Neat-looking problem!

Okay, so we are choosing digits from this list: {4, 5, 6, 7, 8, 9}

5 unique digits which sum to 30.

The numbers in the list sum to 39, so we must omit 9.

Our number must include the digits 4, 5, 6, 7, 8 in some order.

Even, so cannot end in 5 or 7.

Let's see what else we can figure out.

"The digit in the hundreds place is 2 less than the digit in the thousands place."

We have one of these:

A75DE or

A86D4 or

A64D8

What does that allow for the tens place?

~~A6438 ~~<-- omitted!

A754E or

A8654 or

Three final options to select from:

87546

~~67548~~ <-- omitted; n is divis. by 4, so n+6 must not be.

~~78654~~ <-- omitted; digit in ten-thousands is not 1 more than adjacent digit.

Confiming that 87546 + 6 is divisible by 9 and 4:

87552 <-- Sum of digits is 17+10 = 27; Last 2 digits = 52 = (4)(13) --> DONE!

87546 <-- Answer!

Did you or someone you know make this problem up yourself? If so, it's pretty neat!