## Task A-2.5.

Let $A$ be the number of six-digit numbers whose product of digits is 105, and $B$ be the number of six-digit numbers whose product of digits is 147. Determine the ratio $A: B$.
## First solution.

Let's factorize the given products into prime factors: $105=3 \cdot 5 \cdot 7$ and $147=3 \cdot 7 \cdot 7$.

The products of any two of the numbers $3, 5, 7$ are two-digit numbers, so any six-digit number whose product of digits is 105 has digits 1, 1, 1, 3, 5, 7 in some order.

1 point

Similarly, any six-digit number whose product of digits is 147 has digits $1, 1, 1, 3, 7, 7$ in some order.

1 point

If we select one of the two 7s in a number whose product of digits is 147 and replace it with the digit 5, we will get a number whose product of digits is 105. Therefore, the number $A$ is twice as large as the number $B$, i.e., $A: B=2: 1$.