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...