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# How many different factors does the integer n have? (1) n =

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How many different factors does the integer n have? (1) n = [#permalink]

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12 Nov 2008, 09:07
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How many different factors does the integer n have?
(1) n = a^4b^3, where a and b are different positive prime numbers.
(2) The only positive prime numbers that are factors of n are 5 and 7.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

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12 Nov 2008, 10:19
It must be A

(1) n = a^4b^3, where a and b are different positive prime numbers.

n will have factors

1,a, b , a*b, a^2*b,a^3 *b,a^4 *b
a*b^2, a*b^3,a2*b^2, a2*b^3,a3*b^2, a3*b^3,a^4*b^2, a^4*b^3

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12 Jan 2009, 18:28
vishy007 wrote:
It must be A

(1) n = a^4b^3, where a and b are different positive prime numbers.

n will have factors

1,a, b , a*b, a^2*b,a^3 *b,a^4 *b
a*b^2, a*b^3,a2*b^2, a2*b^3,a3*b^2, a3*b^3,a^4*b^2, a^4*b^3

although its already at many places .. but just for sake of if some one reads this post

number of factors can be calculated easily by =>

Number of factors = (4+1)(3+1)

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14 Jan 2009, 00:04
gorden wrote:
How many different factors does the integer n have?
(1) n = a^4b^3, where a and b are different positive prime numbers..

Can u be bit clear in ur question .
Is it $$a[m]{4b}$${3}[/m]

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14 Jan 2009, 00:12
mar2hathoda wrote:
vishy007 wrote:
It must be A

(1) n = a^4b^3, where a and b are different positive prime numbers.

n will have factors

1,a, b , a*b, a^2*b,a^3 *b,a^4 *b
a*b^2, a*b^3,a2*b^2, a2*b^3,a3*b^2, a3*b^3,a^4*b^2, a^4*b^3

although its already at many places .. but just for sake of if some one reads this post

number of factors can be calculated easily by =>

Number of factors = (4+1)(3+1)

This is a very useful shortcut.

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14 Jan 2009, 03:15
No. of factors can be calculated very easily....

suppose there is n= a^m * b^n where a and b are prime numbers..

then the no. of factors can be calculated as (m+1)*(n+1).......

I hope its clear.......

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14 Jan 2009, 19:09
why cannot the answer be "D"

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14 Jan 2009, 21:50
gorden wrote:
How many different factors does the integer n have?
(1) n = a^4b^3, where a and b are different positive prime numbers.
(2) The only positive prime numbers that are factors of n are 5 and 7.

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Statement 1: since a and b a different primes and we know there powers therefore we know that they have (4+1)(3+1)= 20 different factors.
Statement 2: could have endless solutions. Therefore insufficient.

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Re: prime number   [#permalink] 14 Jan 2009, 21:50
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