Maths problems (interesting ones)

[FizMan]

Originally Posted by Denman

There is a five-digit number having digits 1-9, no two digits being identical. Two are prime numbers, two are square numbers, and one is neither. The third digit is twice the fifth, the fourth is six greater than the second and the last is three less than the first.
This is really hard actually, i havn't worked it out yet lol

73894

Here's my notepad logic. Then I just fiddled around with numbers for five minutes at the end until I got something that worked.

abcde
c = 2e
d = b + 6
e = a - 3
c = 2a - 6
c == even
c < 10
2a - 6 < 10
2a < 16
a < 8

c < 10
2e < 10
e < 5

d < 10
b + 6 < 10
b < 4

e < 10
a - 3 < 10
a < 13

a is less than 8
e = (<8) - 3
e < 5

abcde
73894