laythesmack23 wrote:
If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =
A. 4
B. 7
C. 8
D. 11
E. 12
The OG Guide and MGMAT Guide both have different solutions, a bit long. Can someone tell me if I'm doing this incorrectly.
If I'm plugging #'s in, I'm getting 2+3+4+5, = 14, but not all can be added, because not all are prime, and some numbers are repeated right, so if I take the sum of all primes in 2+3+4+5, without repeats I'll get 1+3+2+5, then I get 11? is this correct? I know 1 is not prime, and the first 2, and 4 share the same primes, so do I use 1 as a digit for 2, and use 2 as a prime # for 4? to end up with 1+3+2+5?
You cannot plug in numbers. You need to find the values of i, k, m and p.
x = 1*2*3*4*5*6*7*8 = 8!
x = 2^i*3^k*5^m*7^pTo get the value of i, you need to find the number of 2s in x i.e. 8! (including the 2s you get in 4, 6 and 8). You can quickly count - one from 2, two from 4, one from 6 and three from 8 = total seven 2s are there in 8!
To get the value of k, you need to find the number of 3s in 8!. There are two 3s in 8! (one from 3 and another from 6)
It is easy to see that there is only one 5 and one 7 in 8!.
x = 2^7*3^2*5^1*7^1So 7 + 2 + 1 + 1 = 11
Check this post for more on powers in factorials:
http://www.veritasprep.com/blog/2011/06 ... actorials/
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78% (02:11) correct
