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by Jay@ManhattanReview » Tue Jul 25, 2017 8:52 pm
AbeNeedsAnswers 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

A
We have n = 1*2*3*4*5*6*7*8 = 8!

Thus, the different prime factors greater than 1 are; 2, 3, 5, and 7: Four

The correct answer: A

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by Brent@GMATPrepNow » Wed Jul 26, 2017 6:15 am
AbeNeedsAnswers 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

A
n = (1)(2)(3)(4)(5)(6)(7)(8)
= (1)(2)(3)(2)(2)(5)(2)(3)(7)(2)(2)(2)
The prime factors are 2, 3, 5 and 7

Answer: A
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by [email protected] » Wed Jul 26, 2017 6:52 pm
Hi AbeNeedsAnswers,

With these types of Prime Factorization questions, it often helps to 'break down' the math into 'pieces' (since the individual pieces are rarely all that difficult to deal with.

Here, we're asked for the number of different prime factors in the product of 1 to 8, inclusive (essentially 8!). That product would include the following factors:
1
2
3
4 = (2)(2)
5
6 = (2)(3)
7
8 = (2)(2)(2)

Thus, the different primes are: 2, 3, 5 and 7... and there are 4 of them.

Final Answer: A

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by Scott@TargetTestPrep » Mon Aug 14, 2017 11:52 am
AbeNeedsAnswers 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

A
n = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1

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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by Matt@VeritasPrep » Fri Aug 18, 2017 2:15 pm
Count the unique primes in 2 * 3 * 4 * 5 * 6 * 7 * 8
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