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)
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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.

pushpitkc 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 dave13

I'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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