Is the integer y a multiple of 4?

Yes, if a is odd and b is a multiple of 2, then y won't be a multiple of 4. Check this post for complete solution: is-the-integer-y-a-multiple-of-140569.html#p1130605

Hope it helps.

Is there any logical way of concluding instead of plugging numbers say like y=4k and 3y^2 = 3*6t or y^2 = 6t hence not possible
and y = 12s/3u or y= 4* s/u but where I got tempted and forgot that s/u could be and could not be an integer . hence not possible.

Yes, from the way I see it, the solution given above by Bunuel uses the logical approach. He plugs in numbers later to show you examples of how two different answers (Yes and No) can be obtained.


Question: Is the integer y a multiple of 4?
It is asking us whether there are two 2s in y i.e. whether 4 is a factor of y.

(1) 3y^2 is a multiple of 18.
This tells you that 3y^2 is a multiple of 18 (= 3*6) so y^2 is definitely a multiple of 6. This means y^2 must have at least one 2. But since y is an integer, all primes in y^2 must have even powers. Hence y^2 must have at least two 2s. So y must have at least one 2. But is it necessary that y^2 must have at least four 2s? No. Hence we cannot say whether y will have two 2s or not. Hence not sufficient.

(2) y = p/q, where p is a multiple of 12 and q is a multiple of 3.

y = p/q = 12a/3b = 4a/b

y may or may not have 4 as a factor since b divides it. Say if b cancels out the 4, we may not have a 4 in y. If b doesn't cancel out the 4, then y has 4 as a factor. So y may have no 2, one 2, two 2s or more. Not sufficient.

Using both, all we can conclude is that y has at least one 2. We still can't say whether it has two 2s or not.

Answer (E)

till this point correct "2*2*a/b". but think, b is a 2 and a is 3, in this case Y will have just one two's as its prime factor, as such Y is not the multiple of 4.

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