We'll call this number \(A B C D\)
We have \(A=\frac{1}{5}D\)⇒\(D=5A\)
Since \(D\) is divisible by 5, it is either 0 or 5
But it can't be 0, because then \(A\) would be 0
So \(D=5\) and \(A=1\)
Now I'm assuming that the second and third digits are adding to three times the last digit.
So \(B+C=3*5\)⇒\(B+C=15\)
We have a number of options:
\((B, C)=(6, 9)\)
\((B, C)=(7, 8)\)
\((B, C)=(8, 7)\)
\((B, C)=(9, 6)\)
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.)
