Puzzle ZPXQ 

Workings

Hint
Start by labelling the number as ABCDE.

Answer
24,976.

Reasoning
We can start by labelling the digits as ABCDE.

We know that:

(i)  B = 2 x A

and:

     E = A + B

And using (i) we get:

     E = A + (2 x A)
(ii) E = 3 x A

If A = 1, this isn't allowed (as there are no 1's in the puzzle).

If A = 2, then B = 4, and E = 6.

If A = 3, then B = 6, and E = 9, but this isn't allowed (as C has to be the largest digit).

So, A = 2, B = 4, E = 6, and we now have to find C and D.

We also know that:

D = (A + B + C + D + E) ÷ 4

And using (i) and (ii) we get:

D = [A + (2 x A) + C + D + (3 x A)] ÷ 4

so:

3 x D = (6 x A) + C

so:

(iii) D = [(6 x A) + C] ÷ 3

C can only be 7, 8 or 9 (as it's the largest digit, and we've already found 6) and (iii) tells us that it must be a multiple of 3, which means that C = 9. Leaving D = 7.

So the final number is: 24976.

Double-Checking
The answer is 24976.

The fourth digit is a quarter of the total of all of the digits.
   A + B + C + D + E = 2 + 4 + 9 + 7 + 6 = 28, and 28 ÷ 4 = 7.

The second digit is twice the first digit.
   4 = 2 x 2.

The third digit is the largest.
   9 is the largest digit.

The last digit is the sum of the first two digits.
   6 = 2 + 4.