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If n is the product of the integers from 1 to 8, inclusive [#permalink]
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09 Jul 2012, 03:31
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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09 Jul 2012, 03:32



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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11 Jul 2012, 00:16
Difficulty Sub 550
8! has factors of 2, 3, 5 and 7 as 4 and 8 has 2 and six has 3. thus answer is A.



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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19 Jul 2012, 04:11
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n = 1*2*3*4*5*6*7*8 so only Prime numbers which will be factors of n will be 2,3,5,7 (as prime numbers which are greater than 7 will not be there in the product of 1 to 8!) So Answer is A (four) Hope it Helps!
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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25 Jul 2012, 05:46
Is it correct to say prime factors greater than 1? 1 is not a prime factor at all. If one says prime factor greater than 2, then it does make sense. Am I right in making this statement?



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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25 Aug 2012, 10:12
yep, the phrase " different prime factors greater than 1" sounds strange, since in fact, all of these primes are different.furthermore, no need to point out about 1, since 1 is not prime.
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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10 Oct 2013, 10:07
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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10 Oct 2013, 17:08
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n = 8!
Prime factorization of 8! = 1, 2, 3, 2*2, 5, 2*3, 7, 2*2*2 = 2, 3, 5, 7
= 4 prime factors > 1



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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17 Dec 2013, 09:11
8! = 1x2x3x4x5x6x7x8.. Just do prime factorization of each integer alone and then count the number of different primes in total.



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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04 May 2015, 06:42
In this question, I have noticed that many students are prime factorizing every term in the product to find out the answer. But is that necessary? What if the expression was Z = 1*2*3*…*30. Would you have factorized every term? Let's do a quick concept recap. Concept Recap: Primes are the basic building blocks for every positive integer greater than 1. Every positive integer greater than 1 is itself a prime or a product of primes less than the number itself. How is this related to the question?: Take the example of 6!. 6! as we all know is equal to \(1*2*3*4*5*6\). Obviously, we don't need to factorize every element in this expression to find out the different prime factors of 6!. Using the knowledge from the above concept recap that the different prime factors of 6! will be simply the prime numbers less than or equal to 6 itself, we can say the prime factors of 6! are 2, 3 and 5. Therefore 6! has 3 prime factors. Answer for this question: Primes less than 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29 (a total of 10 primes.). Therefore, if Z = 30!, then there would 10 different prime factors for Z. In such questions, where you need to find the number of prime factors of a factorial expression, do not waste your time factorizing every term. Number of prime factors of n! will be simply the number of prime numbers less than n. Footnote for the curious minded: It would have made sense to factorize every term in the expression, if the question had asked the "total number of factors" instead of "number of prime factors". To find the total number of factors, we definitely would need to find the prime factors and their powers in the expression. You can take a stab at the following questions to test your understanding of these concepts. xisthelargestprimenumberlessthanpositiveintegernpisanin197329.htmlpisthesmallestperfectcubegreaterthan197336.htmlRegards, Krishna
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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03 Dec 2016, 19:48
Here is my solution => n=\(8! =>8*7*6*5*4*3*2*1=> 2^7*3^2*5*7\)=> Clearly it has 4 prime factors. Hence A (Additionally it has > 8*3*2*2 =>96 factors )
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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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04 Dec 2016, 11:11
Bunuel wrote: If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have?
(A) four (B) five (C) six (D) seven (E) eight 8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 There can be only 4 different prime factors greater than 1 , as highlighted in Blue above..
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If n is the product of the integers from 1 to 8, inclusive [#permalink]
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11 May 2017, 09:20
Hi eGMATThis Takeway: Number of prime factors of n! will be simply the number of prime numbers less than n: is true if n = even For example: n = 4 => n! = 1 x 2 x 3 x 4 So, different prime factors of 4! are: 2 & 3 which are less than n = 4 BUT once n = odd, Number of prime factors of n! will be simply the number of prime numbers less than OR EQUAL to n. For example: n = 7 => n! = 7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 So different prime factors of 7! = 2, 3, 5, & 7. In this scenario, the prime number 7 is equal to n = 7. So, the Takeaway would be: Number of prime factors of n! will be simply the number of prime numbers less than OR EQUAL to n. Is this revised Takeaway correct? Many thanks



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Re: If n is the product of the integers from 1 to 8, inclusive [#permalink]
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16 May 2017, 18:12
Bunuel wrote: If n is the product of the integers from 1 to 8, inclusive, how many different prime factors greater than 1 does n have?
(A) four (B) five (C) six (D) seven (E) eight n = 8 x 7 x 6 x 5 x 4 x 3 x 2 We can prime factorize and we have: n = 2^7 x 3^2 x 5^1 x 7^1 Thus, n has 4 different prime factors. Alternate solution: In general, the number of distinct prime factors that k! (where k > 1) has is the number of prime numbers less than or equal to k. We have n = 8!, so k = 8; the number of prime numbers less than or equal to 8 is 4, namely, 2, 3, 5 and 7. Answer: A
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