P is a natural number. 2P has 28 divisors and 3P has 30 divisors. How many divisors of 6P will be there?a)35b)40c)45d)48Correct answer is option 'A'. Can you explain this answer?

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Why is it taken as 2^6 and 3^3?

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I didn't get the solution, as to how it is taken as 2^6 * 3^3 ?

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If 2P has 28 divisors then 2P = (2⁶)(3³)

If 3P has 30 divisors then 3P = (2⁵)(3⁴)

Therefore P = (2⁵)(3³)

Then 6P = 2*3*(2⁵)(3³) = (2⁶)(3⁴)

6P will have (6+1)(4+1) = 35 divisors

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ElBorate this answer please ..why it is taken as p=2^5

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Understanding the Problem
We have two expressions involving the natural number $ P $:
- $ 2P $ has 28 divisors.
- $ 3P $ has 30 divisors.
We need to find the number of divisors of $ 6P $.

Divisor Function and Prime Factorization
The number of divisors function $ d(n) $ for a number $ n = p_1^{e_1} p_2^{e_2} \ldots p_k^{e_k} $ is given by:
$$ d(n) = (e_1 + 1)(e_2 + 1) \ldots (e_k + 1) $$
Assuming the prime factorization of $ P $ is:
$$ P = 2^{a} \cdot 3^{b} \cdot p_1^{c_1} \cdots p_k^{c_k} $$

Analyzing $ 2P $ and $ 3P $
For $ 2P $:
- $ 2P = 2^{a+1} \cdot 3^b \cdot p_1^{c_1} \cdots p_k^{c_k} $
- Divisors: $ (a+2)(b+1)(c_1+1)\cdots(c_k+1) = 28 $
For $ 3P $:
- $ 3P = 2^{a} \cdot 3^{b+1} \cdot p_1^{c_1} \cdots p_k^{c_k} $
- Divisors: $ (a+1)(b+2)(c_1+1)\cdots(c_k+1) = 30 $

Setting Up Equations
We can set up the following equations based on the divisor counts:
1. $ (a+2)(b+1)(c_1+1)\cdots = 28 $
2. $ (a+1)(b+2)(c_1+1)\cdots = 30 $

Finding $ 6P $
Now, for $ 6P $:
- $ 6P = 2^{a+1} \cdot 3^{b+1} \cdot p_1^{c_1} \cdots p_k^{c_k} $
The number of divisors will be:
$$ d(6P) = (a+2)(b+2)(c_1+1)\cdots(c_k+1) $$

Calculating $ d(6P) $
Using the equations derived earlier, we find:
Using the values $ (a+2)(b+1) = 28 $ and $ (a+1)(b+2) = 30 $, we can substitute and solve to find $ d(6P) $.
Through calculations, we find:
$$ d(6P) = 35 $$
Thus, the number of divisors of $ 6P $ is $ \textbf{35} $.

Conclusion
The correct answer is option 'A', which is \(

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35

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A

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A) 35

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