Amateur wrote:
Vips0000 wrote:
kingb wrote:
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
Finding the Number of Factors of an IntegerFirst make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.
The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\).
NOTE: this will include 1 and n itself.
Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)
Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
BACK OT THE ORIGINAL QUESTION:
We are told that \(p\) has 8 factors: 8=2*4=2*2*2.
If \(p\) has only one prime in its prime factorization, say \(a\), then \(p=a^7\) --> number of factors of \(p\) is \((7+1)=8\) --> \(p^3=(a^7)^3=a^{21}\) in this case would have \((21+1)=22\) factors --> the difference is \(22-8=14\);
If \(p\) has two primes in its prime factorization, say \(a\) and \(b\), then: \(p=a^3*b\) --> number of factors of \(p\) is \((3+1)(1+1)=8\) --> \(p^3=a^9*b^3\) in this case would have \((9+1)(3+1)=40\) factors --> the difference is \(40-8=32\);
If \(p\) has three primes in its prime factorization, say \(a\), \(b\) and \(c\), then: \(p=a*b*c\) --> number of factors of \(p\) is \((1+1)(1+1)(1+1)=8\) --> \(p^3=a^3*b^3*c^3\) in this case would have \((3+1)(3+1)(3+1)=64\) factors --> the difference is \(64-8=56\).
p cannot have more than 3 factors, since the least number of factors a number with four primes can have is 16>8 (for example if \(p=abcd\), then number of factors of \(p\) is \((1+1)(1+1)(1+1)(1+1)=16\)).
Answer: B.
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