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Why we have to consider factor power as 1 and 7 and not the other number. Answer comes different using other factor power.
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Hovkial
A number has 16 factors. How many factors will the square of N have?

(A) 32
(B) 40
(C) 45
(D) 48
(E) 56

\(factors(n)=16…n=[a^{15}…a^3b^3…abcd…ab^7]\)
\(n=a^{15}…n^2=a^{30}…factors(n)=30+1=31\)
\(n=a^3b^3…n^2=a^6b^6…factors(n)=(6+1)(6+1)=49\)
\(n=abcd…n^2=(abcd)^2…factors(n)=(2+1)(2+1)(2+1)(2+1)=3^4=81\)
\(n=ab^7…n^2=a^2b^{14}…factors(n)=(2+1)(14+1)=3*15=45\)

Answer (C)
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Hovkial
A number has 16 factors. How many factors will the square of N have?

(A) 32

(B) 40

(C) 45

(D) 48

(E) 56

Bunuel

I think this question is incorrect.
16=1*16=2*8=4*4=2*2*4=2*2*2*2

Every case will have a different answer.

If will is replaced with may, then the question may become correct

Posted from my mobile device
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Hovkial
A number has 16 factors. How many factors will the square of N have?

(A) 32

(B) 40

(C) 45

(D) 48

(E) 56

Let N= X^15
N^2=X^30
31 factors
N=X^7 Y^1
N^2=X^14Y^2
45 factors
I think it should be reworded in such a way that it says "Square of N is a perfect square" Then among choices, C sits perfectly as perfect squares have odd number of factors
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Hovkial
A number has 16 factors. How many factors will the square of N have?

(A) 32

(B) 40

(C) 45

(D) 48

(E) 56

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