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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 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
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Originally posted by subhajeet on 11 May 2012, 06:35.
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
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11 May 2012, 06:38
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. Answer: B.
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Re: If the integers a and n are greater than 1 and the product
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21 Jan 2013, 23:36
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 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 AnsweR: B
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Re: If the integers a and n are greater than 1 and the product
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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



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Re: If the integers a and n are greater than 1 and the product
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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. 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 Check Number Theory chapter of Math Book: mathnumbertheory88376.htmlHope it helps.
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Re: If the integers a and n are greater than 1 and the product
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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
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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
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10 May 2013, 01:17



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Re: If the integers a and n are greater than 1 and the product
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24 Oct 2017, 20:30
I don't think this question is written well. I get this part: Quote: If the integers a and n are greater than 1 So there are two integers, a and n. They are both greater than 1, Got it. The problem is I don't get this part: Quote: and the product of the first 8 positive integers is a multiple of a^n The product of the first 8 positive integers OF WHAT? What are we multiplying? a and n together? Is there a set we are looking at?



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Re: If the integers a and n are greater than 1 and the product
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24 Oct 2017, 23:29



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If the integers a and n are greater than 1 and the product
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26 Oct 2017, 13:29
Bunuel wrote: joondez wrote: I don't think this question is written well. I get this part: Quote: If the integers a and n are greater than 1 So there are two integers, a and n. They are both greater than 1, Got it. The problem is I don't get this part: Quote: and the product of the first 8 positive integers is a multiple of a^n The product of the first 8 positive integers OF WHAT? What are we multiplying? a and n together? Is there a set we are looking at? The product of the first 8 positive integers is 1*2*3*4*5*6*7*8. Check completer solution here: https://gmatclub.com/forum/iftheinteg ... l#p1084330In that case the question should be: Quote: If the integers a and n are greater than 1 and the product of the first 8 positive integers of all real numbers is a multiple of a^n, what is the value of a?
(1) a^n = 64
(2) n = 6



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Re: If the integers a and n are greater than 1 and the product
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26 Oct 2017, 21:36
joondez wrote: In that case the question should be: Quote: If the integers a and n are greater than 1 and the product of the first 8 positive integers of all real numbers is a multiple of a^n, what is the value of a?
(1) a^n = 64
(2) n = 6 The question is correct as it is. The first 8 positive integers are 1, 2, 3, 4, 5, 6, 7, and 8. What does "of all real numbers" has to do here? Also, notice that this is an official question from GMAT Prep, so it's as correct as it gets.
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Re: If the integers a and n are greater than 1 and the product
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27 Oct 2017, 10:05
When the question asked for the first 8 positive numbers, it was clearly describing a set. However, it wasn't clear to me what set they were describing. I spent time wondering if they were describing a set of "a" multiple, "n" multiples, or "a^n" multiples. There are many other GMAT questions were the wording is extremely specific but for this question you had to assume that we were looking at all real numbers, which is not always the case for other GMAT questions. So to me this is a very poorly worded question. You are very good at GMAT questions so you may have been able to recognize the question pattern right away, but for someone who is unable to make assumptions based on the question maker's mind, this question was not solvable. And the fact that this is a GMATPrep question does not make it infallible, in fact I have done other GMATPrep questions where even you yourself have stated that they made a mistake



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Re: If the integers a and n are greater than 1 and the product
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27 Oct 2017, 10:09
joondez wrote: When the question asked for the first 8 positive numbers, it was clearly describing a set. However, it wasn't clear to me what set they were describing. I spent time wondering if they were describing a set of "a" multiple, "n" multiples, or "a^n" multiples. There are many other GMAT questions were the wording is extremely specific but for this question you had to assume that we were looking at all real numbers, which is not always the case for other GMAT questions. So to me this is a very poorly worded question. You are very good at GMAT questions so you may have been able to recognize the question pattern right away, but for someone who is unable to make assumptions based on the question maker's mind, this question was not solvable. And the fact that this is a GMATPrep question does not make it infallible, in fact I have done other GMATPrep questions where even you yourself have stated that they made a mistake Sorry but this does not make sense. "The product of the first 8 positive integers" cannot possible mean anything but 1*2*3*4*5*6*7*8.
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Re: If the integers a and n are greater than 1 and the product
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12 Jun 2018, 04:23
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 Given \(a, n > 1\), \(8! = a^n * K\) so we have \(2^7 * 3^2 * 5 * 7 = a^n * K\).......(i) Statement 1: \(a^n = 64\), hence , we have \(a^n = 2^6\) or \(8^2\) or \(4^3\), therefore \((2^6) * 2 * 3^2 * 5 * 7 = a^n * K\), \(a = 2, n = 6\) \((8^2) * 2 * 3^2 * 5 * 7 = a^n * K\), \(a = 8 , n = 2\) Hence, Statement 1 is Not Sufficient. Statement 2: \(n = 6\), hence we have from (i), that on factorization of \(8!\) the only integer greater than 1 & can accommodate \(6\) as its power is \(a = 2\). Hence, Statement 2 is Sufficient. Answer B. Thanks, GyM
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Re: If the integers a and n are greater than 1 and the product
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28 Aug 2018, 09:46
BunuelWe know that a^n is a factor of 8!. Hence you wrote, a^n (k) =8!=2^7 * 3^2 * 5 * 7 & according to st 2 as n = 6 we said that 2 in the only factor of 8! with power >=6. But my question is, what about the multiple "k"? What is for example, k=3^4 In that case, in a^n, the power of 3 will also be=6.



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Re: If the integers a and n are greater than 1 and the product
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28 Aug 2018, 10:40
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: \(a^n*k=8!=2^7*3^2*5*7\). 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. Answer: B. Beautiful. I did the same way to solve this question, though it took me awhile to realize that prime factorization is the quickest way ?
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