
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process.

[ASSESS]

[ADVANCE]

[VERIFY]

[SIMPLIFY]

[SYNTHESIZE]

[PIVOT]

[OUTPUT]

You should strictly follow the format below:

[ACTION NAME]

# Your action step 1

# Your action step 2

# Your action step 3

...

Next action: [NEXT ACTION NAME]


system
Write a natural number \(a\) in canonical form:

\[ a = p_{1}^{s_{1}} \cdot p_{2}^{s_{2}} \cdot \ldots \cdot p_{n}^{s_{n}} \]

where \( p_{1}, p_{2}, \ldots, p_{n} \) are pairwise distinct prime numbers, and \( s_{1}, s_{2}, \ldots, s_{n} \) are natural numbers.

It is known that the number of natural divisors of the number \(a\), including 1 and \(a\), is equal to \((s_{1}+1)(s_{2}+1) \cdot \ldots \cdot (s_{n}+1)\). According to the problem statement, this number is 101. Since 101 is a prime number, all the terms in the product are 1 except for one term that is 101. Thus, the number \( a = p^{100} \), and its divisors are \( d_{1}=1, d_{2}=p, d_{3}=p^{2}, \ldots, d_{101}=p^{100} \).

Calculate the sum of the divisors:

\[ \Sigma_{d} = 1 + p + p^{2} + \ldots + p^{100} = \frac{p^{101}-1}{p-1} = \frac{a \sqrt[100]{a} - 1}{\sqrt[100]{a} - 1} \]

Calculate the product of the divisors:

\[ \Pi_{d} = p^{1+2+\ldots+100} = p^{50 \cdot 101} = \sqrt{a^{101}} \]

Present the answer in LaTex format: \boxed{Your answer}
user