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If the integers a and n are positive and the product of the first

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If the integers a and n are positive and the product of the first [#permalink]

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New post 24 Aug 2016, 02:49
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Difficulty:

  85% (hard)

Question Stats:

45% (01:36) correct 55% (02:53) wrong based on 69 sessions

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If the integers a and n are positive and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6




After doing the above question Do try your hands on this Official GMAT PREP Question => if-the-integers-a-and-n-are-greater-than-1-and-the-product-58703.html

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Re: If the integers a and n are positive and the product of the first [#permalink]

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New post 24 Aug 2016, 04:01
[quote="stonecold"]If the integers a and n are positive and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6

The product of first 8 positive intergers are 1.2.3.4.5.6.7.8 = 64*144*5

Stat 1: a^n = 64 ( here 64 can be written 2^6 , 4^3 and 8^2 ) then a can be 2,4 and 8 and n 6,3 and 2...Insufficient..no unique

Stat 2: only n given, we don't have info about a..Insufficient.

Stat 1+2 : when n = 6 then a = 2... (2^6 case ).

Hence option C..
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Re: If the integers a and n are positive and the product of the first [#permalink]

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New post 24 Aug 2016, 04:15
msk0657 wrote:
stonecold wrote:
If the integers a and n are positive and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6

The product of first 8 positive intergers are 1.2.3.4.5.6.7.8 = 64*144*5

Stat 1: a^n = 64 ( here 64 can be written 2^6 , 4^3 and 8^2 ) then a can be 2,4 and 8 and n 6,3 and 2...Insufficient..no unique

Stat 2: only n given, we don't have info about a..Insufficient.

Stat 1+2 : when n = 6 then a = 2... (2^6 case ).

Hence option C..


Although your answer is correct but i am not satisfied your explanation for statement 2
You forgot to mention that A=1 and N=6 is a case here too
If i write that A>1 then your answer would be "incorrect"

Take a note
:)
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Re: If the integers a and n are positive and the product of the first [#permalink]

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New post 24 Aug 2016, 05:20
stonecold wrote:
msk0657 wrote:
stonecold wrote:
If the integers a and n are positive and the product of the first 8 positive integers is a multiple of a^n, what is the value of a ?

(1) a^n = 64
(2) n=6

The product of first 8 positive intergers are 1.2.3.4.5.6.7.8 = 64*144*5

Stat 1: a^n = 64 ( here 64 can be written 2^6 , 4^3 and 8^2 ) then a can be 2,4 and 8 and n 6,3 and 2...Insufficient..no unique

Stat 2: only n given, we don't have info about a..Insufficient.

Stat 1+2 : when n = 6 then a = 2... (2^6 case ).

Hence option C..


Although your answer is correct but i am not satisfied your explanation for statement 2
You forgot to mention that A=1 and N=6 is a case here too
If i write that A>1 then your answer would be "incorrect"

Take a note
:)


You forgot to mention that A=1 and N=6 is a case here too
we can't take exactly a = 1 then we can take even 2 ...still 2^6 is a factor...

if there is such condition that a > 1 then only 2^6 is apt.
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Re: If the integers a and n are positive and the product of the first [#permalink]

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New post 24 Aug 2016, 08:06
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Re: If the integers a and n are positive and the product of the first [#permalink]

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New post 13 Apr 2017, 06:17
Here is why I chose C. stonecold can rip it apart.

1. \(a^{n}\) = 64

\(2^{6}\) or \(4^{3}\); different values. statement 1 is insufficient

2. n=6

If I look at the product then it can be deduced to prime numbers as \(2^{7}\) * \(3^{2}\) * 5 * 7.

40320 =k * \(a ^{6}\)

\(2^{7}\) * \(3^{2}\) * 5 * 7 = k * \(a ^{6}\)

a can take different values based on multiplier value; a=2 or a=1

statement 2 is insufficient

Both together -> n=6 and \(a ^{6}\) = 64; then a=2.

C is sufficient.
Re: If the integers a and n are positive and the product of the first   [#permalink] 13 Apr 2017, 06:17
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If the integers a and n are positive and the product of the first

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