Answer :
Answer:
To determine the number of ways to select the digit \( b \) in the number 3a29b041 so that it is divisible by 9, we need to use the rule for divisibility by 9. A number is divisible by 9 if the sum of its digits is divisible by 9.
Let's denote the sum of the digits of the number 3a29b041 as \( S \).
The digits are:
\[ 3, a, 2, 9, b, 0, 4, 1 \]
The sum \( S \) can be written as:
\[ S = 3 + a + 2 + 9 + b + 0 + 4 + 1 \]
Simplifying the known values:
\[ S = 19 + a + b \]
For 3a29b041 to be divisible by 9, \( S \) must be divisible by 9. Therefore:
\[ 19 + a + b \equiv 0 \pmod{9} \]
We need to find valid values of \( b \) that satisfy this equation for all possible values of \( a \).
Let's consider \( a \) ranging from 0 to 9 and find corresponding values of \( b \).
First, let’s analyze the equation:
\[ 19 + a + b \equiv 0 \pmod{9} \]
This simplifies to:
\[ a + b \equiv -19 \pmod{9} \]
Since \(-19 \equiv 8 \pmod{9} \) (because \(-19 + 27 = 8\)):
\[ a + b \equiv 8 \pmod{9} \]
Now, we will find pairs \((a, b)\) that satisfy this equation where both \( a \) and \( b \) are digits from 0 to 9.
\[
\begin{aligned}
&\text{If } a = 0, & b = 8 \\
&\text{If } a = 1, & b = 7 \\
&\text{If } a = 2, & b = 6 \\
&\text{If } a = 3, & b = 5 \\
&\text{If } a = 4, & b = 4 \\
&\text{If } a = 5, & b = 3 \\
&\text{If } a = 6, & b = 2 \\
&\text{If } a = 7, & b = 1 \\
&\text{If } a = 8, & b = 0 \\
&\text{If } a = 9, & b = 8 \\
\end{aligned}
\]
Each pair \((a, b)\) listed above is valid for \( a \) ranging from 0 to 9. Therefore, there are 10 valid pairs that satisfy the condition \( a + b \equiv 8 \pmod{9} \).
Hence, there are \(\boxed{10}\) ways to select the pair of numbers \( a \) and \( b \) such that the number 3a29b041 is divisible by 9.