-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!

• A few hours? A 4 digit password can be brute-forced by hand faster than that :-)
– Bass
Aug 31, 2020 at 20:13
• @Bass She took a nap before working it out by hand! Aug 31, 2020 at 20:19
• @Bass it's been a few hours, where's the answer? ;) Aug 31, 2020 at 23:19

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 (:

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...

• 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... Sep 1, 2020 at 15:59