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Let’s express x as a product of its prime factors:

\(x = 2^a*3^b*5^c*…\)

Total number of its positive divisors will be

\((a+1)*(b+1)*(c+1)*… = y\)

Now we’ll do the same for the 4x

\(4x= 2^2*2^a*3^b*5^c*… = 2^{a+2}*3^b*5^c…\)

Total number of positive divisors of 4x will be

\((a + 3)*(b+1)*(c+1) … = y’\)

For easier calculations let’s take \((b+1)*(c+1)*… = z\)

We have: \((a + 3)*z = y’\)

\((a + 1 + 2)*z = y’\)

\((a+1)*z + 2*z = y’\)

\(y + 2*z = y’\)

So, as we can see, in order to find out the total # of positive divisors of 4x we need to know # of odd divisors of x, which is not given in the question.

Answer E.
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'x' has 'y' positive integral factors including '1' and the number itself.
Thus x should be a prime number.And none of the options fit in except E.
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x=2, factors=1,2, y=2
4x=8, factors=1,2,4,8, y=4=twice of above



x=6,factors=1,2,3,6, y=4
4x=24, factors=1,2,3,4,6,8,12,24. y=8 (4 times above)


Hence y will changes as per x, (of form 2^k or 2^k p^z)

E)
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gmatpapa

Given: An even positive integer 'x' has 'y' positive integral factors including '1' and the number itself.
Asked: How many positive integral factors does the number 4x have?

4x = 2^2x

If x = 16=2^4; y = 5; 4x=64=2^6; Number of integral factors of 4x = 7
But if x= 9=3^2; y = 3; 4x=36=2^2*3^2; Number of integral factors of 4x = 3*3=9

IMO E
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