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13 Jul 2018, 00:49
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B
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If x and y are integers \(x = \frac{(2)(3)(4)(5)(7)(11)(13)}{39y}\) and which of the following could be the value of y ?
A) 15
B) 28
C) 38
D) 64
E) 143
A) 15
B) 28
C) 38
D) 64
E) 143
13 Jul 2018, 01:39
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Bunuel
If x and y are integers \(x = \frac{(2)(3)(4)(5)(7)(11)(13)}{39y}\) and which of the following could be the value of y ?
A) 15
B) 28
C) 38
D) 64
E) 143
A) 15
B) 28
C) 38
D) 64
E) 143
The first step is to prime factorize 39, which is 3*13. Now, the
remaining integers contained in y can only be a multiple of 2,4,5,7, or 11.
Evaluating answer options to check which is a possible value of y
A) 15 = 3*5 (Not possible)
B) 28 = 4*7 (Possible)
C) 38 = 2*19 (Not possible)
D) 64 = 4^2 (Not possible)
E) 143 = 13*11 (Not possible) (Option B)
General Discussion
13 Jul 2018, 03:37
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Bunuel
If x and y are integers \(x = \frac{(2)(3)(4)(5)(7)(11)(13)}{39y}\) and which of the following could be the value of y ?
A) 15
B) 28
C) 38
D) 64
E) 143
A) 15
B) 28
C) 38
D) 64
E) 143
Given
\(\frac{2*3*4*5*7*114*13}{39y}\)
= \(\frac{2*3*4*5*7*11*13}{3*13 *y}\)
= \(\frac{2*4*5*7*11}{y}\)
Thus y = 28. (28 = 4*7)
The best answer is B.
so that after division we have integer ? can you explain please 
