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If y^4 is divisible by 60, what is the minimum number of dis

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Re: If y^4 is divisible by 60, what is the minimum number of dis  [#permalink]

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New post 04 Sep 2017, 20:11
mbah191 wrote:
Hi all,

I read through all responses but I'm still having some trouble understanding this. Going by the formula... Number of distinct factors = (p+1)(q+1)(r+1)... I have 2^2, 3^1, 5^1 - where p =2, q = 1, and r = 1. My initial reaction is to plug in and get 12. I understand that we need y, not y^4, but why does that change things?

I guess my question is, why exactly are we making each power 1 (changing 2^2 to 2^1) in this case? When would we do this vs when would we not do this? I need to understand what makes this problem unique so that on test day, I don't get a question like this wrong.

If it was y is divisible by 60 (instead of y^4), then would the answer be 12? Also, if it was y^2 or y^3 instead of y^4, would that change things? :|

Thank you very much to whoever can set me straight on this! :-D

Best,
Max

Hi y^4 is divisible by 60
We have to find the minimum value
So prime factors of 60 are 2^2*3*5
Now the tricky part in this question is that y has a power of 4
So we have to have 2*3*5 in Y because to divide and get the integer we should have common factors.
And we should have only one 2 in numerator because this that 2 will rise to power 4 in y^4 so ultimately it will be divisible by 60 so minimum factors are 8
Now coming in to your second question it y is divided by 60 , the the situation changes we should have all the powers of prime factors in order to divide
Y/60
Y/(2^2*3*5)
In this the factors are 12
Hope it helps

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Re: If y^4 is divisible by 60, what is the minimum number of dis  [#permalink]

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Re: If y^4 is divisible by 60, what is the minimum number of dis &nbs [#permalink] 03 Oct 2018, 11:35

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