Which integer values of b would give the number `2002 div 10^(-b)` a value between 1 and 100?

Answer

To solve the problem of finding integer values of \( b \) such that the expression \( \frac{2002}{10^{-b}} \) lies between 1 and 100, we can follow these steps:

### Step 1: Rewrite the Expression
The expression can be rewritten using the property of exponents:
\[ \frac{2002}{10^{-b}} = 2002 \times 10^{b} \]
This means we need to find \( b \) such that:
\[ 1 < 2002 \times 10^{b} < 100 \]

### Step 2: Break Down the Inequality
We can break this down into two inequalities:
1. \( 2002 \times 10^{b} > 1 \)
2. \( 2002 \times 10^{b} < 100 \)

### Step 3: Solve the First Inequality
For the first inequality \( 2002 \times 10^{b} > 1 \):
\[ 10^{b} > \frac{1}{2002} \]
Taking the logarithm (base 10) of both sides:
\[ b > \log_{10}\left(\frac{1}{2002}\right) \]
Calculating \( \log_{10}(2002) \) gives approximately \( 3.301 \), thus:
\[ b > -3.301 \]

### Step 4: Solve the Second Inequality
For the second inequality \( 2002 \times 10^{b} < 100 \):
\[ 10^{b} < \frac{100}{2002} \]
Taking the logarithm (base 10) of both sides:
\[ b < \log_{10}\left(\frac{100}{2002}\right) \]
Calculating \( \log_{10}(100) = 2 \) and \( \log_{10}(2002) \approx 3.301 \):
\[ b < 2 - 3.301 \implies b < -1.301 \]

### Step 5: Combine the Results
From the inequalities, we have:
\[ -3.301 < b < -1.301 \]
The integer values of \( b \) that satisfy this inequality are \( b = -3 \) and \( b = -2 \).

### Conclusion
Thus, the integer values of \( b \) that give the number \( \frac{2002}{10^{-b}} \) a value between 1 and 100 are:
\[ \boxed{-3 \text{ and } -2} \]