The number of ways to choose 7 distinct natural numbers from the first 100 natural numbers such that any two chosen numbers differ atleast by 7 can be expressed as `""^(n)C_(7)`. Find the number of divisors of n.

Answer

To solve the problem of choosing 7 distinct natural numbers from the first 100 natural numbers such that any two chosen numbers differ by at least 7, we can follow these steps:

### Step 1: Understanding the Problem
We need to select 7 distinct numbers from the set {1, 2, ..., 100} with the condition that the difference between any two selected numbers is at least 7.

### Step 2: Adjusting for the Minimum Difference
If we select a number \( x_1 \), the next number \( x_2 \) must be at least \( x_1 + 7 \). This means that for every selected number, we need to account for the 6 numbers that we cannot choose after selecting a number.

### Step 3: Transforming the Selection
To simplify the selection, we can define new variables:
- Let \( y_1 = x_1 \)
- Let \( y_2 = x_2 - 6 \)
- Let \( y_3 = x_3 - 12 \)
- ...
- Let \( y_7 = x_7 - 36 \)

This transformation effectively reduces the problem to selecting 7 numbers \( y_1, y_2, ..., y_7 \) from a smaller set.

### Step 4: Finding the Range for \( y_i \)
The largest possible value for \( x_7 \) is 100. Thus:
\[ y_7 = x_7 - 36 \leq 100 - 36 = 64 \]

So, we need to choose 7 distinct numbers \( y_1, y_2, ..., y_7 \) from the set {1, 2, ..., 64}.

### Step 5: Counting the Selections
The number of ways to choose 7 distinct numbers from 64 is given by the combination formula:
\[ \binom{64}{7} \]

### Step 6: Relating to the Given Format
According to the problem, this can be expressed as \( \binom{N}{7} \). Therefore, we have:
\[ N = 64 \]

### Step 7: Finding the Number of Divisors of \( N \)
To find the number of divisors of \( N = 64 \), we first express 64 in its prime factorization:
\[ 64 = 2^6 \]

The formula for finding the number of divisors from the prime factorization \( p_1^{e_1} \times p_2^{e_2} \times ... \times p_k^{e_k} \) is:
\[ (e_1 + 1)(e_2 + 1)...(e_k + 1) \]

For \( 64 = 2^6 \):
- \( e_1 = 6 \)

Thus, the number of divisors is:
\[ (6 + 1) = 7 \]

### Final Answer
The number of divisors of \( N \) is **7**.

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