The digits to be inserted into the blanks are denoted by $a, b$, and $c$, such that the resulting number has the digit representation $43 a 1 b 5 c$.

A number is divisible by 75 if and only if it is divisible by both 3 and 25, since 3 and 25 are coprime. It is divisible by 25 exactly when $c=0$.

It is divisible by 3 exactly when its digit sum $(4+3+a+1+b+5+c=) 13+a+b$ is divisible by 3.

Since $0 \leq a \leq 9$ and $0 \leq b \leq 9$, it follows that $0 \leq a+b \leq 18$. Therefore, for the mentioned digit sum, $13 \leq 13+a+b \leq 31$. It is thus divisible by 3 exactly when it is one of the numbers 15, 18, 21, 24, 27, 30, i.e., exactly when the sum $s$ of the two digits $a$ and $b$ is one of the numbers 2, 5, 8, 11, 14, 17.

The following table contains all digit pairs $(a ; b)$ that have one of these sums $s=a+b$:

| Digit sum $s$ | All digit pairs $(a ; b)$ with $a+b=s$ | Number of digit pairs |
| :---: | :--- | :---: |
| 2 | $(0 ; 2),(1 ; 1),(2 ; 0)$ | 3 |
| 5 | $(0 ; 5),(1 ; 4),(2 ; 3),(3 ; 2),(4 ; 1),(5 ; 0)$ | 6 |
| 8 | $(0 ; 8),(1 ; 7),(2 ; 6),(3 ; 5),(4 ; 4),(5 ; 3),(6 ; 2),(7 ; 1),(8 ; 0)$ | 9 |
| 11 | $(2 ; 9),(3 ; 8),(4 ; 7),(5 ; 6),(6 ; 5),(7 ; 4),(8 ; 3),(9 ; 2)$ | 8 |
| 14 | $(5 ; 9),(6 ; 8),(7 ; 7),(8 ; 6),(9 ; 5)$ | 5 |
| 17 | $(8 ; 9),(9 ; 8)$ | 2 |

Therefore, the sought number of digit pairs is $3+6+9+8+5=33$.
## Task 5 - 240735

In the schema 43_1_5_, each of the blanks _ is to be filled with a digit so that the resulting seven-digit number is divisible by 75.

State how many seven-digit numbers can arise in this way!