We'll call this number [tex]A B C D[/tex]
We have [tex]A=\frac{1}{5}D[/tex]⇒[tex]D=5A[/tex]
Since [tex]D[/tex] is divisible by 5, it is either 0 or 5
But it can't be 0, because then [tex]A[/tex] would be 0
So [tex]D=5[/tex] and [tex]A=1[/tex]
Now I'm assuming that the second and third digits are adding to three times the last digit.
So [tex]B+C=3*5[/tex]⇒[tex]B+C=15[/tex]
We have a number of options:
[tex](B, C)=(6, 9)[/tex]
[tex](B, C)=(7, 8)[/tex]
[tex](B, C)=(8, 7)[/tex]
[tex](B, C)=(9, 6)[/tex]
So the possibilities are:
1695
1785
1875
1965
(If there were another detail we could narrow down the choices further, but otherwise we have all these options.)
