Let `log_2N=a_1+b_1,log_3N=a_2+b_2` and `log_5N=a_3+b_3`, where `a_1,a_2,a_3notin1` and `b_1,b_2 b_3 in [0,1)`.
If `a_1=6,a_2=4` and `a_3=3`,the difference of largest and smallest integral values of N, is

Options

A. 2
B. 8
C. 14
D. 20

Answer

To solve the problem, we need to find the integral values of \( N \) based on the logarithmic expressions provided. We will analyze each logarithmic equation step by step.

### Step 1: Analyze \( \log_2 N = a_1 + b_1 \)

Given:
- \( a_1 = 6 \)
- \( b_1 \in [0, 1) \)

Using the property of logarithms, we can rewrite this as:
\[ N = 2^{a_1 + b_1} = 2^{6 + b_1} = 2^6 \cdot 2^{b_1} = 64 \cdot 2^{b_1} \]

Since \( b_1 \) can vary from \( 0 \) to just below \( 1 \):
- Minimum value of \( N \):
\[ N_{\text{min}} = 64 \cdot 2^0 = 64 \]
- Maximum value of \( N \):
\[ N_{\text{max}} < 64 \cdot 2^1 = 128 \]

Thus, \( N \) can take values in the interval \( [64, 128) \).

### Step 2: Analyze \( \log_3 N = a_2 + b_2 \)

Given:
- \( a_2 = 4 \)
- \( b_2 \in [0, 1) \)

Using the property of logarithms:
\[ N = 3^{a_2 + b_2} = 3^{4 + b_2} = 3^4 \cdot 3^{b_2} = 81 \cdot 3^{b_2} \]

Since \( b_2 \) can vary from \( 0 \) to just below \( 1 \):
- Minimum value of \( N \):
\[ N_{\text{min}} = 81 \cdot 3^0 = 81 \]
- Maximum value of \( N \):
\[ N_{\text{max}} < 81 \cdot 3^1 = 243 \]

Thus, \( N \) can take values in the interval \( [81, 243) \).

### Step 3: Analyze \( \log_5 N = a_3 + b_3 \)

Given:
- \( a_3 = 3 \)
- \( b_3 \in [0, 1) \)

Using the property of logarithms:
\[ N = 5^{a_3 + b_3} = 5^{3 + b_3} = 5^3 \cdot 5^{b_3} = 125 \cdot 5^{b_3} \]

Since \( b_3 \) can vary from \( 0 \) to just below \( 1 \):
- Minimum value of \( N \):
\[ N_{\text{min}} = 125 \cdot 5^0 = 125 \]
- Maximum value of \( N \):
\[ N_{\text{max}} < 125 \cdot 5^1 = 625 \]

Thus, \( N \) can take values in the interval \( [125, 625) \).

### Step 4: Find the intersection of the intervals

Now we have the following intervals for \( N \):
1. From \( \log_2 N \): \( [64, 128) \)
2. From \( \log_3 N \): \( [81, 243) \)
3. From \( \log_5 N \): \( [125, 625) \)

To find the common values of \( N \), we need the intersection of these intervals:
- The intersection of \( [64, 128) \) and \( [81, 243) \) is \( [81, 128) \).
- The intersection of \( [81, 128) \) and \( [125, 625) \) is \( [125, 128) \).

### Step 5: Determine the integral values of \( N \)

From the interval \( [125, 128) \), the only integer values are:
- \( 125 \)
- \( 126 \)
- \( 127 \)

### Step 6: Calculate the difference between the largest and smallest integral values of \( N \)

The largest integral value is \( 127 \) and the smallest integral value is \( 125 \). Thus, the difference is:
\[ \text{Difference} = 127 - 125 = 2 \]

### Final Answer:
The difference of the largest and smallest integral values of \( N \) is \( 2 \).

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