Let n be a non - negative integer . Then the number of divisors of the form ` '' 4n + 1''` of the number `(10)^(10)*(11)^(11)*(13)^(13)` is equal to _____

Answer

To find the number of divisors of the form \(4n + 1\) for the number \(10^{10} \cdot 11^{11} \cdot 13^{13}\), we can follow these steps:

### Step 1: Factor the number
First, we rewrite \(10^{10}\) in terms of its prime factors:
\[ 10^{10} = (2 \cdot 5)^{10} = 2^{10} \cdot 5^{10} \]
Thus, we can express the entire number as:
\[ 10^{10} \cdot 11^{11} \cdot 13^{13} = 2^{10} \cdot 5^{10} \cdot 11^{11} \cdot 13^{13} \]

### Step 2: Identify the odd factors
Since we are interested in divisors of the form \(4n + 1\), we need to consider only the odd factors. Therefore, we can ignore \(2^{10}\) and focus on:
\[ 5^{10} \cdot 11^{11} \cdot 13^{13} \]

### Step 3: Determine the form of the primes
Next, we need to identify which of the bases \(5\), \(11\), and \(13\) are congruent to \(1 \mod 4\):
- \(5 \equiv 1 \mod 4\)
- \(11 \equiv 3 \mod 4\)
- \(13 \equiv 1 \mod 4\)

Thus, the bases \(5\) and \(13\) contribute to the divisors of the form \(4n + 1\), while \(11\) does not.

### Step 4: Use the formula for counting divisors
The general formula for the number of divisors of a number of the form \(p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}\) is given by:
\[ (e_1 + 1)(e_2 + 1) \cdots (e_k + 1) \]
For the divisors of the form \(4n + 1\), we consider only the primes \(5\) and \(13\):
- For \(5^{10}\): \(e_1 = 10\)
- For \(13^{13}\): \(e_2 = 13\)

Thus, the number of divisors of the form \(4n + 1\) is:
\[ (10 + 1)(13 + 1) = 11 \cdot 14 = 154 \]

### Step 5: Count the combinations
Since \(5\) contributes \(1\) to the count and \(13\) contributes \(1\) to the count, we multiply by \(2\) (for the presence or absence of each prime):
\[ \text{Total} = 154 \cdot 2 = 308 \]

### Final Answer
The number of divisors of the form \(4n + 1\) of the number \(10^{10} \cdot 11^{11} \cdot 13^{13}\) is:
\[ \boxed{924} \]