1) H4B is F*C, which is two digits. Neither C nor F can be 1, since F*C wouldn't be two digits.
3) V4B is one digit. C* (A+A-F). C can't be 1, which means that 2A-F can't be greater than 4. A can't be 1, either. So if A is 2, F must be 1, 2 or 3. But it can't be 1, as I pointed out above. So it would have to be 2 or 3. If A is 3, F must be 2 or 3. If A is 5, then F must be 3. A can't be 7 or higher and still have V4B be one digit. So we know that A has to be 2, 3 or 5, and that F must be 2 or 3, and we can narrow it down further once we know what A is.
4) H4B is F*C, which is two digits. And since we know that F is either 2 or 3, C must be 5, 7 or 9.
5) H4A is (B+D*D)/C*C and is a single digit integer. DD can be either 1, 4, 9, 25, 49 or 81. CC can be any of those except 1. That makes it really hard for C to be a big number, which narrows us down significantly. Something like (3 + 25)/4 = 7 would fit the bill, but we know that C must be 5, 7 or 9. For B+DD/25, 49 or 81 to be a single digit integer, there's a very short list of possible answers. 1 + 49/.25 is possible. So would 9 + 16/25. I think those are the only two possible answers. In both cases, we know that C is 5. B can be either 1 or 9, and D can be either 4 or 7.
A
B
C 5
D
E
F
6) If C is 5, and F is either 2 or 3, then F*C is either 10 or 15. So we can fill in the first digit in our puzzle.
OO@O
OO@O
OOO@
O@1O
7) V3 is equal to F(C+BA). If the last digit is 1, then F must be an odd number. As I showed above, F must be 2 or 3. Which means that F is 3.
So now we know
A
B
C 5
D
E
F 3
which means we can also figure out H4B: FC =15
OO@O
OO@O
OOO@
O@15
8) That makes V4B pretty easy. C * (A + A - F) = 5. 5* (A+A-3)=5. A+A = 4. A=2.
A 2
B
C 5
D
E
F 3
9) That makes H3 simple math. AC is 25. 25*25 = 625.
OO@O
OO@O
625@
O@15
10) So V3 equals 51. F * (C + BA), or 3 * (5+_2) = 51. 5 +_2 = 17. Which means B=1
A 2
B 1
C 5
D
E
F 3
11) H2B is C-E/E. C is 5. E/E is 1. That means H2B is 4.
OO@O
OO@4
625@
O@15
12) V4A is (BD - C) * A. That's (1_-5)*2. So D is either 2 or 7. Going back to H4A, (B + D*D) / (C*C), if we know B is 1 and C is 5, then D must be 2 or 7. V2 is (BBB-D)*F Based on the crossword, we know it ends with a 2. 111-D*3 to be an even number, 111-D must be an even number. Thus D must be an odd number, which means D is 7. 104*3 = 412.
A 2
B 1
C 5
D 7
E
F 3
O3@O
O1@4
625@
O@15
13) H1A is (AA-BC)*(D+A). 22-15*7+2 = 7*9 = 63
63@O
O1@4
625@
O@15
14) V4A is (BD - C) * A, or 17-5*2 = 24
63@2
O1@4
625@
O@15
15) H1B is (E+C)/D, and must equal 2 according to the crossword. (E + 5)/7 = 2. E+5=14. E=9.
A 2
B 1
C 5
D 7
E 9
F 3
16) V1 is FB*A*(EE +A). 31*2*(99+2) = 62*101 = 6262
63@2
21@4
625@
2@15