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-What is your password?- my niece asks me.

-It is a four digit number.

-I know that.

-It is divisible by precisely three primes.

-Tell me more.

-It has at least one common divisor greater than 1 with precisely eight of the other 23 4-digit numbers that can be formed with those very same digits.

-More.

-I was alive in that particular year.

A few hours later:

-Now I know!

What is my password?

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  • 1
    $\begingroup$ A few hours? A 4 digit password can be brute-forced by hand faster than that :-) $\endgroup$ – Bass Aug 31 '20 at 20:13
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    $\begingroup$ @Bass She took a nap before working it out by hand! $\endgroup$ – Bernardo Recamán Santos Aug 31 '20 at 20:19
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    $\begingroup$ @Bass it's been a few hours, where's the answer? ;) $\endgroup$ – TCooper Aug 31 '20 at 23:19
  • $\begingroup$ @WhatsUp I have added additional information to make answer unique. $\endgroup$ – Bernardo Recamán Santos Sep 1 '20 at 3:19
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EDIT: with extra information provided by the OP, now we could say that the answer is

1976

and this also spoils a lower bound of the age of the OP, assuming he's not lying (:

Original answer:


I don't know what I'm missing, but here's what I got.

Clarifications:

  • I assume that "the other 23 4-digit numbers" implies that the number consists of four different non-zero digits.
  • I assume that "at least one common divisor" means "one common divisor greater than $1$".

With these assumptions, I got:

$1435 = 5 \times 7 \times 41$
$1495 = 5 \times 13 \times 23$
$1976 = 2^3 \times 13 \times 19$
$2135 = 5 \times 7 \times 61$
$2431 = 11 \times 13 \times 17$
$3145 = 5 \times 17 \times 37$
$3196 = 2^2 \times 17 \times 47$
$3289 = 11 \times 13 \times 23$
$3514 = 2 \times 7 \times 251$
$3598 = 2 \times 7 \times 257$
$4697 = 7 \times 11 \times 61$
$5423 = 11 \times 17 \times 29$
$6149 = 11 \times 13 \times 43$
$6391 = 7 \times 11 \times 83$
$6475 = 5^2 \times 7 \times 37$
$6479 = 11 \times 19 \times 31$
$6935 = 5 \times 19 \times 73$
$7385 = 5 \times 7 \times 211$
$7469 = 7 \times 11 \times 97$
$7843 = 11 \times 23 \times 31$
$7931 = 7 \times 11 \times 103$
$8435 = 5 \times 7 \times 241$
$9361 = 11 \times 23 \times 37$
$9581 = 11 \times 13 \times 67$
$9823 = 11 \times 19 \times 47$
$9835 = 5 \times 7 \times 281$

And I don't see any information coming from the sentences of the niece (i.e. "I know that" and "Tell me more" don't give extra information here, as far as I can tell).

Thus I'm confused at this point.

Some analysis here:

There are $1104$ numbers with four different non-zero digits that have three prime divisors.
Among them, we found the above $26$ solutions, which is about $1/42$ of all the $1104$ candidates.

This is still lower than a naive estimation of $1 / 24$ (assuming that the number of permutations with common divisors distributes uniformly in $\{0,1,\dots, 23\}$).

Thus it is quite reasonable to expect such a situation that we have many solutions. That is, if this is just a mathematical puzzle...

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  • $\begingroup$ smh. I wrote a script to find this and could never identify the final number by the divisor rule because I didn't exclude 1... $\endgroup$ – TCooper Sep 1 '20 at 15:59

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