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# If the integers a and n are greater than 1 and the product

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If the integers a and n are greater than 1 and the product [#permalink]

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11 May 2012, 06:35
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55% (02:08) correct 45% (00:48) wrong based on 395 sessions

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If the integers a and n are greater than 1 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
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Aug 2015, 16:11, edited 1 time in total.
Edited the question.
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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11 May 2012, 06:38
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subhajeet wrote:
If the integers a and n are greater than 1 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

Can anyone help me with this question. How is B the correct answer.

Prime factorization would be the best way to attack such kind of questions.

Given: $$a^n*k=8!=2^7*3^2*5*7$$.
Question: $$a=?$$

(1) $$a^n=64=2^6=4^3=8^2$$, so $$a$$ could be 2, 4, or 8. Not sufficient.

(2) $$n=6$$ --> the only integer (more than 1), which is a factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

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Re: If the integers a and n are greater than 1 and the product [#permalink]

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14 May 2012, 02:41
Bunuel wrote:
subhajeet wrote:
If the integers a and n are greater than 1 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

Can anyone help me with this question. How is B the correct answer.

Prime factorization would be the best way to attack such kind of questions.

Given: $$a^n*k=8!=2^7*3^2*5*7$$. Q: $$a=?$$

(1) $$a^n=64=2^6=4^3=8^2$$, so $$a$$ can be 2, 4, or 8. Not sufficient.

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

Dear Bunuel,
OA is B but why B?
Kindly tell the source from where you get all these number properties/Prime no. properties?
If its your Brain then only the explaination for the above will do !
All DS are on Number Properties and i am doing silly mistakes. i opted 'C'
Thanx
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Kudos [?]: 76348 [0], given: 9973

Re: If the integers a and n are greater than 1 and the product [#permalink]

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14 May 2012, 02:48
Expert's post
kashishh wrote:
Bunuel wrote:
subhajeet wrote:
If the integers a and n are greater than 1 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

Can anyone help me with this question. How is B the correct answer.

Prime factorization would be the best way to attack such kind of questions.

Given: $$a^n*k=8!=2^7*3^2*5*7$$. Q: $$a=?$$

(1) $$a^n=64=2^6=4^3=8^2$$, so $$a$$ can be 2, 4, or 8. Not sufficient.

(2) $$n=6$$ --> the only integer (more than 1), which is the factor of 8!, and has the power of 6 (at least) is 2, hence $$a=2$$. Sufficient.

Dear Bunuel,
OA is B but why B?
Kindly tell the source from where you get all these number properties/Prime no. properties?
If its your Brain then only the explaination for the above will do !
All DS are on Number Properties and i am doing silly mistakes. i opted 'C'
Thanx

Check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Kudos [?]: 372 [2] , given: 11

Re: If the integers a and n are greater than 1 and the product [#permalink]

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21 Jan 2013, 23:36
2
KUDOS
This one is a fun question (a good practice of our understanding of factors)!

Analyze the given first before delving into the statements.
$$8*6*7*5*4*3*2*1 = a^n * R$$

Statement (1): a^n = 64
$$8*6*7*5*4*3*2*1 = 64 * R$$

64 could be 2^6 or 8^2.
In 8!, it contains at least 6 factors of 2. In 8!, it also contains 2 factors of 8. Thus, a could be 2 or 8.
Thus, INSUFFICIENT.

Statement (2): n = 6
Let us analyze 8! = 8*7*6*5*4*3*2*1. How many prime factors have at least 6 factors in 8!.
Let us then try a=3. NO!

We are certain that 2 has the most number of factors in 8! and it has at least 6. SUFFICIENT.
a = 2

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Re: If the integers a and n are greater than 1 and the product [#permalink]

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22 Jan 2013, 13:22
product of the first 8 integers is: (2^8)(3^2)(5)(7)

Statement 1: a^n = 64. this means a could equal 2,4, or 8 because we don't know what n is. Not sufficient.
Statement 2: if n = 6 the only possible value for a is 2 as the product of the first 8 integers does not include any other number raised to the 6th power. Sufficient
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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09 May 2013, 18:15
i have a question here, what if n = 2 or 3? would the answer be E? or C?
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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10 May 2013, 01:17
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supreetb wrote:
i have a question here, what if n = 2 or 3? would the answer be E? or C?

(1)+(2) a^n = 64 and n=2 --> a^2=64 --> a=8 (discard a=-8 since we know that a is a positive integer). Sufficient.

Hope it's clear.
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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10 Aug 2014, 11:34
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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16 Aug 2015, 16:09
Request you not to write your queries/answers/opinions in question window. It prevents ppl from analysing the question. The whole purpose of GMAT Club forum goes wasted by doing so.

You have response windows to do all such things.
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Re: If the integers a and n are greater than 1 and the product [#permalink]

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16 Aug 2015, 16:12
Expert's post
sagarbuss wrote:
Request you not to write your queries/answers/opinions in question window. It prevents ppl from analysing the question. The whole purpose of GMAT Club forum goes wasted by doing so.

You have response windows to do all such things.

Edited the original post. Thank you.
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Re: If the integers a and n are greater than 1 and the product   [#permalink] 16 Aug 2015, 16:12
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