A is a prime number (A>2). If C = A^3, by how many : PS Archive
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# A is a prime number (A>2). If C = A^3, by how many

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SVP
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A is a prime number (A>2). If C = A^3, by how many [#permalink]

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14 Aug 2005, 19:44
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A is a prime number (A>2). If C = A^3, by how many different integers can C be equally divided?

(a) 3.
(b) 4.
(c) 5.
(d) 6.
(e) 7.

(Edit: I've edited the question stem to add the power sign. Himalaya, please edit the post back if the edit does not reflect the correct version. - Hong)
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14 Aug 2005, 23:19
I think there is a flaw in the stem. Since both 3 and A are prime numbers, C is divisible by

1
3
A
3A

This way it seems that the correct answer is (B). But if A=3, two of the divisors are the same, and we have only 3 different divisors.
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15 Aug 2005, 18:54
Looks like A^3 only have 1, A, A^2 and itself as its divisors.
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15 Aug 2005, 19:14
HongHu wrote:
A is a prime number (A>2). If C = A^3, by how many different integers can C be equally divided?

(Edit: I've edited the question stem to add the power sign. Himalaya, please edit the post back if the edit does not reflect the correct version. - Hong)

Thank you honghu, that's correct and doesnot change anything.
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15 Aug 2005, 19:44
If a positive integer N can be represented in terms of prime factors as:

N = A^p * B^q * C^r....

the total number of factors that it has is (p+1)(q+1)(r+1)....

Answer here is 3+1 = 4
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15 Aug 2005, 21:16
N = A^p * B^q * C^r....

the total number of factors that it has is (p+1)(q+1)(r+1)....

That is a good formula !

Looks like for A^p (factor can be A^0 ... A^p) => (p+1)
for B^q, similarly => (q+1)
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16 Aug 2005, 18:33
the question is asking how many factors does C=A^3 have? the factors are:
C= A^3 = 1 x A^3 = A x A^2 = A^2 x A = A^3x1
so 1, A, A^2 and A^3, altogather 4.

OA is B.
Re: PS: I   [#permalink] 16 Aug 2005, 18:33
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