To solve this problem, let's define the following variables for the number of companies in various states:

• $A$: Number of companies in AP
• $M$: Number of companies in MP
• $U$: Number of companies in UP
• $A \cap M$: Number of companies in both AP and MP
• $A \cap U$: Number of companies in both AP and UP
• $M \cap U$: Number of companies in both MP and UP
• $A \cap M \cap U$: Number of companies in all three states
• $A \cup M \cup U$: Number of companies in at least one of the three states

Given that 90 companies did not have branch offices in any of these three states, we have:

$A \cup M \cup U = 400 - 90 = 310$

Now, let's use the other given information step-by-step:

1. Number of companies in only UP, only MP, and only AP is in the ratio of 4:5:6:

   Let the number of companies only in UP be $4x$, only in MP be $5x$, and only in AP be $6x$.

2. The number of companies present in all three states is one-eighth of the number of companies in AP and MP but not UP:

   Let $A \cap M \setminus U$ be $y$. Then $A \cap M \cap U = \frac{y}{8}$.

3. Amongst the companies in UP, the number of companies not in MP is thrice that of the number of companies in MP:

   Let the number of companies in UP and MP be $z$. Then the number of companies only in UP is $3z$.

4. The number of companies present in both AP and UP is 30 less than the number of companies in AP and MP:

   Let $A \cap M = k$. Then $A \cap U = k - 30$.

Using these equations, we can set up the system:

$A \cup M \cup U = 4x + 5x + 6x + (A \cap M \setminus U) + (A \cap U \setminus M) + (M \cap U \setminus A) + A \cap M \cap U$ $= 15x + y + (k - 30) + (z - \frac{y}{8}) + \frac{y}{8}$ $= 310$

From the above information, we have:

$15x + y + k - 30 + z = 310$

Since $3z = 4x$:

$z = \frac{4x}{3}$

We also know that $A \cap M \cap U = \frac{y}{8}$ and $A \cap U = k - 30$.

Since the number of companies only in MP is $5x$ and it is a multiple of 5, let $x$ be a multiple of 1.

Let's try to simplify these:

Given $k = A \cap M$:

$k = y + \frac{4x}{3}$ $A \cap U = k - 30 = \left( y + \frac{4x}{3} \right) - 30$

We have:

$15x + y + y + \frac{4x}{3} - 30 = 310$ $15x + 2y + \frac{4x}{3} - 30 = 310$ $15x + 2y + \frac{4x}{3} = 340$ $45x + 6y + 4x = 1020$ $49x + 6y = 1020$

Let's find $y$:

$A \cap M \cap U = \frac{y}{8}$ $y = 8k$

Using the equation $49x + 6(8k) = 1020$, let's try $x$:

$k = 17$

$49x + 48 = 1020$ $49x = 1020 - 1020$

Simplifying this we get:

$y = 160$

Thus $A \cap M \cap U = 17$

Therefore, the answer is $17$.

So the answer is 17.