Largest 3 digited number following some criteria!!

It is easy to get the number with the form 1AB $1AB \times 5 = CDE$ E can be 0 or 5. if E=5, can't find this number with C>D>5. if E=0: $\Rightarrow$ the number is $\boxed{168}$

ABC × 5 = DEF
A must be 1 or the product would exceed 1000 and becomes a 4 digit number. F can be 5 or 0, but if it's 5, E = { 6, 7, 8 }. This will make our C = { 3, 5, 7 } respectively.

Possible ABCs ending in 5 :
123
125
127
135
137
145
147
157
167

None of the above listed numbers have a product with descending digits, and since D ≥ 7 making ABC > 140, we only need to check the last 4.
145 X 5 = 725
147 X 5 = 735
157 X 5 = 785
167 X 5 = 835

So F = 0 and C must be even. C = { 4, 6, 8 }, so we can try with 178, then 168, and so on. 168 X 5 = 840 ✓, while the biggest possibility yield 178 X 5 = 890.