Let's find all solutions using formal logic.
Given a logic statement of the form "if $A$ then $B$", we can express it symbolically as $A \implies B$ or its equivalent, $\bar A \lor B$.
Numbering the given statements 1,2,3 respectively, we have
$A_1$ = the truth value of this number is a multiple of 5,
$B_1$ = the truth value of this number would lie between 1 to 19,
and so on.
Let $E$ be (the truth value of) the conjunction of the 3 statements. Then:
E = & (\bar A_1 \lor B_1) (\bar A_2 \lor B_2) (\bar A_3 \lor B_3) \\
= & \bar A_1 \bar A_2 \bar A_3 \lor \bar A_1 \bar A_2 B_3
\lor \bar A_1 B_2 \bar A_3 \lor \bar A_1 B_2 B_3 \\
& \lor B_1 \bar A_2 \bar A_3 \lor B_1 \bar A_2 B_3
\lor B_1 B_2 \bar A_3 \lor B_1 B_2 B_3
Now, since every multiple of 5 is also a multiple of 10, the conjunction $A_1 A_3$ must be false because it asserts that the number is a multiple of 5 and not a multiple of 10.
Also, $B_i B_j$ is false if $i \neq j$ because the intervals being disjoint means that the number cannot be simultaneously a member of two intervals.
We can therefore simplify to get $E = \bar A_1 \bar A_2 B_3 \lor B_1 \bar A_2 \bar A_3$.
The first term requires a multiple of 8 from the interval 30 to 39 that is not a multiple of 5, so 32 is a solution.
The second term requires a multiple of 8 and 10 from the interval 1 to 19. There aren't any.
Hence 32 is the only solution.