o=3............................eq1
h*t*3=30
ht=10..................eq2
h=t+o
h=t+3..................eq3
sustituting for h in eq2 we get
(t+3)t=10
t^2+3t-10=0
t^2+5t-2t-10=0
t(t+5)-2(t+5)=0
(t-2)(t+5)=0
t=2 (or t=-5.......omit since digit cannot be -ve)
h=t+3
h=5
The number is hto=523