Claudiu2803 wrote:
If x and y are integers and x= [(2)(3)(4)(5)(7)(11)(13)][/39y], which of the following could be the value of y?
a)15
b)28
c)38
d)64
e)143
Can someone please explain me the steps in this exercise? I have an explanation in the book but I don´t understand how they came to the answer.
Thank you!
Please note that this is a Problem Solving question and should go as a new post in that forum.
gmat-problem-solving-ps-140/\(x = \frac{(2)(3)(4)(5)(7)(11)(13)}{39y}\)
\(x = \frac{(2)*(3)*(4)*(5)*(7)*(11)*(13)}{3*13y}\)
3 and 13 get cancelled out and you are left with
\(x = \frac{(2)*(4)*(5)*(7)*(11)}{y}\)
Cross multiply to get:
\(y = \frac{(2)*(4)*(5)*(7)*(11)}{x}\)
Since y is an integer, whatever factors x has, they must get cancelled out by the factors in the numerator.
e.g. If x = 15 (i.e. 3*5)
\(y = \frac{(2)*(4)*(5)*(7)*(11)}{3*5}\)
\(y = \frac{(2)*(4)*(7)*(11)}{3}\)
This is not an integer so x cannot be 15.
If x = 28 (i.e. 4*7)
\(y = \frac{(2)*(4)*(5)*(7)*(11)}{4*7}\)
4 and 7 get cancelled out.
\(y = (2)*(5)*(11)\)
This is an integer. So x can be 28.
Answer (B)
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