dave13 wrote:
pushpitkc wrote:
Bunuel wrote:
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
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) hello there world traveler
39 has 3 and 13 as prime numbers
in numerator we have (2)(3)(4)(5)(7)(11)(13)
i thought we need such number that would fill in missing prime numbers i mean 39 has 3 and 13 and correct answer choice 28 contains 4 and 7 ...BUT how about 5, and 11 shouldnt they be part of denominator ??
so that after division we have integer ? can you explain please
Hey
dave13I'm no world traveler - I was hoping you would be gracious enough
to sponsor my trip around the world
Now coming back to the problem.
The simple reason the numbers 5 and 11 are not contained in the
numerator is the reason we can't have 5 and 11 in the denominator.
The problem statement gives us that both x and y must be integers.
Therefore, if y = 5 or y = 11 the number x will become a non-integer
Let me explain by a small example:
Let the numerator be 546(which is prime-factorized as 2*3*7*13)
If we have 5 or 11 in the denominator, this will make the value of
the fraction a non-integer.
Hope this clears your confusion!
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80% (01:35) correct 
