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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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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

Kudos [?]: 104 [0], given: 11

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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, 02:48
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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Re: If the integers a and n are greater than 1 and the product [#permalink]

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21 Jan 2013, 23:36
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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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14 Sep 2017, 10:41
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Re: If the integers a and n are greater than 1 and the product   [#permalink] 14 Sep 2017, 10:41
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