or }13$, $\alpha=16\text{ or }18$, $\gamma=28\text{ or }24$, giving the password $16528511$ or $18524513$, of which the latter is not prime but the former is.

Final solution for the password

$16528511$.

Answer 3

OP's Answer

The answer is, as shown by @MacGvyer88,

SEVEN

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

Word of length five.

Well, intrinsically, you can say that

we only need to consider "THREE" "SEVEN"

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

A B C D E or F.

And in second position it is very likely a

vowel. so A or E are most likely.

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

+a_+_ or +e_+_ or _a_a_ or _e_e_

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

SEVEN.

---

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.

Answer 4

Answer:

Alpha = 16, Beta = 5, Gamma = 28, Delta = 5, Epsilon = 11

Checking

1) 16+28+11 = 11*5
2) 28 is unique largest
3) 11+5=16
4) 5 >= 5
5) 5+5 < 11
6) Beta and Delta are both 5
7) 16528511 is prime

Partial explanation (to be continued):

Under the assumption that the Greek letters denote single digits, we quickly see that there is no solution. By trial and error, we show that Beta = 2, 3 and 4 won't work, and Beta = 5 yields the answer given above. Perhaps the larger value for Beta can give different solution, so this is not unique...