Let `log_(a)N=alpha + beta` where ` alpha ` is integer and ` beta =[0,1)`. Then , On the basis of above information , answer the following questions.
The difference of largest and smallest integral value of N satisfying ` alpha =3 and a =5` , is

Options

A. 499
B. 500
C. 501
D. 502

Answer

To solve the problem, we need to find the difference between the largest and smallest integral values of \( N \) satisfying the given conditions. Let's break down the steps:

### Step 1: Understand the given equation
We are given:
\[ \log_a N = \alpha + \beta \]
where \( \alpha \) is an integer, and \( \beta \) is in the range \([0, 1)\).

### Step 2: Substitute the values
We have \( \alpha = 3 \) and \( a = 5 \). Thus, we can rewrite the equation as:
\[ \log_5 N = 3 + \beta \]

### Step 3: Convert the logarithmic equation to exponential form
Using the properties of logarithms, we can convert this to:
\[ N = 5^{3 + \beta} \]
This can be further expressed as:
\[ N = 5^3 \cdot 5^{\beta} \]
Calculating \( 5^3 \):
\[ N = 125 \cdot 5^{\beta} \]

### Step 4: Determine the range of \( N \)
Since \( \beta \) ranges from \( 0 \) to just below \( 1 \) (i.e., \( \beta \in [0, 1) \)), we can find the minimum and maximum values of \( N \).

- **Minimum value of \( N \)** occurs when \( \beta = 0 \):
\[ N_{\text{min}} = 125 \cdot 5^0 = 125 \]

- **Maximum value of \( N \)** occurs when \( \beta \) approaches \( 1 \):
\[ N_{\text{max}} = 125 \cdot 5^{1} = 125 \cdot 5 = 625 \]
However, since \( \beta \) cannot actually equal \( 1 \), we consider the largest integer less than \( 625 \), which is \( 624 \).

### Step 5: Find the difference between the largest and smallest integral values of \( N \)
Now we can find the difference:
\[ \text{Difference} = N_{\text{max}} - N_{\text{min}} = 624 - 125 \]
Calculating this gives:
\[ \text{Difference} = 499 \]

### Final Answer
Thus, the difference of the largest and smallest integral values of \( N \) satisfying the conditions is:
\[ \boxed{499} \]