Number Tiles Distribution

Let $A$ 's tiles be $a_1<a_2$ , etc. The clues are $\begin{aligned} 3a_1 a_2&=c_1 c_2 \\ 4\left(b_1+b_2 \right)&=d_1 d_2 \\ c_1+c_2&=d_1+d_2 \\ a_1 a_2 + b_1 b_2 + c_1 c_2 + d_1 d_2 &= 109 \end{aligned}$

Substituting for the products of $C$ 's and $D$ 's tiles, $a_1 a_2 + b_1 b_2 + 3a_1 a_2 + 4\left(b_1+b_2 \right)=109$ ie $4a_1 a_2 + b_1 b_2 + 4\left(b_1+b_2 \right)=109$

Considering this modulo $4$ , we have $b_1 b_2 \equiv 1 \pmod4$

This means both of $B$ 's tiles are congruent to $1$ or both are congruent to $3$ modulo $4$ . The only possibilities are $(b_1,b_2)=(5,9)$ or $(b_1,b_2)=(3,7)$ .

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Case $(b_1,b_2)=(5,9)$ : $4\left(b_1+b_2 \right)=56=d_1 d_2$ so $D$ 's tiles are $7$ and $8$ . But $c_1+c_2=d_1+d_2=15$ doesn't have a solution without reusing the $9$ .

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Case $(b_1,b_2)=(3,7)$ : As above, we find $D$ 's tiles are $5$ and $8$ ; we can carry on to find $C$ 's tiles are $4$ and $9$ and $A$ 's tiles are $2$ and $6$ . It's easy to check all the conditions work out, so the answer is $3+7=\boxed{10}$ .

Make the matrix like this:

$\begin{bmatrix}A\\B\\C\\D\end{bmatrix}=\begin{bmatrix}A_1&A_2\\B_1&B_2\\C_1&C_2\\D_1&D_2\end{bmatrix}$

According to clues, we have these equations:

$\begin{cases} & 3(A_1A_2)=C_1C_2\\ & 4(B_1 + B_2)=D_1D_2\\ & C_1+C_2=D_1+D_2\\ & A_1A_2+B_1B_2+C_1C_2+D_1D_2=109 \end{cases}$

Then we got

$\begin{bmatrix}A\\ B\\ C\\ D\end{bmatrix}=\begin{bmatrix}2&6\\ \blue{3}&\blue{7}\\ 4&9\\ 5&8\end{bmatrix}$

Therefore, the sum of $B$ 's tiles is $\boxed{10}$ .

1) The product of A's tiles is one-third of C's product.
2) The sum of B's tiles is one-fourth of D's product.
3) The sum of C's tiles is equal as D's sum.
4) The total of group's tiles product is 109.

Let's have the tiles be A, a, B, b, C, c, D and d.
Aa = (Cc) / 3 ==> Cc = 3Aa
B + b = (Dd) / 4 ==> Dd = 4(B + b)

We already have a great number theory solution by Chris, so I'm just presenting a logic based one with basic even odd for school children (no higher mod here, even though I know even-odd parity IS a tale of mod 2).

Looking at the sum of
As + Bb + Cc + Dd = 109,
we can see that there must exist at least a set with two odd numbers (say we call this wholly odd set) to get an odd sum of 109, and that there must be an odd number of wholly odd set at that. Since there exists only 4 odd numbers to be distributed and no sharing allowed, it's obvious we can only have one wholly odd set at most.

Now with Cc = 3Aa, since it involves two products here, we know that relatively big primes such as 5 and 7 which are alone in the selection (for small primes like 2 and 3, in a consecutive selection of range 8 from 2 ~ 9, they got 'friends' around with 9/2 = 4.5 ≥ 2 and 9/3 = 3 ≥ 2), both can neither be a member of A set nor C set, so the two would have to be in B set of D set, together or separately remain to be seen.

Looking at Dd = 4(B + b) now, there must be at least a factor of 4 in Dd, and that means one of two things could happen here :

1) one of D set tiles is either 4 or 8,
OR
2) both of D set tiles are even ==> B must take both 5 and 7 ==> with a sum of two relatively big primes, D would also be forced to take the relatively big even numbers to get a quadruple product, and those would be 6 and 8 ((6 x 8)/4 = 12 = 5 + 7) ==> then, D set's sum would also be relatively big with the largest 2 even number members at 6 + 8 = 14 with no equal other than reusing B's 5 for C's supposedly biggest number 9.

By now, we've seen the impossibility of that route, so it must be that D got either 4 or 8 to be paired of to either 5 or 7 (since the two together spells trouble with their relatively large sum for B set, and they cannot be together within the D set since one space is already reserved for that multiple of 4). Logically, D cannot take the smaller numbers for both options, because if D take the smaller 5 AND 4, then the bigger 7 in B cannot reach a sum of 5 with whatever positive numbers it was paired to in B.

Anyway, the formula for that other tile in set B is this : b = (1 or 2) x D - B
D & B is 5 or 7 each, and 1 or 2 depending on the 4 or 8 in D set.

So if we use 1, then it should be D > B, but 2 has no limit though b < 10.

b = 1 x 5 - 7 = -2 (impossible as discussed earlier) or
b = 1 x 7 - 5 = 2 or
b = 2 x 7 - 5 = 9 or
b = 2 x 5 - 7 = 3 or

Now, let's return to the sets A & C. 5 & 7 excluded, most are also possible for D and B sets, except 6. We know that between A & C, one has an extra factor of 3 than the other. With 6 already a confirmed member of either A or C set, they would need an additional even factors of 3, if any. But since there is only one b in existence, b can only choose one of the 3 options between 2 or 3 or 9. But 3 and 9 together has an odd factor of 3, so b has to take the odd one and return the even factors of 3 to A-C coalition along with 2 (returning both 3 AND 9 is impossible because A-C already have 6 on their side, thus if they had to take extra factors of 3, those have to come in pairs --> to even things out).

Now we're almost done by knowing that 3 and 8 (by the use of 2 in 3 = b = 2 x 5 - 7) and 5 and 7 are in the same B-D coalition. Referring back to our formula,
b = (1 or 2) x D - B
we see that 3 is paired to 7 in B set and 5 to 8 in D set.

Answer = 3 + 7 = 10