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Vips0000
kingb
If a natural number ‘p’ has 8 factors, then which of the following cannot be the difference between the number of factors of p3 and p
a. 14
b. 30
c. 32
d. 56
e. None of these
p has 8 factors. That gives following possiblities:

a) p has one prime factor with power 7
=> p^3 will have 22 factors. Diffrerence with number of factors of p = 22-8 = 14: A is ok
b) p has two prime factors with powers 3 and 1
=> p^3 will have 40 factors. Difference with number of facors of p =40-8 = 32 : C is ok
c) p has three prime factors with powers 1 each
=> p^3 will have 64 factors. Difference with number of factors of p=64-8 = 56 : D is ok

There are no other possiblities. Hence remaining answer choice B is not possible.

Ans B it is!
i didnot understand a bit of your explanation.... can you please explain it? thank you
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Since p has 8 total factors, we can assume that the prime factorization of p will take three forms based on the principle that the total number of factors of an integer can be derived from the prime factorization of that integer where the integer's total factors = prime factors (a +1) * (b +1):

p --> x^7 (where 7+1 = 8 total factors)

p -->x^3 * y^1 (or y^3 * x^1) (where (3+1)*(1+1) = 8 total factors)

p -->x^1 * y^1 * z^1 (where (1+1) * (1+1) * (1 +1) = 8 total factors)

Now, we can address the possible total factors of (p^3 - p) as follows:

(x^7)^3 = (x)^21 + 1 = 22 total factors --> 22 - 8 = 14

(x^3)^3 * (y^1)^3 = (9+1) * (3+1) --> 40 - 8 = 32

(x^1)^3 * (y^1)^3 * (z^1)^3 = (3+1) * (3+1) * (3+1) = 64 - 8 = 56

Thus, (B) is the only answer that is not possible.

Great question. Source?
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Good explanation bunuel.
I was calculating for 3p and p.
The way question is posted
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From hardest questions set.

Bumping for review and further discussion.
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