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21 Jan 2013, 15:11
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D
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Difficulty:
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Question Stats:
84% (01:37) correct
16%
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wrong
based on 1687
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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
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?
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?
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30 Dec 2018, 10:44
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feellikequitting
GIVEN: x = (8)(7)(6)(5)(4)(3)(2)(1)
Let's find the prime factorization of x by rewriting each value as the product of primes.
We get: x = (2)(2)(2)(7)(2)(3)(5)(2)(2)(3)(2)
Rearrange to get: x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(5)(7)
Rewrite as powers to get: \(x = (2^7)(3^2)(5^1)(7^1)\)
We're told that \(x = (2^i)(3^k)(5^m)(7^p)\)
This means: i = 7, k = 2, m = 1 and p = 1
So, i + k + m + p = 7 + 2 + 1 + 1 = 11
Answer: D
Cheers,
Brent
21 Jan 2013, 15:29
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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 \(p=1\).
Therefore, \(i + k + m + p=7+2+1+1=11\).
Answer: D.
Hope it's clear.
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 \(p=1\).
Therefore, \(i + k + m + p=7+2+1+1=11\).
Answer: D.
Hope it's clear.
