Joined: 15 Sep 2009
[, given: 0] 0
Integers a and n are greater than 1, and the product of the [
17 Oct 2009, 05:57
Could anyone help me solve this one?
Integers a and n are greater than 1, and the product of the 1st 8 positive integers is a multiple of a^n. What is the value of a? A) a^n = 64. B) n = 6. The answer was b. I'm confused, particularly since it says: the product of the 1st 8 positive integers, but it does not say what sequence the first 8 integers relates to. Thanks in advance, Kusum.
Joined: 31 Aug 2009
Location: Sydney, Australia
[, given: 0] 20
The first 8 positive integers multiplied means 1x2x3x4x5x6x7x8
This equals \(2^7*3^2*5*7\). For simplicity lets call this number X. So the question stem says the above number is a multiple of a^n. Statement 1) a^n = 64 64 can be 2^6 or 4^3. Both of these are factors of the first X (as they are the same number). The question asks what is a. A could be both 2 or 4. Not sufficient. Statement 2) This says n=6. If you look at X, the only possible exponent that could be a factor of X is 2^6. This means a=2. The other prime factors to the power of 6 cannot possibly be factors. Hence ANS = B. Hope that helps.