If y^4 is divisible by 60, what is the minimum number of distinct factors that y must have
A) 2
B) 6
C) 8
D) 10
E) 12

My take-

As y^4/60 = k where k is an integer. means, y^4 is completely divisible by 60. Therefore y must have all the prime factors as those of number 60.
That is - 60 = 2*2*3*5
Also, as y contains power of 4... therefore y must have the following factors-
2^4,3^4 and 5^ 4.

Now, what to do next? How to find the minimum number of distinct factors of y? Please clarify
Thanks

This is an interesting question... Let us see...

Question says y^4 is divisible by 60.

But, we know, prime factors of 60 are 2*2*3*5.

But, these prime factors are present in y^4, not in y as such. So, to get the minimum number of factors, we have to take the first occurrence of every prime number and ignore the rest. (We ignore the repetition because we do not know whether there were 2 occurrence of that factor in y itself or it is occurred due to multiplication, like y*y)

So, minimum y has 2, 3, 5 as prime factors. But, we need to identify factors (not prime factors).
So, there will be at least
2*3 = 6
2*5 = 10
3*5 = 15
2*3*5=30

Now that adds up to 7 factors (2,3,5,6,10,15,30)...

Are we done? NOooo! Don't forget the universal factor "The 1"

That adds up to 8 factors (1,2,3,5,6,10,15,30)

ANSWER C

[However, there is a catch in this question. It didn't state that y is an integer. Without that statement, this question is wrong].

Thanks for providing your inputs. However, I am not sure whether I understood the above part. Pls clarify.

Thanks
Himanshu

We are looking for the minimum factors of y. For that, we have to know the minimum prime factors of y.

What we know is that y^4 is divisible by 60. That means, y^4 contains at least has 2, 2, 3, & 5 as prime factors.

Now the questions is what are the (minimum) prime factors of y? Is it all of 2,2,3,5 or just 2,3,5?

Now consider the number 30, which is 2*3*5

There is only one 2 in 30. 30 is not divisible by 60 either. However, 30^4 is divisible by 60 because
30^4 = 30*30*30*30 = (2*3*5)*(2*3*5)*(2*3*5)*(2*3*5) = (2*3*5*2)*(3*5)*(2*3*5)*(2*3*5) = 60*(3*5)*(2*3*5)*(2*3*5)

The second 2 is part of the second 30.

Similarly, we do not know whether the repeating 2 in y^4 is part of first y or second y. So, to get the minimum factors, we have to ignore the repetitions.

[However, as I mentioned earlier, this question is wrong if y is not stated as an integer. One can find y with lesser factors, but with y^4 divisible by 60.]

Wow! Nice. Please let us know if more help is needed.

_________________
Jamie Nelson
ManhattanGMAT Instructor