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If x is the product of the positive integers from 1 to 8, in [#permalink]
21 Jan 2013, 14:11

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Difficulty:

15% (low)

Question Stats:

79% (02:06) correct
21% (02:09) wrong based on 80 sessions

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?

Re: If x is the product of the positive integers from 1 to 8, in [#permalink]
21 Jan 2013, 14:29

Expert's post

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

Given that \(x=8!=2^7*3^2*5*7\). Hence, \(x=2^7*3^2*5^1*7^1=2^i * 3^k * 5^m * 7^p\), since i, k, m, and p are positive integers, then we can equate the exponents, so we have that \(i=7\), \(k=2\), \(m=1\), and \(p=1\).

Re: If x is the product of the positive integers from 1 to 8, in [#permalink]
21 Jan 2013, 19:56

Expert's post

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!

\(x = 2^i*3^k*5^m*7^p\)

To 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!.

If x is the product of the positive integers from 1 to 8, in [#permalink]
11 Aug 2013, 00:44

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 \(x = 2^i3^k5^m7^p\), then i + k + m + p = A 4 B 7 C 8 D 11 E 12 _________________

Re: If x is the product of the positive integers from 1 to 8, in [#permalink]
11 Aug 2013, 00:48

Expert's post

Stiv 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 \(x = 2^i3^k5^m7^p\), then i + k + m + p = A 4 B 7 C 8 D 11 E 12

Merging similar topics. Please refer to the solutions above. _________________

Re: If x is the product of the positive integers from 1 to 8, in [#permalink]
11 Aug 2013, 00:51

Stiv 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 \(x = 2^i3^k5^m7^p\), then i + k + m + p = A 4 B 7 C 8 D 11 E 12

\(X= 1*2*3*4*5*6*7*8\) OR =\(1*2*3*2^2*5*(2*3)*7*2^3\) =\(1*2^7*3^2*5*7\) THEREFORE \(i + k + m + p = 7+2+1+1 = 11\)

HENCE D _________________

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