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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.

Answer: B.

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

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.

Answer: B.

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

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 start with a = 2. YES! 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

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

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

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10 Aug 2014, 11:34

Hello from the GMAT Club BumpBot!

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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.

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]

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08 Sep 2016, 13:54

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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